Contents
1 Book Compilation
3
2 Kinematics of Particles
2.1 Straight line motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
5
3 Kinematics of Particles
3.1 Straight line motion .
3.2 Cartesian Coordinates
3.3 Polar . . . . . . . . . .
3.4 Path . . . . . . . . . .
3.5 Rel Motion . . . . . .
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1 Homework Solutions
2 Kinematics of Particles
2.1 Straight-Line Motion . . . . . . . .
2.2 Cartesian Coordinates . . . . . . .
2.3 Polar and Cylindrical Coordinates
2.4 Path Coordinates . . . . . . . . . .
2.5 Relative Motion and Constraints .
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1
Full file at https://testbanku.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-2nd-Edition-by-Tongue
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29
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70
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CONTENTS
CONTENTS
2
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Full file at https://testbanku.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-2nd-Edition-by-Tongue
Chapter 1
Book Compilation
3
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CHAPTER 1. BOOK COMPILATION
4
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Chapter 2
Kinematics of Particles
2.1
Straight line motion
Homework Problems
2.1.1: [Easy] The 2007 BMW Z4 Coupe 3.0si can accelerate from 0 to 60 mph in 5.6 s, and its
maximum speed is ẋmax = 155 mph. Assuming the car accelerates from 0 to 60 mph at a
constant rate and that it can continue to accelerate at this rate without redlining, find how
far the car travels and how long it takes to reach its maximum speed.
2.1.2: [Easy] A map tells you to go 3 paces east, 2 paces south, 5 paces east, 7 paces north, and
4 paces west. Plot the complete path as well as the overall vector from the initial to final
position. Then plot the same thing but in reverse order, that is, 4 paces west, 7 paces north,
and so on. Do you end up at the same position?
2.1.3: [Easy] A particle moves as shown. What is the average speed over the illustrated time
interval? Estimate the maximum and minimum speeds from the graph.
2.1.4: [Easy] Find ẋ(0) such that ẋ(2.5) = 0 m/s for a particle whose acceleration is given below.
2.1.5: [Easy]
A ball bearing is accelerated from rest by means of an electromagnet according
to s̈ = as + becs , where a = 3 s−2 , b = 1 ft/s2 , and c = 0.2 ft−1 . How fast is the ball bearing
moving after having traveled d = 10 ft?
2.1.6: [Easy] One of the severe dynamics realities of aircraft operations on an aircraft carrier is
the need to stop the fighter plane from airspeed to zero in a very short distance. To do this
the jet actually trails a hook behind it which catches onto a cable that very decisively brings
the jet to a stop.
Assume that a TOPGUN jet touches down at 170 mph and its speed is reduced to zero in 240
ft. Assuming a constant deceleration, calculate the time elapsed and the acceleration acting
on the jet (expressed both in ft/s2 and g’s).
2.1.7: [Easy] Taking off from an aircraft carrier is a non-trivial operation. Basically, there’s not
enough room for the jet to taxi normally and thus an additional means of getting the jet up
to takeoff speed is needed. What’s commonly used is a steam-driven catapult. Large pistons
5
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2.1. STRAIGHT LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
move within their cylinders, driven by steam pressure at one end, and tow the jet forward.
Assume that the jet is brought from a stationary state to 165 mph in 2.23 s and that the
acceleration was constant. How much deck does this imply must be in front of the jet?
2.1.8: [Easy]
The displacement of a particle is given by
x(t) = a1 + a2 t + a3 t3
where a1 = 10 m, a2 = −16 m/s, and a3 = 31 m/s3 . Plot the speed and acceleration for
0 ≤ t < 6 s. Find the value of t for which ẋ = 0 m and indicate it on your plot of ẋ and ẍ.
2.1.9: [Easy]
The displacement of a car is given by
x(t) = a1 + a2 t + a3 t2
where a1 = 100 ft, a2 = −88 ft/s, and a3 = 10 ft/s2 .
a. What is the car’s speed at t = 2 s?
b. What is the car’s acceleration at t = 10 s?
2.1.10: [Easy] What constant acceleration is needed to get a car from 0 to 60 mph in 5 s? Express
your result in both ft/s2 and g’s (1 g = 32.2 ft/s2 ).
2.1.11: [Easy] Assume that the driver of a car, traveling at 70 mph, has to panic brake to a dead
stop. If the car is capable of sustaining a 1 g deceleration, how long will it take and how far
will the car travel before stopping?
2.1.12: [Easy]
The 2008 Audi TT Coupe can accelerate from 0 to 60 mph in 6.5 s and has a
maximum speed of ẋmax = 147 mph. Assume that the car accelerates from 0 to 60 mph at a
constant rate and that it can continue to accelerate at this rate without redlining.
a. How far does the car travel before reaching its maximum speed?
b. Suppose a competitor’s car needs 0.312 mi to accelerate to its maximum speed (which is
slightly higher) under the same assumptions. How does the TT compare?
c. What change in the 0-to-60 time is needed so that both vehicles reach their respective
maximum speed in the same length of road?
2.1.13: [Easy] A favorite sportscar of mine has an engine for which the bore/stroke is 3.40/3.53
in. This means that the piston moves 3.53 in. (and then back again) during one cycle of
operation. Let’s assume that it does this in a sinusoidal fashion:
x(t) =
3.53 in.
sin(ωt)
2
The engine operates at a maximum of 7000 rpm. What is the maximum speed of the piston,
and what is its maximum acceleration (in g’s)?
2.1.14: [Easy] Here’s a nice exercise that has some important real-world applications. Safety
experts are always telling us that it’s important to buckle up, but how important is it really?
I mean, come on—moving at city speeds can’t be all that dangerous, can it? You can get a
physical feel for the answer to this question by asking how far you would have to fall vertically
in order for your contact speed with the ground to equal 35 mph (a “typical” speed in an
automotive collision). You should picture yourself falling from that height, face first, onto a
6
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT LINE MOTION
hard floor because that’s essentially what will be happening if you get into a crash without
a fastened seatbelt. The car will smash into whatever it’s colliding with and essentially stop
immediately. Your body, meanwhile, having nothing to restrain it, will continue moving at
35 mph until you smash into the dash. Find the height h from which you will contact the
floor at 35 mph.
2.1.15: [Easy] The force generated by air resistance that opposes the motion of a car is given by
Fa = Av 2 where A is a constant that depends on the density of the air and the shape of the
car and v is the car’s speed. When the car is traveling as fast as its engine can drive it, there
exists a balance between the drag force and the drive force due to the car’s engine: Fd = Fa .
By what percentage must the car’s drive force be increased to increase the car’s maximum
speed by 10%? What does this tell you about the ease (or difficulty) of increasing a car’s top
speed?
2.1.16: [Easy] Two bicycle racers are racing along a straight course. Bicyclist A has a lead of 250
yards and has a mile to go until he reaches the finish line. He maintains a constant speed of
18 mph. At what constant speed will Bicyclist B need to travel to catch Bicyclist A at the
finish line?
2.1.17: Computational [Easy] A body traveling in a straight line starts from rest and experiences
an acceleration equal to a0 − a1 v 3 where v is the body’s speed. a0 = 80,000 ft/s2 and a1 =
0.01 s/ft2 . Determine the body’s terminal speed vterm (speed as time approaches infinity).
How long does it take for the body to reach 95% of vterm ?
2.1.18: [Moderate] A plot of acceleration versus time for a particle is shown. Estimate the
position of the particle at t = 3 seconds if it starts at x = 1 m with a speed of ẋ = 0 m/s.
2.1.19: [Moderate] A plot of acceleration versus time for a particle is shown. What’s the
difference between its position at t = 4 s and t = 0 s if ẋ(0) = −4 m/s?
2.1.20: [Moderate]
A ball is launched vertically with an initial speed of ẏ0 = 50 m/s, and its
acceleration is governed by ÿ = −g − cD ẏ 2 , where the air drag coefficient cD is given by
cD = 0.001 m−1 . What is the maximum height that the ball reaches? Compare this to the
maximum height achieved when air drag is neglected.
2.1.21: [Moderate] An important area in controls research is to determine minimum time solutions to a problem. Consider the following case. You’re asked to start from rest, go a distance
L = 105 m, and end up at rest once again, all in minimum time. Your vehicle has a maximum
acceleration
capability of a1 and a maximum braking capability of a2 . The minimum time solution is
to use maximum braking and acceleration. Determine t1 , t2 and the car’s maximum speed.
a1 = 7.5 m/s2 and a2 = −10 m/s2 .
2.1.22: [Moderate] In this problem we’ll analyze a car collision. Assume that both Car A and
B are initially traveling to the right at 98 ft/s. Car A is tailgating, leaving very little room
between the front (F ) of his car and the rear (R) of car B. In spite of this, the driver in car
A starts fiddling with the radio and stops paying any attention to Car B. At this instant, the
driver in Car B notices an accident ahead of him and starts to brake, causing a deceleration of
7
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2.1. STRAIGHT LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
20 ft/s2 . By the time the driver of Car A looks up the gap between the two cars has narrowed
to 22.5 ft and the speed of Car B has dropped to 78 ft/s.
Assume that the driver of Car A immediately reacts and tries to apply maximum braking.
The driver of Car A has good reflexes and it therefore “only” takes 0.5 s for him to hit the
brakes. Assume that his car has better brakes than the one in front and that his car begins
to decelerate at 30 ft/s2 upon contact with the brake pedal (at t = 0.5 s). Calculate when the
collision occurs and the relative speed of the collision.
2.1.23: [Moderate] In this problem we’ll examine a similar two-car system to that of Exercise
2.2.22. Again, both Car A and B are initially traveling to the right at 98 ft/s. In this case the
following car has worse brakes than the car in front - thus making it more difficult to avoid
a collision when the lead car starts braking. At t = 0 the driver in Car B sees an accident
taking place ahead of him and jams on his brakes, causing a deceleration of 35 ft/s2 . It takes
1 s for the driver of Car A to react to this and begin braking with a deceleration of 30 ft/s2 .
The initial separation between the cars is 30 ft. Determine whether or not Car A collides with
Car B. If it does, what is the relative collision speed?
2.1.24: [Moderate] Mercedes Benz offers what they call “Brake Assist” on their cars, a technology
that’s meant to increase safety. The idea is that people, when in an emergency situation, might
not brake to the fullest capabilities of the car, thus increasing the likelihood of a crash. To
evaluate how well such a system might work, we’ll consider two scenarios. In the first, the
driver is initially traveling at 60 mph and at t = 0 begins to brake. We’ll assume that the
driver initially only uses part of the car’s full braking, causing a deceleration of 20 ft/s2 . After
1 s, we’ll assume that the driver then jams to brakes on fully, achieving a braking deceleration
of 30 ft/s2 .
Compare the absolute and percentage difference between this scenario and one in which the
Active Braking system ensured maximum braking (30 ft/s2 ) for the entire braking interval.
2.1.25: [Moderate] A Separatist Droid Army battlestation is damaged during an intense battle
over Coruscant, and it begins to fall toward the planet with an acceleration of
s̈ = −
g0 R2
s2
where s is the battlestation’s distance from Coruscant’s center and R is Coruscant’s radius.
g0 = 30 ft/s2 , R = 4000 mi and the initial height above the planet is H = 400 mi. What is
the battlestation’s impact speed?
2.1.26: [Moderate]
is given by
A particle is initially at x0 ≡ x(0) = 20 m, and its time history after t = 0
x(t) = [x0 + c1 t + c2 t3 ]
where c1 = 180 m/s, c2 = −30 m/s3 , and t is expressed in seconds. If you differentiate this
to solve for the time at which the speed equals zero, you will obtain two results, one positive
and the other negative. What’s the meaning behind the negative result?
2.1.27: [Moderate] A rock that’s dropped from the top of a cliff will experience an acceleration
due to gravity, along with a deceleration due to drag. The total downward acceleration is
g − cd ṡ2 , where g = 32.2 m/s2 , cd = 0.01 m−1 , and the downward speed ṡ is in ft/s. If the
rock is dropped from 100 ft up, calculate the rock’s impact velocity.
8
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT LINE MOTION
2.1.28: [Moderate] In the year 2137 a hovering spacecraft releases a probe in the atmosphere
far above the surface of New Earth, a planet orbiting Alpha Proximi. The gravitational
acceleration g1 is constant and equal to 7.5 m/s2 . How far will the probe have traveled when
it has reached 98% of its terminal speed? The probe’s speed is governed by
ÿ = −g1 + cẏ 2
where y is the probe’s position above the planet’s surface (expressed in m). c = 1.2 × 10−4
m−1 .
2.1.29: Computational [Difficult] This exercise is a variation on Zeno’s Paradox. (For those of
you who haven’t run across this, run a search on www.google.com with the keywords “Zeno’s
paradox.” You’ll find a lot of information.) In our case we have a particle that follows the
illustrated speed versus displacement plot. You will note that the speed drops off linearly
and hits zero when the particle reaches s = 8 m.
a. Use a numerical integrator (see Appendix A) to determine how fast it is going and how
far it has traveled if it starts from ẋ = 10 m/s and travels for 5 seconds. How about after 10
seconds?
b. Determine how long it takes to get to 7.99 m, 7.999 m, and 7.9999 m. (Note that each
time we’re getting 10 times closer to 8.0000 meters than the time before.)
c. What’s the relationship between these successive jumps toward x = 8.0000 m? Can you
analytically predict the change in time associated with one of these 90% jumps?
9
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2.1. STRAIGHT LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
10
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Chapter 3
Kinematics of Particles
3.1
Straight line motion
3.2
Cartesian Coordinates
Homework Problems
2.2.1: [Easy]
⇀
⇀
Determine the coordinate transformation array between ⇀
c 1, ⇀
c 2 and b 1 , b 2 .
2.2.2: [Easy] The position
of a particle P moving in the horizontal plane is described by r⇀(t) =
⇀
⇀
2
0.1t
5t ı + 2te  m. Find the magnitude of the particle’s position, velocity, and acceleration
at time t = 2 s.
2.2.3: [Easy] The figure contains three sets of unit vectors. ⇀
ı,⇀
 are fixed with respect to ground.
⇀ ⇀
b 1 , b 2 are fixed with respect to a gauge which is itself fixed to an airplane which has begun
a roll maneuver. The final two unit vectors ( ⇀
c 1, ⇀
c 2 ) are fixed with respect to the indicator
needle within the gauge.
⇀
⇀
Determine the coordinate transformation array between b 1 , b 2 and ⇀
c 1, ⇀
c 2 and between ⇀
ı,⇀

⇀ ⇀
and c 1 , c 2 .
2.2.4: [Easy] A sailor is stuck on a tropical island and emergency supplies are airlifted to him
before a ship can come to his rescue. The supply plane (at an altitude of H = 150 ft) is
moving horizontally and approaching the island at a speed of v = 200 mph. How far before
the island must the supply package be released if it is to just reach the island?
2.2.5: [Easy] A particle P is moving in the horizontal plane according to r⇀(t) = 0.3t3 ⇀
ı + 0.4t2 sin 2t ⇀
 ft.
Find the magnitude of the particle’s position, velocity, and acceleration at time t = 1 s.
⇀
⇀
2.2.6: [Easy] Construct a coordinate transformation array from ⇀
ı, ⇀
 to b 1 , b 2 and express the
⇀
⇀
⇀
⇀
⇀
⇀
⇀
◦
vector p = 4 ı − 8  in terms of b 1 , b 2 for θ = 130 . The b 1 , b 2 unit vectors are attached to
the illustrated link AB.
⇀
⇀
2.2.7: [Easy] Construct a coordinate transformation array from ⇀
ı, ⇀
 to b 1 , b 2 and express the
⇀ ⇀
⇀ ⇀
⇀
⇀
⇀
◦
vector p = 3 ı + 4  in terms of b 1 , b 2 for θ = 53 . The b 1 , b 2 unit vectors are attached to
the pivoting bar.
11
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3.2. CARTESIAN COORDINATES
CHAPTER 3. KINEMATICS OF PARTICLES
2.2.8: [Easy] A child tosses a very strong magnet m out the back of a station wagon, which is
subsequently attracted by the sheet metal of the car. Will the magnet return and
√ stick to
ı m/s.
the back of the car or fall to the ground? The initial velocity of the magnet is v⇀ = 5 ⇀
Treat the magnet/car interaction as an acceleration that acts on the magnet equal to −am ⇀
ı.
am = 10 m/s2 and L = 1 m. If the magnet hits the car at a point more than L below its
starting height, it will encounter the plastic bumper and thus fail to stick.
2.2.9: [Moderate] In some parts of the country, pumpkin launching is a favorite pastime during
Halloween. You’ve designed a pumpkin launcher that launches the pumpkin at a height
of h = 3 ft and want the pumpkin to impact the ground a distance d = 200 ft downfield.
The launch speed is v0 = 100 ft/s. What is the minimum needed launch angle and the
corresponding total time of flight?
2.2.10: [Moderate] Flugtag is a popular competition in which contestants “fly” their homemade
aircraft by pushing it off a ramp and into the water. Assume that the ramp is set at 45◦ , the
height above the water at release is h = 20 ft and the speed at launch is v0 = 25 ft/s. How
long will the aircraft be in the air, what maximum height does it reach and how far does it
travel horizontally?
2.2.11: [Moderate]
Kyle (A) throws a ball from the left bank of a ravine (H = 30 ft) to his
buddy Clem (B) who’s on the (lower) right bank (h = 10 ft). Kyle throws the ball up at a
30◦ angle with a release speed of v = 20 ft/s. Ignore the height of the two people.
(a) What’s the needed horizontal separation s between Kyle and Clem for Clem to catch the
ball without moving?
(b) If Clem is actually at point C, a distance d = 20 ft further than the distance calculated
in part (a), how fast must he run, at a constant speed, in order to catch the ball?
2.2.12: [Moderate] Greg has been playing skee ball all afternoon at the local arcade, but much
to his dismay, he has not been able to get a single ball into the 100-point hole. The 100-point
hole is located L = 3 ft from the base of the skee ball machine’s backboard, which is angled
at β = 20◦ with respect to the horizontal. If the end of the launch ramp is h = 2 ft above the
backboard’s base and oriented at θ = 45◦ to ground, how fast should Greg project a ball up
the launch ramp so that it lands in the 100-point hole? How long does it take for the ball to
reach the hole?
2.2.13: [Moderate]
A football is kicked from the ground at an angle of θ = 50◦ and speed
v0 = 40 ft/s. A player downfield (Karl) can jump up with initial speed vj = 9 ft/s. Karl can
reach a height of 7 ft without jumping and catches the football at the top of a 1 ft jump.
(a) How long after kickoff did Karl initiate his jump in order to catch the football?
(b) How far away was Karl from the kickoff position?
2.2.14: [Moderate] A ball is launched with a speed v0 at an angle β = 50◦ with respect to the
horizontal toward point A, which is located d = 5 m along a surface inclined at θ = 15◦ to
ground. If the launch site is s = 4 m from the base of the incline, at what speed v0 should
the ball be launched so that it lands at A? How long is the ball airborne?
⇀
⇀
2.2.15: [Moderate] Construct coordinate transformation arrays from ⇀
ı, ⇀
 to b 1 , b 2 and from
⇀
⇀
⇀
⇀
⇀ ⇀
ı ,  to c 1 , c 2 for the illustrated mechanism. Use these to express b 1 and b 2 in terms of
⇀ ⇀
⇀ ⇀
c 1 , c 2 . Show that the overall transformation array between ⇀
c 1, ⇀
c 2 and b 1 , b 2 can be expressed
in terms of sin(θ − φ) and cos(θ − φ).
12
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CHAPTER 3. KINEMATICS OF PARTICLES
3.2. CARTESIAN COORDINATES
2.2.16: [Moderate] Ronald, a clever kid, has designed a snowball launcher to get his snowballs
over a neighbor’s fence and into their yard. The fence is 17.58 m high (these neighbors are
very unsociable), the launcher is positioned 67.5 m from the wall, and the neighbor’s son,
Rupert, is sitting on the ground 79.5 m from Ronald (12 m from the wall). What is the only
allowable launch angle that, with a sufficient launch velocity, will cause the snowball to just
miss hitting the top of the wall and then land on Rupert?
2.2.17: [Moderate] A dried pea is projected upward at an angle of η degrees with speed v.
Derive the formula that will let you solve for the h (the distance up the slope at which the
pea contacts the surface) as a function of η, g, and v.
2.2.18: [Moderate] A cricket wants to leap onto a spot that’s 0.97 feet higher than the spot it’s
currently on. It’s currently
resting on a 30◦ slope. Assuming that it can launch itself at an
√
angle, η, with speed 8 2 ft/s, determine the necessary value of η for it to attain its goal.
2.2.19: [Moderate] A watermelon launcher is designed to launch watermelons short distances
so as to gauge the watermelon’s ability to withstand rough handling during shipment. If the
melon is launched at a 45◦ angle and has to land 10 feet along a 35◦ slope, how fast must the
watermelon be launched?
2.2.20: [Moderate] A tennis ball is hit onto the illustrated inclined surface and rebounds with a
velocity v⇀1 . Once it bounces, it experiences a constant downward acceleration of 9.81 m/s2 .
How far down the slope will it impact? Solve by finding the intersection of the sloped surface
with the ball’s parabolic trajectory. kv⇀1 k = 25 m/s and β = 30◦ .
2.2.21: [Moderate] Imagine you’re at a party, talking with a friend, when a person suddenly
trips against a nearby table, causing a glass to fall off the edge. An excellent human reaction
time is 0.25 s. In that time, how far will the glass fall? Assume that you immediately (after
the 0.25 s have elapsed) begin to accelerate your hand so that you grab the glass when it is
6 in. from the floor. What constant acceleration was necessary, and how fast was your hand
traveling when you contacted the glass? Assume that your hand moved in a straight line.
(Incidentally, this actually once happened to me and, yes, I actually caught the glass.)
2.2.22: [Moderate] During action movies, cars often leave the ground for a variety of dramatic
reasons. In the scene under examination, the stuntdriver has been asked to get the car up to
a sufficient speed so that it can go up a 20 degree incline, launch into the air, and land on a
platform located ahead of the car. The relevant dimensions are shown in the sketch. Treat
the car as a mass particle and solve for the minimum speed v that will allow the stunt to be
successfully undertaken.
2.2.23: [Difficult] Ernst drops a ball off of a bridge a distance H = 70 ft from the river below
and watches it fall. When it reaches the point A (50 ft above the river) a strong gust of wind
pushes it to the right with a speed of vw = 80 ft/s.
(a) How long does the ball take to hit the ground?
(b) Where does it hit?
(c) What is the impact angle and speed?
2.2.24: [Difficult] You already know that acceleration is the time derivative of velocity, but did
you know that the time derivative of acceleration is known as “jerk?” Yep, that’s the real
word for it. Imagine what it feels like to be in a car in which the driver alternately mashes
the accelerator and then jams on the brakes. That’s what jerk measures. Calculate how the
acceleration and jerk in the vertical direction (of the wheel’s center) are affected by speed
13
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3.3. POLAR
CHAPTER 3. KINEMATICS OF PARTICLES
when a sharp-edged bump is encountered. Assume that the wheel is attached to a car that is
traveling to the right at a speed v0 . As the wheel climbs up the bump, the horizontal velocity
component of the wheel’s center stays constant and equal to v0 .
2.2.25: [Difficult] A paintball is fired from a starting position x = y = 0 with speed v0 and
inclination θ, the intention being to hit a car when the car has traveled a distance L up a
slope. The car travels at a constant speed vC and starts from (x1 , y1 ) at the same instant
that the paintball is fired. Find an expression for the necessary launch angle θ in terms of φ,
d, h, L, g, and vC .
2.2.26: [Difficult] A simple batting game allows the user to fire a mass m from a tube when a
button is pressed. Initially, the bat is horizontal. The bat has a length L and θ varies as
θ̈ = αt, α a constant. At what speed must the mass be fired so that it strikes the middle of
the bat when θ = π2 rad? The launch angle φ is fixed. Let the time of impact be denoted by
t∗ .
3.3
Polar
Homework Problems
2.3.1: [Easy] A young boy named Andy is located at the center of a circular turntable of radius
10 m. The turntable is rotating in a counter-clockwise direction at 0.4 rad/s. His brother Ben
is located on the outer edge of the turntable. At t = 0 Andy begins to run directly toward
Ben, moving with a constant acceleration of 3 m/s2 . Ultimately he collides with Ben and they
both fall off the turntable. At the instant of collision, what was the absolute value of Andy’s
velocity and acceleration with respect to ground?
2.3.2: [Easy] Bill is initially at the outer edge (A) of a carousel of radius 40 ft which is rotating
in a counter-clockwise direction at 0.6 rad/s. He begins to move across the carousel and at
the moment he reaches the center O he has a speed of 5 ft/s and a constant acceleration of
2 ft/s2 (t = 0 is referenced to the point at which he reaches the center). At t = 0 the carousel
begins to decelerate at a constant rate of 0.1 rad/s2 . What are Bill’s velocity and acceleration
with respect to ground when he reaches point B?
2.3.3: [Easy] A ceiling fan is moving at a constant rotation rate and makes one revolution in 1.3
s. The distance from the fan’s center to the outermost tip of a blade is 1 m. A fly F is resting
right at the tip. What is the fly’s acceleration?
2.3.4: [Easy]
Give an example of a motion of point A for which
dk r⇀A k
=0
dt
and
kv⇀A k = 1
14
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CHAPTER 3. KINEMATICS OF PARTICLES
3.3. POLAR
2.3.5: [Easy] The lower portion of a fire ladder (OA) rotates about its hinge O at θ̇ = 0.05 rad/s
and θ̈ = 0.04 rad/s2 . The upper portion extends out from the lower portion such that ċ = 0.4
m/s and c̈ = 0.1 m/s2 . Determine the velocity and acceleration of B with respect to e⇀r , e⇀θ
as well as ⇀
ı, ⇀
 for θ = π6 rad and c = 2 m.
2.3.6: [Easy] A hydraulic piston controls the opening of a car’s trunk lid. What is the magnitude
of the velocity and acceleration of the end B if θ̇ = −0.683 rad/s, θ̈ = −0.1585 rad/s2 ,
d2 ⇀
d ⇀
2
k r B/ k = −0.0549 m/s, and dt
k r⇀B/ k = 0.3 m, dt
2 k r B k = 0.0375 m/s ?
/
O
O
O
2.3.7: [Easy] You’re tracking an airplane from the ground. The aircraft is at a constant height
h from the ground, a distance r0 from you at the illustrated instant, and at an inclination
θ. The aircraft’s speed is constant at 1200 km/h. Find the rate at which your tracking dish
must rotate if r0 = 3 km and θ = 30◦ . Does the acceleration of gravity make any contribution
to your answer?
2.3.8: [Easy] Car B is driving straight toward the point O at a constant speed v. An observer,
located at A, tracks the car with a radar gun. What is the speed |ṙB/ | that the observer at
A
A records?
2.3.9: [Easy]
Explain why
dr(t)
dt
d ⇀
dt k r (t)k
is not in general equal to
equal to
d ⇀
dt ( r (t))
2.3.10: [Easy]
When is
2.3.11: [Easy]
If ṙ = 3 ft/s and rθ̇ = −3 ft/s, will
d ⇀
dt k r (t)k.
?
d ⇀
dt ( r )
= 0?
2.3.12: [Moderate]
In this exercise we’ll see how path curvature can affect the acceleration
felt by an object. In case (a) the path traveled by a particle A is described by a sinusoid:
y = r0 − r1 cos(2ωt). In case (b) we have the same variation of height but in this case we
vary the radial variable r rather then the height measure y. The particle travels around at a
constant rate (θ = 2ωt, with ω constant) and r = r0 − r1 cos(θ). Calculate and compare the
acceleration in each case, ÿ for case (a) and ar for case (b).
2.3.13: [Moderate] When I’m out photographing birds in flight, I often have a problem keeping
the bird in view. Let’s analyze this problem. Assume that I’ve spotted a Violet-Green Swallow
that’s moving in a straight line at a constant speed, as shown. My camera’s telephoto lens
captures 40 inches of target image when the object is 80 feet away. Thus a 40 inch long bird
would go from end to end in the picture. Luckily, swallows are smaller than this, being only
8 inches long. Assume that the bird is at its closest position to me (θ = 90◦ , h = 80 ft) and
traveling at 40 mph.
(a) At what angular speed will I have to rotate my camera to keep the bird centered?
(b) What is the acceleration of the far end of the lens? Assume a distance of 17 inches from
the far end to the center of my head (the assumed center of rotation).
2.3.14: [Moderate] In this problem we’ll look more deeply into the problems of photographing
moving objects. Consider Exercise 13, in which we’re trying to photograph a moving swallow.
One of the key problems is that a slight change in speed (either on the part of the swallow of
the photographer) will cause the bird to disappear from view. Let’s determine the accuracy
that one needs in order for this not to happen. A good human reaction time is 0.5 s so we’ll
take that as our time interval. Assume that an 8 inch bird (traveling at 40 mph at a constant
distance of 80 feet from the camera) which is initially centered in the viewfinder, moves to
15
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3.3. POLAR
CHAPTER 3. KINEMATICS OF PARTICLES
the left and completely out of view in 0.5 s. Determine what the constant angular velocity of
the camera needs to be for this to occur. Compare this angular speed to the angular speed
needed to perfectly track the bird and determine the percent variation from the ideal case.
2.3.15: [Moderate]
In this problem we’ll continue to examine the problems of photographing
moving objects. Consider Exercise 13, in which we’re trying to photograph a moving swallow.
One of the problems with tracking a moving bird is that the rate at which one has to rotate
the camera isn’t constant, there exists a significant angular acceleration. Assume that a
swallow is moving at a constant 40 mph in a straight line, as shown. The swallow starts off
far off to the right (θ ≈ 0) and then proceeds to move to the left. Calculate at what angle θ
the camera’s angular acceleration is a maximum. What is its angular acceleration at θ = 0
and θ = 90◦ ?
2.3.16: [Moderate]
A Republic laser weapon targets a Trade Federation battleship fleeing to
space at a constant velocity of v⇀ = 25 ⇀
 m/s. If the length of the laser beam and its orientation
with respect to the ground are described by r and θ, respectively, find what r, ṙ, r̈, θ̇, and θ̈
are when θ = 50◦ . The horizontal distance between the laser weapon and battleship is 300 m.
2.3.17: [Moderate] Sarah has attached a resistance cable to her foot to give her leg a workout.
Let r be the length of cable from the spool to Sarah’s foot and θ be the angle the cable makes
with the ground. At a certain instant, the cable is 2 ft long and is angled at 15◦ , and Sarah’s
foot has a velocity and acceleration of v⇀ = (4.33 ⇀
ı + 2.5 ⇀
 ) ft/s and a⇀ = (0.87 ⇀
ı + 0.5 ⇀
 ) ft/s2 ,
respectively. Calculate ṙ, r̈, θ̇, and θ̈ at this instant.
2.3.18: [Moderate]
The length of a robotic arm to its end effector varies according to r(θ) =
(0.5 + 0.25[1 − cos(2θ)]) m. Suppose the robotic arm is rotating at a constant angular speed
of 0.5 rad/s. Determine the magnitude of the velocity and acceleration for the robotic arm’s
end effector when θ = 45◦ .
2.3.19: [Moderate] A light, rigid pole with a heavy lumped mass on its end is being hoisted up
with a constant angular speed of ω = 0.25 rad/s by means of a rope being reeled into a spool
to the left of the pole’s base. The rope is attached to the lumped mass on the pole’s end, and
let r describe the length of rope from the spool to the lumped mass. At a certain instant the
rope makes an angle of 40◦ with the ground, and the pole is vertical. If the pole is 20 ft long,
find what r, ṙ, r̈, θ̇, and θ̈ are at this instant.
2.3.20: [Moderate] A little kid places his RC toy car at the center of a merry-go-round at the
local playground. He then spins the merry-go-round to a constant angular speed of 3 rad/s
and proceeds to drive the RC car directly outward to A on the merry-go-round’s edge with
an acceleration of 3 ft/s2 relative to the merry-go-round. What is the magnitude of the toy
car’s velocity and acceleration at A? The radius of the merry-go-round is 5 ft.
2.3.21: [Moderate] A dolphin, being tracked by sonar, produces readings of r = 300 m, vr = 10
m/s, vθ = 17.3 m/s, ar = 0 m/s2 , and aθ = 0 m/s2 . What are the corresponding values of ṙ,
r̈, θ̇, and θ̈?
2.3.22: [Moderate] Imagine that you’re a poor graduate student who has been given the job of
determining how fast a gecko (a small lizard) can climb walls. Since your research budget is
small, all you have available is a spool of thread and a lightweight spinometer. The spinometer
records the rate at which it spins and from that calculates the rate at which the thread leaves
the spool. You put the spool of thread in the spinometer and attach the thread’s free end to
the gecko’s neck, as shown. When the gecko is 2 m up the wall, your spinometer reads 0.78
m/s. What is the gecko’s speed?
16
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CHAPTER 3. KINEMATICS OF PARTICLES
3.3. POLAR
2.3.23: [Moderate] The position of a charged particle is given by r = at − bt3 and θ = ce−t .
Derive expressions for the magnitudes of the particle’s velocity and acceleration.
2.3.24: [Moderate] A stunt plane is at the bottom of a circular loop. The loop’s radius is 1000 feet,
its center is directly above O, and the plane is traveling at a constant speed. At the illustrated
instant it is 100 feet above an observer at O and traveling at 300 mph. The acceleration of the
plane is oriented upward (toward the center of the loop) and has a magnitude of 193.6 ft/s2
(the centripetal acceleration of the plane moving in a circle about C). Is r̈ equal to 193.6 ft/s2
as well? θ = π2 rad.
2.3.25: [Moderate] A rope runs from a reel at O to an eyelet A attached to a vertically operating
door. Ignore the dimensions of the reel. Assume that rope is being reeled in at a constant
rate of 2 m/s and at the illustrated instant r = 2 m and θ = 25◦ . What are the velocity and
acceleration of A at this instant?
2.3.26: [Moderate] A dual axis accelerometer at the end of an extensional robotic arm gives a
reading of a⇀ = (4 e⇀r + 5 e⇀θ ) m/s2 . At this instant r = 1.5 m, r̈ = 10 m/s2 , θ = 45◦ , and θ̈ =
−0.5 rad/s2 .
a. Express the acceleration in the form a1 ⇀
ı + a2 ⇀
.
b. Determine θ̇ and ṙ.
2.3.27: [Moderate] Car C is driving on a circular section of track and at the position shown has
a velocity of v⇀ = −50 ⇀
 mph. At the illustrated instant the car is accelerating at 0.3 g in the
direction of travel.
a. What are the acceleration components of the car in the ⇀
ı and ⇀
 directions?
b. What is L̈ equal to?
2.3.28: [Moderate] Let the distance from O to A be given by r = aθ, with a equal to 10 ft/rad.
If θ̇ = 10 rad/s, what are v⇀A and a⇀A as functions of time?
2.3.29: [Moderate] What is the acceleration a⇀A of point A on the periphery of a drill bit that’s
rotating at a constant 5000 rpm? d = 6 mm. What would the drill bit’s constant, negative
angular acceleration need to be for the magnitude of the e⇀θ component of A’s total acceleration
to equal the value of ||a⇀A || you just calculated? How long would it take for the drill to spin
down to rest at this value of acceleration?
2.3.30: [Moderate] A laser pointer, located at O, is aimed at the top of a tunnel, the shape of
which is such that rB/ = L(1 + sin θ). The pointer is swung in a counterclockwise manner
E
such that θ̇ = (2.0 rad/s2 )t. Assume that the swing began at θ = 0 when t = 0 s. What are
v⇀A and a⇀A when θ = π2 rad, where A is the point on the tunnel where the laser beam hits?
2.3.31: [Moderate] This exercise is an extension of Exercise 2.3.22. As in Exercise 2.3.22,
you’re trying to analyze the motion of a gecko on a wall. A spool of thread is fed from a
spinometer, a handheld device that records the rate at which the thread unreels. When the
gecko is in the position shown in E2.3.22, the thread is feeding out from the reel at a speed
of 0.78 m/s and acceleration of −1.2 m/s2 . What is the gecko’s speed? Can you determine
its acceleration as well?
17
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3.3. POLAR
CHAPTER 3. KINEMATICS OF PARTICLES
2.3.32: [Moderate] A Confederation battlestation has captured a derelict Xiyalian transport
vessel in its tractor beam, causing the distance from the battlestation to the transport to
decrease as r = a(t0 −t). The Xiyalian vessel’s speed in the e⇀θ direction is constant at b. What
are the velocity and acceleration of the Xiyalian transport (before it hits the battlestation)
as a function of time?
2.3.33: [Difficult]
A car B is traveling as illustrated, and follows a path given by
r⇀B/ = bt2 er ,
θ = at2
O
What are r⇀B/ , v B , and aB when θ =
O
π
2
rad?
2.3.34: [Difficult] Two tracking stations are observing a satellite S. Station B is directly under
the satellite, and Station A lies under the satellite’s path at an angle of 45 degrees from
the satellite’s current position. Assume that both Stations A and B are positioned as shown,
both on the same level. This neglects the earth’s curvature, a reasonable assumption for these
distances. Station B records the values β = 90◦ , β̇ = −4.88 × 10−2 rad/s, r2 = 100 miles,
and ṙ2 = 0 mph. The total acceleration of the satellite is a⇀ = −30.7 e⇀r2 ft/s2 .
a. Find r̈2 and β̈.
b. Find r1 , ṙ1 , r̈1 , θ̇, and θ̈.
2.3.35: [Easy] The illustrated device is a robotic probe that tests the surface integrity of curved
panels. The arm’s length L is constant, but the height h of the probe tip and the orientation
θ of the arm are variable and given by
θ(t) = a[1 − cos(ω1 t)]
h(t) = b[1 − cos(ω2 t)]
What is the total acceleration felt by the probe tip?
2.3.36: [Moderate] Consider a carousel ride. Each carved horse is attached to a fixed vertical
post that’s fixed to the carousel’s floor. As the carousel rotates, the horse moves up and
down the post. We will analyze the motion as the carousel ride is slowing to a stop. Let the
vertical motion of the horse be given by z = z0 + z1 sin(ω2 θ), where θ is the angular rotation
of the floor. The distance from the carousel’s axis of rotation and the fixed post is rh , and
the carousel’s rotational speed is given by ω1 e−bt . What are the velocity and acceleration of
the horse?
2.3.37: [Moderate] The Guggenheim Museum in New York is designed so that the floor is a
continuously rising spiral, letting people view the artwork as they slowly climb the ramp.
Assume that an art thief has nabbed a painting and is running down toward the main door.
Given the illustrated dimensions and assuming a running speed of 12 mph, determine the
magnitude and direction of the thief’s velocity and acceleration.
2.3.38: [Moderate] A wingnut is shown, attached to a fixed, threaded shaft. The wingnut is spun
rapidly (assume a constant rotation rate) and moves down 0.9 inches in 1.2 s. The acceleration
of point A, located 0.8 in. from the shaft’s centerline, is measured to be 2.193 × 103 in./s2 .
a. How many revolutions of the wingnut took place during this time?
b. By what percentage does point A’s speed increase due to the wingnut moving down as
compared to the case in which it turns at the same rate but also stays at the same height?
18
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CHAPTER 3. KINEMATICS OF PARTICLES
3.4
3.4. PATH
Path
Homework Problems
2.4.1: [Easy]
The acrobatic airplane (P ) is currently flying level at a speed of 300 ft/s. What
is the normal acceleration felt by the airplane if the pilot pulls into an upward loop with a
radius of 1300 ft?
2.4.2: [Easy]
A car P is moving along the track and the on-board accelerometer indicates an
acceleration of a⇀P = (−7.0 e⇀t + 5.0 e⇀n ) m/s2 . At this instant it’s midway through a curve
having a radius of 100 m. What is the car’s speed?
2.4.3: [Easy] A marble M is constrained to move in the horizontal plane along a wire whose shape
is as depicted below. At point A, the marble is traveling at a constant speed vA = 5 m/s, and
the radius of curvature at that location is rA = 2 m. When the marble reaches point B, it has
a speed of vB = 3 m/s and a total acceleration of ka⇀B k = 10 m/s2 . The tangential component
of the marble’s acceleration at B is aB,t = 2 m/s2 . What is the marble’s total acceleration at
A and its radius of curvature at B?
2.4.4: [Easy] A GPS tracking system determines that the race car P in a curve of radius r = 90 m
has an acceleration of a⇀P = (0 e⇀t + 0 e⇀n ) m/s2 . What is its speed?
2.4.5: [Easy]
A motorcyclist enters a curve whose radius of curvature is described by rc =
25 − 0.002s3 ft. Suppose the motorcycle can accelerate at 30 ft/s2 before it begins to slip,
and slipping occurs when the motorcycle is 20 ft into the curve. At what constant speed was
the motorcyclist driving with through the curve?
2.4.6: [Easy] A person P on a Ferris wheel travels in a circle with a 30-foot radius and experiences
an acceleration of magnitude 0.33 ft/s2 . The person’s speed is constant during the ride. What
is the Ferris wheel’s rotational speed?
2.4.7: [Easy] A car enters a circular offramp (radius of 180 m) at 30 m/s. The car’s on-board accelerometers sense a total acceleration magnitude of 7.07 m/s2 . What is the car’s acceleration
magnitude at (the component tangent to its path)?
2.4.8: [Easy]
Consider a marble M constrained to move in the horizontal plane along the
illustrated wire. When the marble arrives at point A, its total acceleration is given by
ka⇀A k = 14 ft/s2 . The radius of curvature at A is rA = 0.7 ft. At point B, the radius of
curvature is rB = 0.4 ft, and the marble is traveling with a speed and total acceleration of
vB = 2.5 ft/s and ka⇀B k = 18 ft/s2 , respectively. Find the marble’s constant speed at A and
the tangential component of its acceleration at B.
2.4.9: [Easy] When designing a rollercoaster, the designer wants to ensure that the passengers
survive the ride. At the bottom of a circular loop, the designer wants the cars to move at 60
mph, and safety considerations demand that the normal acceleration not exceed 3.5 g. What
is the minimum allowable radius r?
2.4.10: [Easy] A Ferris wheel with radius 30 feet is decelerating such that at t = 0 s the tangential
3
speed of a point P on the rim is v = 10 ft/s and dv
dt = ct. c = −0.4 ft/s . What is the
acceleration of P at t = 3 s?
19
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3.4. PATH
CHAPTER 3. KINEMATICS OF PARTICLES
2.4.11: [Easy]
Suppose two cars are traveling on a wide road at constant speeds when they
encounter a U-turn. Car A is in the inside lane, moving at vA = 24 m/s in a semicircular
path of radius rA = 90 m. Car B has a speed of vB = 28 m/s in the outside lane, and
the radius of its path is rB = 95 m. Assume that the cars can handle a maximum normal
acceleration of an,max = 0.75g before their tires lose traction and they slide off the road. Will
either vehicle make it around the turn without sliding off the road, and if so, how long does
it take?
2.4.12: [Easy] A skier using the illustrated ski jump starts from rest at A and accelerates at a
constant 28 ft/s2 to B. Just after B the straight slope changes to a curve having a radius of
curvature equal to 220 ft. What is the magnitude of the total acceleration felt by the skier at
this point? Assume that the tangential acceleration doesn’t immediately change. Comment
on the magnitude.
2.4.13: [Easy] The small particle b slides to the right at a constant speed vb toward A, the end
of the table. When it passes A, it begins to accelerate downward at 32.2 ft/s2 . Thus there
is a very marked difference in its acceleration just before and just after reaching A. Is the
same true for the velocity? What are the direction and magnitude of the velocity vector the
instant before reaching A as well as the instant after?
2.4.14: [Moderate] The sports car A starts from rest on a circular track (radius 100 m) and its
speed is given by v = v0 1 − e−at where v0 = 60 m/s and a = 0.3 s−1 . How much time must
elapse before the sports car’s tangential acceleration is 10% of its normal acceleration?
2.4.15: [Moderate] A particle P is moving along a curve defined by y = ax2 , where a = 6 ft−1 .
ẏ = bx, b = 12 s−1 . Determine where along the curve P ’s speeds in the ⇀
ı and ⇀
 direction are
equal in magnitude and calculate the normal and tangential acceleration at that point.
2.4.16: [Moderate] A particle P is moving along a sinusoidally varying path, it’s position defined
π.
1 ft−1 and a = 80 ft. a = (8.0 ⇀
ı +2.0 ⇀
 ) ft/s2 when x = 2λ
by y = a[1−cos(λx)] with λ = 320
P
Can you determine what the particle’s speed is equal to at this instant?
2.4.17: [Moderate]
A car is traveling at a constant speed vA = 25 mph when it encounters a
90◦ bend to the right at point A, where the path is given by an arc of radius r1 = 70 ft. The
car exits the bend at point B and then travels a straight-line distance of d = 40 ft before
encountering another 90◦ bend (this time to the left) at point C, for which the path’s radius
is r2 = 80 ft.
a. If the car can handle a maximum normal acceleration of an,max = 0.75g before its tires
lose traction and it slides off the road, what is its maximum constant acceleration from B
to C so that it just makes it around the bend from C to D while traveling at a constant
speed?
b. How long does it take for the car to go from A to D under these conditions?
2.4.18: [Moderate]
A snowball is careening down a hill whose surface can be approximated
by y = 0.25x2 m. At x = 4 m, the snowball has a tangential velocity and acceleration of
10 m/s and 2 m/s2 , respectively. Find the magnitude of the snowball’s total acceleration at
this position.
2.4.19: [Moderate] The end of a test tube in a centrifuge is located 0.75 ft from the center. If the
end of the test tube has a tangential velocity and acceleration of 10 ft/s and 10 ft/s2 , respectively, at a particular instant, calculate the magnitude of the total acceleration it experiences
in g’s. Also, find the angular velocity and acceleration at this instant.
20
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CHAPTER 3. KINEMATICS OF PARTICLES
3.4. PATH
2.4.20: [Moderate] A snowboarder is on the right edge of a halfpipe, just about to propel himself
into the air. Immediately before leaving the halfpipe’s edge, the snowboarder experiences a
tangential acceleration of 10 ft/s2 and a total acceleration of 20 ft/s2 . What is his tangential
speed and normal acceleration at this instant? The halfpipe has a radius of 20 ft. How high
does the snowboarder get in the air, and how long is he airborne?
2.4.21: [Moderate] A car (at point A) is 100 m away from point B along a circular track (distance measured along track’s perimeter) and begins to accelerate from rest. What constant
speed increase is needed for the total acceleration magnitude to be 8 m/s2 when it reaches
B? The radius of the track is 150 m.
2.4.22: [Moderate] A bicyclist comes around a decreasing radius turn and maintains a constant
speed of 20 mph. The radius of curvature varies as rC = a + bs2 , where s indicates motion
along the curve expressed in feet (a = 50 ft and b = −0.0025 ft−1 ). How long will it be until
the bicycle slips if the maximum acceleration it can sustain without slip is 28 ft/s2 ?
2.4.23: [Moderate] A Ferris wheel is started up from rest such that its rotational speed is given
by θ̇ = ct, where θ measures the rotational position of the wheel (c = 0.05 rad/s2 ). What are
the magnitude of the velocity and acceleration of a point P on the wheel’s rim at t = 5 s?
The distance from the center of rotation to the rim is 35 feet.
2.4.24: [Moderate] A car is traveling along a horizontal race track. From A to B the track is
straight, and from B to D it is in the form of a circular arc. The car is accelerating from A to
C with a constant tangential acceleration of 5 ft/s2 . Its velocity at A is 80 ft/s. The distance
from A to B is 300 feet and from B to C the distance is 300 feet as well. The magnitude of
the car’s total acceleration (accounting for both normal and tangential components) is 5 ft/s2
at A and 14.6 ft/s2 at C. What is the track’s radius of curvature at C?
2.4.25: [Moderate] When a particular meteoroid enters our atmosphere (thus becoming a meteor and soon to be a meteorite), it encounters aerodynamic drag. At the instant it is first
noticed by a tracking station it’s experiencing a gravitational acceleration of 9.5 m/s2 and
an aerodynamic deceleration of 5 m/s2 . Its speed is 25,000 kph, and its angular heading is 3
degrees down from horizontal.
What is the radius of curvature of the trajectory at this instant? Note that the deceleration
due to aerodynamic drag points in the opposite direction to the meteor’s velocity vector.
2.4.26: [Moderate] At the instant illustrated, a rocket is experiencing a thrust from its engine that is producing a forward acceleration and is also being accelerated downward due to
gravity. The overall acceleration of the rocket is
a⇀ = (5.657 ⇀
ı − 3.843 ⇀
 ) m/s2
The acceleration due to gravity at the rocket’s position has a magnitude of 9.5 m/s2 , and the
rocket’s velocity vector is
v⇀ = (5000 ⇀
ı + 2000 ⇀
 ) m/s
a. Determine the acceleration due to the rocket’s engine.
b. Determine at and an .
c. Determine the radius of curvature of the rocket’s path at this instant.
2.4.27: [Moderate] A particle moves along the path defined by xy = 36 m2 . At time t0 its speed is
constant at 10 m/s, and its velocity has a positive ⇀
ı component. At the illustrated instant it’s
at x = 9 m. What are the acceleration components of the particle in the ⇀
ı and ⇀
 directions?
21
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3.4. PATH
CHAPTER 3. KINEMATICS OF PARTICLES
2.4.28: [Moderate] A car starts from rest on a circular track with a radius of 300 m. It accelerates
with a constant tangential acceleration of at = 0.75 m/s2 . Determine the distance traveled
and the time elapsed when the magnitude of the car’s overall acceleration is equal to 0.9 m/s2 .
2.4.29: [Moderate] A race car C is being tracked via telemetry. The on-board accelerometer records an acceleration level of −0.7g e⇀t + 0.5g e⇀n . If you know that v at this instant is
200 km/hr, what is the track’s radius of curvature at this instant?
2.4.30: [Moderate] A sportscar C is being driven around a decreasing radius turn. On entry, the
car’s speed is 100 mph, and the driver keeps the speed constant throughout the curve. The
radius of curvature at entry is rC1 = 1000 ft, and the radius of curvature at exit is rC2 = 500
ft. Will the car complete the turn without losing traction and sliding? Assume that the tires
can maintain rolling contact up to an acceleration of 35 ft/s2 .
2.4.31: [Difficult]
Roman chariot racer Andronicus is competing in the big race at the Circus
Maximus. Starting at A, Andronicus accelerates to B at 2 m/s2 and then travels through
the first round to C at a constant speed. From C to D, he accelerates at 1 m/s2 and then
returns to A with a constant speed through the second round. The straightaways are 50 m
long, and the rounds have a radius of 15 m. What is Andronicus’s speed at B, and what is his
acceleration through the first round? Also, determine his speed at D and how long it takes
Andronicus to complete one circuit.
2.4.32: [Difficult]
Derive equation (??).
22
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CHAPTER 3. KINEMATICS OF PARTICLES
3.5
3.5. REL MOTION
Rel Motion
Homework Problems
2.5.1: [Easy]
The Red Devils high performance, radio controlled model team is performing at
a local airshow, and two jets are in the middle of a death-defying maneuver. Jet A is in the
bottom of a loop with a radius of 100 ft, and at this instant it has a velocity and acceleration
of v⇀A = 660 ⇀
ı ft/s and a⇀t,A = 50 ⇀
ı ft/s2 , respectively. Just a few feet below jet A, jet B is
traveling at v⇀B = −700 ⇀
ı ft/s with an acceleration of a⇀B = −40 ⇀
ı ft/s2 . What are the velocity
and acceleration of jet A as seen by jet B at this instant?
2.5.2: [Easy] A commuter plane is en route from A to B with a velocity of v⇀p = 200 ⇀
 mph in a
straight line path. At a point d = 100 mi from B, the plane encounters some strong winds of
velocity v⇀w = 80 ⇀
ı mph. If the plane maintains its current speed and direction relative to the
wind, how far over from B will the plane be at the anticipated time of arrival (i.e., with no
wind)? At what speed and angle should the plane fly relative to the wind to actually arrive
at B in this time?
2.5.3: [Easy] In an old James Bond movie, 007 used an ejector seat to remove a bad guy from
his Aston Martin. If 007 was traveling at 30 m/s when he activated the ejector and the
launch velocity relative to the Aston Martin was 10 m/s (oriented vertically with respect to
the Aston Martin), what was the overall velocity of the bad guy as he left the car?
2.5.4: [Easy] A disk of radius R = 7 in. spins at a constant angular speed of ω = 500 rpm in the
clockwise direction. If the disk is released from rest far above the ground, what is the speed
and acceleration of the disk’s top C after it has fallen h = 3 ft?
2.5.5: [Easy] A worker is dragging a large drum of mass m through the use of a pulley arrangement.
If she’s able to move to the left at 2 m/s, how fast will the mass move to the right?
2.5.6: [Easy] Two masses, A and B, are connected by a rope that goes over a central reel C.
The reel is attached to a horizontal bar that rides in two vertical guides. The bar’s motion is
given by x, and the height of A above the ground is given by y. What is the velocity of B in
terms of ẋ and ẏ?
2.5.7: [Easy]
Suppose two ball bearings (A and B) are constrained to move in the horizontal
plane along concentric circular paths, where rA = 0.4 m and rB = 0.9 m. At the illustrated
instant, the two ball bearings are collinear with respect to the vertical, and it is measured
that A is traveling at a constant v⇀A = −5 ⇀
ı m/s and B is moving at v⇀B = 8 ⇀
ı m/s with a
2
⇀
⇀
tangential acceleration of aB,t = 0.7 ı m/s . What are the velocity and acceleration of B as
seen by A?
2.5.8: [Easy]
2.5.9: [Easy]
The free end A of the pulley is pulled down at 2 m/s. What is Block B’s velocity?
A motor at R reels in rope at vR = 3 m/s. What is the velocity vC of block C?
2.5.10: [Moderate] An excursion train, traveling at 80 ft/s, encounters a rain shower. The point
A lies directly eight feet beneath the lip of an overhang in the observation car. The raindrops
can be assumed to be falling at a constant speed of 40 ft/s. Under the current conditions,
a raindrop just clearing the lip would strike the observation car’s floor sixteen feet from the
point A. At what constant rate would the train need to be decelerating (deceleration starting
just as the drop passes the lip) for the drop to hit the floor eight feet from A? What angle
does the raindrop’s velocity vector make with respect to the floor at the time of impact, as
seen by an observer on the train?
23
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3.5. REL MOTION
CHAPTER 3. KINEMATICS OF PARTICLES
2.5.11: [Moderate]
An air force fighter jet on its way back to base with a velocity of v⇀j =
⇀
360 ı mph needs to refuel mid-flight. A refueling plane is called in, deploys its fuel nozzle,
and travels at vp = 350 mph at an angle β = 85◦ to the vertical. Calculate the approach
speed of the fuel nozzle with respect to the jet. Suppose the refueling plane slows down to
340 mph. What speed should the fighter pilot slow down to if he wants to maintain the same
nozzle approach speed?
2.5.12: [Moderate] Fred attaches a decorative windmill to his car’s antenna and then goes out
for a drive. At a certain instant, Fred’s velocity and acceleration are v⇀c = 30 ⇀
ı mph and
2
⇀
⇀
ac = 5 ı ft/s , respectively, and the windmill is spinning at θ̇ = 200 rpm counterclockwise.
Take A to be the edge of the windmill directly above the center of rotation. If the windmill
has a radius of r = 4 in., find the magnitude of the velocity and acceleration of A at the given
instant, for which e⇀t = − ⇀
ı and e⇀n = − ⇀
 . Assume that the antenna doesn’t flex.
2.5.13: [Moderate] A secret agent is running toward the back of a moving bus, his intent being
to jump off the back before the explosive he planted at the front goes off. At the illustrated
instant, the bus has a constant velocity of v⇀bus = 10 ⇀
ı mph. Currently, our agent is 30 feet
from the rear of the bus and moving at zero mph with respect to the bus.
a. What constant acceleration will he need with respect to the bus so that he will leave
the rear of the bus with zero velocity relative to the ground?
b. Where will he land with respect to point A?
2.5.14: [Moderate] Two cars are passing by each other, moving in the directions illustrated. Car
A is moving at a constant 30 mph around a √
circle with radius 100 ft, and Car B is moving
 . What are the velocity and acceleration of
at a constant 60 mph in the direction 0.5 ⇀
ı + 23 ⇀
Car A with respect to Car B? r⇀A/ = 100 ⇀
ı ft and r⇀B/ = 200 ⇀
ı ft.
O
A
2.5.15: [Moderate] A swordfish has been hooked by a fisherman who has 500 m of line remaining
in his fishing reel. The initial position of his boat (B) and the swordfish (S) are shown.
v⇀B = −vB ⇀
 = −3 ⇀
 m/s and v⇀S = −vS ⇀
ı = 10 ⇀
ı m/s, both constant.
a. How long before he runs out of line?
b. What are v⇀S/ and a⇀S/ ?
B
B
2.5.16: [Moderate] An excursion train, traveling at 30 mph from left to right in the figure,
encounters an unexpected rain shower. Point A lies directly 7 ft beneath the lip of an overhang
in the observation car. How far toward the rear of the train will the raindrops reach if they’re
falling at 25 mph? What is the velocity of a raindrop with respect to an observer in the train?
2.5.17: [Moderate] Assume you’re out paddling in a river that has a 3 m/s current. You’re
midstream, 11 meters from either shore. If you paddle toward the left shore at 4 m/s and at
a 45 degree angle ( ⇀
ı +⇀
 direction), where along the shore will you land?
2.5.18: [Moderate] You’re out paddling a raft R in the middle of the river and hear a friend
(located at F ) call from the shore. In what direction should you paddle to arrive at F in 2
minutes? The river’s current has a velocity 1 ⇀
ı m/s.
2.5.19: [Moderate] A rod, B, is free to slide vertically within the slot cut into board A. Another
board, C, with an inclined slot cut into it, is placed over the first board so that the rod is
held within both slots. Determine the velocity of the rod, relative to board C, if A is held
stationary and C is moved to the right at 1 m/s.
24
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CHAPTER 3. KINEMATICS OF PARTICLES
2.5.20: [Moderate]
velocity?
3.5. REL MOTION
The free end A of the pulley is pulled down at 3 m/s. What is Block B’s
2.5.21: [Moderate] A motor at O reels in the illustrated rope at 14 in/s. An observer measures
the absolute velocity of Block B and finds it to be −2 ⇀
 in./s. What is the absolute velocity
of Block A?
2.5.22: [Moderate] A motor at A reels in the illustrated rope at 0.4 m/s. What is the absolute
velocity of Block D?
2.5.23: [Moderate]
The free end B is given a velocity of −4 ⇀
 ft/s. What is the velocity of A?
2.5.24: [Moderate] A motorized reel R pulls in rope at a rate of 10 in./s. The rope goes around
pulley B, up and around pulley C, and then terminates at the center axle of pulley B. What
is the velocity of D, a point on the rightmost piece of rope connecting C and B?
2.5.25: [Moderate] Rope is drawn into the motorized reel at A at a rate of v0 . How fast does
the weight B rise above the floor (ẏ)?
2.5.26: [Moderate] If the free end of the pulley rope is pulled down with an acceleration of
4 ft/s2 , what will the acceleration of m be? For simplicity, assume that all the straight rope
segments are vertical.
2.5.27: [Moderate]
What is ẏ equal to if ẋ = 10 m/s and the rope remains taut?
2.5.28: [Moderate] A complicated system of pulleys and motors is shown. At A is a motor that
reels in the pulley rope at a rate of 20 in./s. Another motor, at G, reels in pulley rope at
10 in./s. What is Block B’s velocity?
2.5.29: [Moderate] The free end D of the pulley is pulled down with a speed of 10 in./s. What
is Block A’s velocity? For simplicity, assume that all the straight rope segments are oriented
vertically.
2.5.30: [Moderate]
If the free end at A is pulled down at a speed vA , what is the resultant
velocity of the Block B?
2.5.31: [Moderate] The reels R1 and R2 are taking in rope at speeds v1 and v2 , respectively.
Determine ẏ3 and ẏ4 in terms of v1 and v2 .
2.5.32: [Moderate]
Block A is observed to be dropping down at a steady 0.9 ft/s. At what
velocity must the free end of the pulley rope be moving?
2.5.33: [Moderate] Derive an expression for the velocity vA of block A in terms of the velocity
vR at which rope is reeled in by a motor at R.
2.5.34: [Moderate] Derive an expression for the velocity vB of block B in terms of the velocity
vA at which rope is pulled down at A.
2.5.35: [Moderate]
A motor at R reels in rope at 3 ft/s. What is the velocity of block A?
2.5.36: [Moderate]
B.
A person pulls down the rope at A at 2 ft/s. Determine the velocity of block
25
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3.5. REL MOTION
CHAPTER 3. KINEMATICS OF PARTICLES
2.5.37: [Difficult] Today, Judy is kayaking across a 70 m-wide pond from A to B with a velocity
of v⇀k = 2 ⇀
 m/s. In the first 20 m stretch of the pond, the water has a current of velocity
v⇀w = 3 ⇀
ı m/s, and she winds up at C. How far over is Judy at C? If the remainder of the
pond is still water, what angle does Judy need to travel at to get to B? Suppose Judy was
kayaking in the pond yesterday, and the entire pond was still. How much longer did it take
her to get from A to B today as compared to yesterday?
2.5.38: [Difficult] A student is out in her canoe one day and suddenly realizes that she is
midstream and heading for a waterfall. It’s 10 m to shore and 100 m to the waterfall. The
river’s current is 4 m/s. If she wants to reach the left (with respect to her) shore 5 m before
the waterfall’s edge, in what direction should she paddle the boat? Assume that she can
paddle at 3 m/s.
2.5.39: [Difficult] A single length of rope runs from B, up around two pulley wheels, and terminates at C. Derive expressions for the velocity and acceleration of A if the end C is pulled
horizontally such that v⇀C = ẋ ⇀
ı and a⇀C = ẍ ⇀
ı.
2.5.40: [Difficult] Two cars are traveling as illustrated, with v⇀A = vA ⇀
ı (vA a constant and car
starting from O) and Car B following a path given by
r⇀B/ = bt2 e⇀r ,
θ = at2
O
a. How long does it take for Car B to reach a position of
b. What are r⇀B/ , v⇀B/ , and a⇀B/ at that time?
A
A
π
2
rad?
A
26
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Chapter 1
Homework Solutions
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CHAPTER 1. HOMEWORK SOLUTIONS
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Chapter 2
Kinematics of Particles
29
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CHAPTER 2. KINEMATICS OF PARTICLES
DYNAMICS: Analysis and Design of
Systems in Motion
Benson H. Tongue/Sheri D. Sheppard
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1
2.1. STRAIGHT-LINE MOTION
Straight-Line Motion
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.1
GOAL: Determine the distance and time needed for the car to reach its maximum speed.
GIVEN: The 2007 BMW Z4 Coupe 3.0si can accelerate from 0 to 60 mph in 5.6 s, and its maximum
speed is ẋmax = 155 mph. Assume that it accelerates from 0 to 60 mph at a constant rate and that
this acceleration is maintained as the vehicle pushes toward its maximum speed.
DRAW:
FORMULATE EQUATIONS:
Since we’re assuming that the car’s acceleration is constant, we can say that
ẍ =
∆ẋ
∆t
(1)
ẋ2 − ẋ20 = 2ẍ∆x
(2)
SOLVE:
To go from 0 to 60 mph in 5.6 s at a constant rate, the car’s acceleration needs to be
(1) ⇒
ẍ =
(60 mph)
hr
3600 s
5.6 s
5280 ft
mi
ẍ = 15.7 ft/s2
If the car continues to accelerate at this rate, then it will reach its maximum speed after traveling
(2) ⇒
∆xmax
∆xmax =
h
ẋ
= max
2ẍ
(155 mph)
2
hr
3600 s
2(15.7 ft/s2 )
5280 ft
mi
i2
∆xmax = 1.64 × 103 ft = 0.311 mi
32
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
The time it takes for the car to attain maximum speed is
(1) ⇒
∆tmax =
∆tmax =
(155 mph)
ẋmax
ẍ
hr
3600 s
15.7 ft/s2
5280 ft
mi
∆tmax = 14.5 s
33
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.2
GOAL: Plot the path of someone following a map’s directions.
DRAW:
Figure 1: Paths from map
SOLVE: Figure 1 shows the two cases. (a) shows the map’s original directions and (b) shows the
overall change in position. (c) shows what we get by reversing the directions and (d) shows the
starting position to finishing position vector. As can be seen from the figure, the overall start to
finish vector is the same in each case. This illustrates how the order in which vector components
are added won’t affect the final answer.
34
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.3
GOAL:
Using the given information, find average speed and approximate instantaneous speed.
GIVEN: Graph of position vs time.
DRAW:
FORMULATE EQUATIONS: We’ll use the graph and divide the distance traveled by the time
taken in order to estimate the average speed and estimate the slope at a point in order to estimate
the instantaneous speed.
SOLVE:
You can see from the graph of displacement versus time that the particle travels 8 meters in 20
8
seconds. Thus, v = 20
m/s = 0.4 m/s .
The maximum slope occurs at t = 10 seconds. Estimating the slope with a straight-edge provides an estimate for the maximum speed of 1 m/s .
The minimum slope occurs at t = 0 and t = 20 seconds. Estimating this slope provides you
with an estimate for the minumum speed of 0.1 m/s .
35
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.4
GOAL: Find the appropriate initial speed so that the speed at t = 2.5 seconds is equal to zero.
GIVEN: Plot of acceleration vs. time.
DRAW:
FORMULATE EQUATIONS: We’ll need to integrate the acceleration to find speed and then
adjust our initial condition so that at t = 2.5 s it’s equal to zero.
SOLVE:
Because the acceleration after t = 2 is equal to 0, the velocity does not change over the period
2 ≤ t ≤ 2.5, so we can solve the problem on the period 0 ≤ t ≤ 2. For this time interval, the
acceleration is 50m/s2 . The velocity is therefore given by
v(t) = a t + v0
Using v(2.5 s) = 0, a = 50, and t = 2 in (1) and solving for v0 shows that v0 =-100 m/s .
36
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(1)
CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.5
GOAL: Determine the ball bearing’s speed v after traveling the given distance.
GIVEN: The ball bearing’s acceleration is described by s̈ = as + becs , where a = 3 s−2 , b = 1 ft/s2 ,
and c = 0.2 ft−1 . The ball bearing travels d = 10 ft, and it starts from rest.
DRAW:
FORMULATE EQUATIONS:
Since the ball bearing’s acceleration is a function of its position,
s̈ =
dṡ
dṡ ds
dṡ
=
·
= ṡ
dt
ds dt
ds
s̈ds = ṡdṡ
SOLVE:
(as + becs ) ds = ṡdṡ
(1) ⇒
Z d
(as + becs ) ds =
ṡdṡ
0
0
Z v
1 2 b cs
as + e
2
c
d
0
1
= ṡ2
2
v
0
1
1 2 b cd
ad +
e − 1 = v2
2
v=
v=
s
c
s
(3 s−2 )(10 ft)2
2
ad2 +
2b cd
(e − 1)
c
2(1 ft/s2 ) (0.2 ft−1 )(10 ft)
e
−1
+
−1
0.2 ft
v = 19.1 ft/s = 13.0 mph
37
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(1)
2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.6
GOAL: Find the constant acceleration a0 that brings a jet from 170 mph to 0 mph in 240 ft and
the elapsed time ∆t.
GIVEN: Distance needed to go from landing speed to zero.
FORMULATE EQUATIONS: Because the acceleration is constant we can use
v22 − v12 = 2a0 (x2 − x1 )
(1)
v2 − v1 = a0 ∆t
(2)
where a0 is the constant acceleration, ∆t denotes the elapsed time, and the subscripts 1,2 indicate
initial and final conditions, respectively.
SOLVE: First we’ll convert 170 mph to ft/s:
(170 mi) (5280 ft) (1 hr)
= 249.3 ft/s
(1 hr) (1 mi) (3600 s)
Using this, along with the known distance traveled, in (1) gives us
0 − (249.3 ft/s)2 = 2a0 (240 ft)
a0 = −129.5 ft/s2
Because 1 g is equal to 32.2 ft/s2 we have
a0 = −129.5 ft/s2
(1 g)
2
(32.2 ft/s )
= −4.02 g
We can then use (2) to find the time ∆t taken for the jet to come to a halt:
0 − (249.3 ft/s) = (−129.5 ft/s2 )∆t
∆t = 1.93 s
38
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.7
GOAL: Find the needed deck space to allow a fighter jet to take off from an aircraft carrier.
GIVEN: The jet is brought from 0 to 165 mph in 2.23 s.
DRAW:
ASSUME: We’ll assume that the acceleration is constant.
FORMULATE EQUATIONS: We’ll use the equation relating speed and acceleration, when
starting from rest under a constant acceleration:
v = at
and the expression for distance traveled under a constant acceleration:
1
x = at2
2
SOLVE:
First we’ll convert 165 mph to ft/s:
(165 mi) (5280 ft) (1 hr)
= 242 ft/s
(1 hr) (1 mi) (3600 s)
The acceleration is therefore
a=
v
242 ft/s
=
= 109 ft/s2
t
2.23 s
The needed space is given by
1
x = (109 ft/s2 )(2.23 s)2
2
x = 270 ft
39
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.8
GOAL: Construct plots of speed and acceleration for the given position function.
GIVEN: Differentiate to find speed and then differentiate again to find acceleration.
FORMULATE EQUATIONS:
1
3
x(t) =
m/s t3 − (16 m/s)t + 10 m
Given:
3
(1)
Differentiating (1):
ẋ = (1 m/s3 )t2 − 16 m/s
(2)
Differentiating (2):
ẍ = (2 m/s3 )t
(3)
SOLVE:
Solving (2) for ẋ = 0 gives t2 = 16 s2 so t = ±4 s. Since t = −4 s occurs before the beginning of the
time interval, the correct answer is
t = 4s .
The associated plots of speed and acceleration are given below, where ẋ has units of m/s and ẍ has
units of m/s2 .
40
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.9
GOAL: Find the speed, ẋ, and the acceleration, ẍ, at prescribed times.
GIVEN: x as a function of time.
GOVERNING EQUATIONS: The position of the car is given by:
x(t) = 100 ft − (88 ft/s)t + (10 ft/s2 )t2
(1)
Differentiating with respect to time, we get:
and
ẋ(t) = −88 ft/s + (20 ft/s2 )t
(2)
ẍ(t) = 20 ft/s2
(3)
SOLVE:
(2) ⇒
ẋ(2) = −88 ft/s + (20 ft/s2 )(2 s) = −48 ft/s
(4)
ẍ(10) = 20 ft/s2
(5)
and
(3) ⇒
41
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.10
GOAL: Find the constant acceleration a0 that brings a car from 0 to 60 mph in 5 seconds.
GIVEN: Time needed to go from zero to 60 mph.
FORMULATE EQUATIONS: Because the acceleration is constant we have
v(t) = v(0) + a0 t
(1)
where a0 is a constant acceleration.
SOLVE: First we’ll convert 60 mph to ft/s:
(60 mi) (5280 ft) (1 hr)
= 88 ft/s
(1 hr) (1 mi) (3600 s)
Using this in (1) gives us
88 ft/s = 0 + a0 (5 s)
a0 = 17.6 ft/s2
Because 1 g is equal to 32.2 ft/s2 we have
a0 = 17.6 ft/s2
(1 g)
2
(32.2 ft/s )
= 0.55 g
42
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.11
GOAL: Determine the time to bring a car to a stop from an initial speed along with the distance
over which stopping occurs.
GIVEN: Initial speed of car and the fact that the car decelerates at 1 g
FORMULATE EQUATIONS:
For a constant acceleration,
v(t) = v(0) + at
x(t) = x(0) + v(0)t +
at2
2
SOLVE:
70 mph =
And for the distance:
70
(88) = 102.6 ft/s
60
v(t∗ ) = 102.6 ft/s − (32.2 ft/s2 )t∗ = 0 ⇒ t∗ = 3.19 s
x(t∗ ) = 0 + (102.6 ft/s)t∗ −
32.2 ft/s2 ∗ 2
t
2
At t∗ = 3.19 s, x = 164 ft
43
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.12
GOAL: Determine the distance the car needs to reach its maximum speed and the change in its
0-to-60 time such that it and a competitor’s car reach their respective maximum speed in the same
distance.
GIVEN: The 2008 Audi TT Coupe can accelerate from 0 to 60 mph in 6.5 s and has a maximum
speed of ẋmax = 147 mph. Assume that it accelerates from 0 to 60 mph at a constant rate and that
this acceleration is maintained as the vehicle pushes toward its maximum speed.
DRAW:
FORMULATE EQUATIONS:
Since we’re assuming that the car’s acceleration is constant, we can say that
ẍ =
∆ẋ
∆t
(1)
ẋ2 − ẋ20 = 2ẍ∆x
(2)
SOLVE:
To go from 0 to 60 mph in 6.5 s at a constant rate, the car’s acceleration needs to be
(1) ⇒
ẍ =
(60 mph)
hr
3600 s
6.5 s
5280 ft
mi
ẍ = 13.5 ft/s2
If the car continues to accelerate at this rate, then it will reach its maximum speed after traveling
(2) ⇒
∆xmax
∆xmax =
h
ẋ
= max
2ẍ
(147 mph)
2
hr
3600 s
2(13.5 ft/s2 )
5280 ft
mi
i2
∆xmax = 1.72 × 103 ft = 0.326 mi
44
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
We find that it takes a longer distance for the TT to reach its maximum speed as compared to its
competitor. For the car to attain maximum speed in 0.312 mi like its competitor, its 0-to-60 time
would need to be
(1) → (2) ⇒
ẋmax
2
t∗ =
t∗ =
2(60 mph)
h
hr
3600 s
2(∆ẋ)(∆x)
t∗
=
2(∆ẋ)(∆x)
ẋmax
(147 mph)
2
5280 ft
mi
hr
3600 s
t∗ = 6.24 s
(0.312 mi)
5280 ft
mi
i2
5280 ft
mi
Thus, the car’s 0-to-60 time would need to decrease by
∆t = 6.5 s − 6.24 s
∆t = 0.26 s
45
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.13
GOAL: Determine a piston’s maximum speed and acceleration.
GIVEN: Position of piston as a function of time.
FORMULATE EQUATIONS: The governing equation of motion is given as
x(t) =
3.53
sin ωt
2
where x is given in inches and ω = 7000 rpm = 733 rad/s.
SOLVE:
v(t) =
vmax =
3.53 in
2
a(t) =
amax =
3.53 in
2
d
x(t) =
dt
3.53 in
ω cos ωt
2
(733 rad/s) = 1.29 × 103 in/s
3.53 in
d
v(t) = −
ω 2 sin ωt
dt
2
(733 rad/s)2 = 9.48 × 105 in/s2 = 2.45 × 103 g
46
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.14
GOAL: Find the height h for which a falling body will contact the ground at 35 mph.
GIVEN: Speed of contact.
DRAW:
ASSUME:
vi = 0
vf = 35 mph = 51.3 ft/s
a = 32.2 ft/s2
(1)
FORMULATE EQUATIONS: We’ll use the formula for the difference in speed due to a constant
acceleration a over a distance h:
(2)
vf2 − vi2 = 2ah
SOLVE:
vf2 − vi2 = 2gh
(1)→(2)⇒
h =
vf2 − vi2
(51.3 ft/s)2 − 0
=
2g
2(32.2 ft/s2 )
h =
40.9 ft
47
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.15
GOAL: Find the percentage increase in drive force that will increase a car’s maximum speed by
10%.
GIVEN: Air resistance Fa = Av 2 where A is a constant
ASSUME: When the car is going as fast as the engine can drive it, Fd = Fa
SOLVE:
From Fdrive = Av 2 we see that if v increases by 10% we have
F = A(1.1v)2 = 1.21Av 2 = 1.21Fdrive
Thus we must increase the force generated by the engine by 21% to support a 10% change in
speed. This tells me that it is very difficult to increase a car’s top speed because the force increases
as the square of the speed.
48
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.16
GOAL: Find the constant speed needed for a pursuing cyclist to catch another cyclist.
GIVEN: Initial positions of the cyclists and the lead cyclist’s speed.
DRAW:
FORMULATE EQUATIONS: We’ll need to use the formula for position as a function of time
due to a constant speed v:
x(t) = x(0) + vt
SOLVE: Bicyclist A has to travel one mile, or 5280 ft. 18 mph corresponds to 26.4 ft/s. Thus we
have
5280 ft = (26.4 ft/s)t ⇒ t = 200 s
Bicyclist B has to travel an additional 750 ft, for a total of 6030 ft and has 200 s to do so. Thus
we have
6030 ft = vB (200 s)
vB = 30.2 ft/s = 20.6 mph
49
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.17
GOAL: Find the terminal speed of an object with a given acceleration and determine at what
time it reaches 95 percent of terminal speed.
GIVEN: a0 = 80, 000 ft/s2 , a1 = 0.01 s/ ft2
FORMULATE EQUATIONS:
The acceleration is given by
s̈ = a0 − a1 ṡ3
SOLVE:
We can numerically integrate using MATLAB with initial conditions of s(0) = 0, ṡ(0) = 0. Using
a time interval from t = 0 to t = 0.01 s yields the following plot:
It can be seen from the plot that the terminal speed is
vterm = 200 ft/s
This result can be seen analytically as well. When vterm is reached, the acceleration s̈ is zero. Using
this in our acceleration equation gives
3
0 = a0 − a1 vterm
which, when solved, returns the result vterm = 200 ft/s.
An examination of the output data allows the time at which the speed reaches 95 percent of its
terminal value (0.95(200 ft/s) = 190 ft/s) to be determined as
tterm = 0.00367 s
50
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.18
GOAL: Find the position of a particle at t = 3 seconds given a plot of the particle’s acceleration.
GIVEN: Plot of acceleration vs. time.
DRAW:
FORMULATE EQUATIONS: We’ll need to integrate the acceleration to find speed and then
integrate the speed to find position.
SOLVE:
The plot of acceleration versus time can be divided into three distinct regions on the time interval
0 ≤ t ≤ 3. For the region where 0 ≤ t ≤ 1, the acceleration is given by
s̈ = 10t
Integrating for velocity gives
ṡ(t) = 5t2 + v0
(1)
5
s(t) = t3 + v0 + s0
3
(2)
Integrating again for position provides
Using the initial condition v0 = 0 in (1) and evaluating at t = 1 gives ṡ(1) = 5. Using the initial
conditions s0 = 1 and v0 = 0 in (2) yields s(1) = 38 .
For 1 < t ≤ 2 the acceleration is constant;
s̈ = 10
Re-initialize the time counter to begin at t = 1, treating this segment as if it’s simply another initial
condition problem for which the initial conditions start at t = 1.
Integrating for velocity gives
ṡ(t) = 10t + v1
(3)
Integrating again for position gives
s(t) = 5t2 + v1 t + s1
(4)
Applying the conditions found over the first time interval into these equations, we find ṡ(2) = 15
and that s(2) = 38
3 .
51
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
For the final interval, 2 < t ≤ 3, the acceleration equals 0, so ṡ(3) = v2 and s(3) = ṡ(2)t + s2 .
Solving these equations shows that
s(3)=27 23 m
52
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.19
GOAL: Determine s(4) − s(0) given s̈(t).
GIVEN: A plot of acceleration vs. time.
DRAW:
FORMULATE EQUATIONS: We’ll have to integrate the acceleration to find speed and then
integrate the speed to find position.
ASSUME: We have three phases of acceleration - constant from 0 to 1 s, linearly decreasing from
1 to 2 s and constant (and zero) beyond 2 s.
SOLVE: For 0 ≤ t < 1 s:
ṡ(t) = ṡ(0) +
Z t
0
20 m/s2 dη = −4 m/s + (20 m/s2 )t
s(t) = s(0) − (4 m/s)t + (10 m/s2 )t2
Thus s(1 s) = s(0) + 6 m, ṡ(1 s) = 16 m/s
For 1 s ≤ t < 2 s:
ṡ(t) = ṡ(1 s) +
Z t
1s
(20 m/s2 )(2 s − η)dη = −14 m/s + (40 m/s2 )t − (10 m/s3 )t2
t
10
m/s3 t3
3
1s
10
10
m − (14 m/s)t + (20 m/s2 )t2 −
m/s3 t3
= s(0) +
3
3
s(t) = s(1 s) + −(14 m/s)η + (20 m/s2 )η 2 −
Evaluating at t = 2 s gives s(2 s) = s(0) + 28 23 m and ṡ(2) = 26 m/s.
Finally, for t = 2 s and beyond we have zero acceleration. Thus the speed is constant at ṡ(2 s) and
the displacement increases linearly.
For 2 s ≤ t:
ṡ(t) = 26 m/s
2
s(t) = s(2 s) + ṡ(2 s)(t − 2 s) = s(0) + 28 m + (26 m/s)(t − 2 s)
3
2
s(4 s) = s(0) + 80 m
3
s(4 s) − s(0) = 80 23 m
53
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.20
GOAL: Determine the maximum height the ball reaches, and compare this to the maximum height
achieved when air drag is neglected.
GIVEN: The ball is launched vertically with an initial speed of ẏ0 = 50 m/s, and the air drag
coefficient cD is given by cD = 0.001 m−1 .
DRAW:
FORMULATE EQUATIONS:
The ball’s acceleration is governed by
ÿ = −g − cD ẏ 2
(1)
The ball’s acceleration is given in terms of its velocity ẏ, and so
ÿ =
dẏ
dẏ dy
dẏ
=
·
=
ẏ
dt
dy dt
dy
dy =
ẏ
dẏ
ÿ
(2)
SOLVE:
We can simply integrate (1) to find the maximum height reached when air drag is considered:
Z ymax
(1) → (2), integrate ⇒
dy =
ẏ
0
y
ymax
0
0
−ẏ
dẏ
g + cD ẏ 2
1
ln(g + cD ẏ 2 )
2cD
0
1
g
ln
2cD
g + cD ẏ02
!
=−
ymax = −
ymax
Z 0
ẏ
0
1
9.81 m/s2
=−
ln
2(0.001 m−1 )
9.81 m/s2 + (0.001 m−1 )(50 m/s)2
!
ymax = 114 m
54
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
When air drag is neglected, the ball’s acceleration is given by ÿ = −g, and hence the maximum
height achieved is
−gdy = ẏdẏ
(2) ⇒
−g
Z ymax
dy =
ẏ
0
−gy
ymax
0
−gymax
ẏdẏ
0
1
= ẏ 2
2
0
ẏ
0
1
= − ẏ02
2
ymax =
ymax =
Z 0
ẏ02
2g
(50 m/s)2
2(9.81 m/s2 )
ymax = 127 m
The percent error e associated with neglecting air drag is
e=
|114 m − 127 m|
× 100%
114 m
e = 12.3%
55
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.21
GOAL: Find t1 (time to switch from acceleration to braking), t2 (time to stop braking), and
maximum speed of a car traveling a set distance.
GIVEN: Constant acceleration (7.5 m/s2 ) and braking (-10 m/s2 ) and total distance traveled (105
m).
DRAW:
FORMULATE EQUATIONS: We’ll use the displacement/velocity expressions for constant
acceleration a0 , going from an initial time tA to a final time tB :
v(tB ) = v(tA ) + a0 [tB − tA ]
(1)
1
(2)
x(tB ) = x(tA ) + v(tA )[tB − tA ] + a0 [tB − tA ]2
2
ASSUME: The speed will be greatest at time t1 , just before braking commences and the speed
will be zero when the car has traveled 105 m.
SOLVE: Starting from zero, we reach a maximum speed at t1 :
v(t1 ) = a1 t1
At this point its position is, from (2)
1
x(t1 ) = a1 t21
2
We can’t evaluate yet because we don’t know t1 .
Moving on, we can find the position and speed at t2 , given the initial position x(t1 ) and speed
v(t1 ). Knowing that the speed must be zero at t2 gives us
v(t2 ) = 0 = v(t1 ) + a2 (t2 − t1 )
= a1 t1 + a2 (t2 − t1 )
a1 t1 = −a2 (t2 − t1 )
(a1 − a2 )t1 = −a2 t2
t2 =
where b =
a −a
2
1
a
2
=
−10−7.5
−10
a2 − a1
t1 = bt1
a2
= 1.75.
We know that the car travels a distance L = 105 m and so we have
56
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(3)
CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
1
1 2
a1 t1 + a1 t1 [t2 − t1 ] + a2 (t2 − t1 )2
2
2
1 2
1
2
=
a1 t1 + a1 t1 (b − 1) + a2 (b − 1)2 t21
2
2
1
2 a2
=
a + (b − 1)a1 + (b − 1)
t21 = L
2 1
2
x(t2 ) =
Using the given values gives us
7.5 m/s2
+ 0.75(7.5 m/s2 ) − (0.75)2 (5 m/s2 ) t21 = 105 m
2
!
t1 = 4 s
We can now find the maximum speed:
v(t1 ) = a1 t1 = (7.5 m/s2 )(4 s) = 30 m/s
(3) ⇒
t2 = 1.75(4) = 7 s
57
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.22
GOAL: Find the time of collision between Car A to hit Car B and their relative speed at collision.
GIVEN: At t = 0 s Car A is traveling at a constant speed of 98 ft/s and Car B is 22.5 ft in front of
Car A, traveling at 78 ft/s and decelerating at 20 ft/s2 . At t = 0.5 s Car A decelerates at a constant
30 ft/s2 . The separation of the two cars is given by h.
DRAW:
FORMULATE EQUATIONS: We’ll use the general expressions for position and speed, given
a constant acceleration and initial position and speed x0 , v0 :
x(t) = x0 + v0 t +
at2
2
v(t) = v0 + at
SOLVE:
At t = 0 s let the position of Car A be xAF = 0 ft and the position of Car B be xBR = 22.5 ft.
After 0.5 s have elapsed the positions and speeds are found from
xAF (0.5 s) = (98 ft/s)(0.5 s) = 49 ft
xBR (0.5 s) = 22.5 ft + (78 ft/s)(0.5 s) − 0.5(20 ft/s2 )(0.5 s)2 = 59 ft
vAF (0.5 s) = 98 ft/s
vBR (0.5 s) = 78 ft/s − (20 ft/s2 )(0.5 s) = 68 ft/s
At this point Car A begins to decelerate at −30 ft/s2 . For convenience we’ll reset time to zero (t
indicates time beyond the 0.5 s needed for Car A’s braking to begin.)
xAF (t) = 49 ft + (98 ft/s)t + 0.5(−30 ft/s2 )t2
xBR (t) = 59 ft + (68 ft/s)t + 0.5(−20 ft/s2 )t2
The separation h is given by
h = xBR − xAF = 10 ft − (30 ft/s)t + (5 ft/s2 )t2
and the collision occurs when h = 0. We’re left with a quadratic to solve:
10 ft − (30 ft/s)t + (5 ft/s2 )t2 = 0
t2 − (6 s)t + 2 s2 = 0
58
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
The relevant solution to this equation is t = 0.354 s. Thus the elapsed time from when the driver
of Car A first decides to brake is given by 0.5 s + 0.354 s = 0.854 s .
The collision speed is found from
−ḣ = (−68 ft/s + 98 ft/s) + (−30 ft/s2 + 20 ft/s2 )(0.354 s) = 26.5 ft/s
Thus we have a collision speed of
26.5 ft/s = 18 mph
59
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.23
GOAL: Determine whether Car A hits Car B and, if so, determine the relative speed of the
collision.
GIVEN: Initially both cars are traveling at 98 ft/s. Car A is initially 30 ft behind Car B. Car B
begins to decelerate at 35 ft/s2 and 1 s later Car A begins to decelerate at 30 ft/s2 .
DRAW:
FORMULATE EQUATIONS:
We’ll use the formulas for motion with constant acceleration a (initial position and speed given by
x0 , v0 ):
v(t) = v0 + at
x(t) = x0 + v0 t +
at2
2
SOLVE:
xAF indicates the front position of Car A and xBR indicates the position of the rear of Car B. h
indicates the separation between the two cars. Initially xAF = 0 and xBR = 30 ft
We’ll break the analysis into two phases. First we’ll determine the position and speed of the vehicles
at t = 1 s. Next, we’ll “reset” our time axis, defining t = 0 as the point at which vehicle A begins
to decelerate. We’ll then find the time to contact and the relative speed at contact.
At t = 0 we have
xAF = 0,
xBR = 30 ft
vA = 98, ft/s vB = 98 ft/s
aA = 0,
aB = −35 ft/s2
At t = 1 s we have
35 ft/s2 (1 s)2
= 110.5 ft
xAF = (98 ft/s)(1 s) = 98 ft, xBR = 30 ft + (98 ft/s)(1 s) −
2
2
vA = 98, ft/s
vB = 98 ft/s − (35 ft/s )(1 s) = 63 ft/s
Note that by the time the driver of Car A notices the problem ahead of him and initiates his
braking (just 1 s delay), his distance from the car ahead of him has been reduced from 30 ft to
(110.5 − 98) ft = 12.5 ft.
We’ll now “reset” our system, starting t from zero again but with the new values of position and
speed, as well as having both cars decelerate (Car A at 30 ft/s2 and Car B continuing at 35 ft/s2 ).
(30 ft/s2 )t2
(35 ft/s2 )t2
, xBR (t) = 110.5 ft + (63 ft/s)t −
2
2
2
2
vA (t) = 98 ft/s − (30 ft/s )t
vB (t) = 63 ft/s − (35 ft/s )t
xAF (t) = 98 ft + (98 ft/s)t −
60
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(1)
(2)
CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
Solving xBR − xAF = 0 will allow us to see whether a collision (separation of the two cars goes to
zero) occurs.
(5 ft/s2 )t∗ 2
12.5 ft − (35 ft/s)t∗ −
=0
(1) ⇒
2
This has only one positive solution: t∗ = 0.348 s . Thus we do have a collision.
Using this value of time and (2) gives us a collision speed of
vcoll = vA − vB = 35 ft/s + (5 ft/s2 )(0.348 s) = 36.7 ft/s
vcoll = 36.7 ft/s = 25 mph
Note that these figures are not unreasonable. People often travel far too close behind the car ahead
of them and this shows that even a very brief lack of attention can easily lead to a collision.
61
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.24
GOAL: Determine the performance difference between manual braking and using Brake Assist.
GIVEN: In each case the car decelerates from 60 mph. Without Brake Assist the car decelerates
at 20 ft/s2 for 1 s and then at 30 ft/s2 for the remaining time. With Brake Assist active the car
decelerates at 30 ft/s2 the entire time. Let x represent the position of the car C.
DRAW:
FORMULATE EQUATIONS:
We’ll use the formulas for motion with constant acceleration a (initial position and speed given by
x0 , v0 ):
v(t) = v0 + at
x(t) = x0 + v0 t +
at2
2
SOLVE:
Case 1: Without Brake Assist. For this case we’ll examine the car’s response in two phases - from
t = 0 s to t = 1 s and then for t > 1 s. We’ll “reset” our time axis for the second phase, remembering
that 1 s has already elapsed.
At t = 0 we have
x = 0 ft
v = 60 mph = 88 ft/s
a = −20 ft/s2
At t = 1 s we have
x = (88 ft/s)(1 s) − 0.5(20 ft/s2 )(1 s)2 = 78 ft
v = 88 ft/s − (20 ft/s2 )(1 s) = 68 ft/s
We’ll now “reset” our system, starting t from zero again but with the new values of position and
speed as initial conditions.
(30 ft/s2 )t2
2
v(t) = 68 ft/s − (30 ft/s2 )t
x(t) = 78 ft + (68 ft/s)t −
Solving v(t∗ ) = 0 will tell us when the car’s speed goes to zero:
(30 ft/s2 )t∗ = 68 ft/s ⇒ t∗ = 2.27 s
(2) ⇒
(1) ⇒
x(2.27 s) = 78 ft + (68 ft/s)(2.27 s) −
(30 ft/s2 )(2.27 s)2
= 155 ft
2
Case 2: With Brake Assist we have
62
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(1)
(2)
CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
(30 ft/s2 )t2
x(t) = (88 ft/s)t −
2
v(t) = 88 ft/s − (30 ft/s2 )t
(3)
(4)
Solving for v(t∗ ) = 0 gives us
(4) ⇒
(3) ⇒
(30 ft/s2 )t∗ = 88 ft/s ⇒ t∗ = 2.93 s
x(2.93 s) = (88 ft/s)(2.93 s) −
(30 ft/s2 )(2.93 s)2
= 129 ft
2
What we see from comparing these two results is that without Brake Assist the car traveled 26 ft
farther, an increase of 20% .
63
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.25
GOAL: Determine the impact speed.
GIVEN: The battlestation falls from rest at a height H = 400 mi. The planet has radius R =
4000 mi and a gravitational acceleration of g0 = 30 ft/s2 .
DRAW:
FORMULATE EQUATIONS:
The battlestation falls according to
s̈ = −
g0 R 2
s2
(1)
The acceleration is given in terms of the position s, so
s̈ =
dṡ
dṡ ds
dṡ
=
·
= ṡ
dt
ds dt
ds
ṡdṡ = s̈ds
(2)
SOLVE:
Z ṡf
(1) → (2), integrate ⇒
0
1 2
ṡ
2
ṡf
0
ṡdṡ = −
ṡf =
s
mi
2(30 ft/s )
5280 ft
2
H+R
= −g0 R2
1
(ṡ )2 =
2 f
ṡf
Z R
=
g0 R2
ds
s2
Z R
H+R
ds
s2
R2 R
g0
s
s
3600 s
hr
H+R
2g0
2
R2
1
1
−
R H +R
(4000 mi)2
1
1
−
4000 mi 400 mi + 4000 mi
ṡf = 7318 mph
64
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.26
GOAL: Explain the significance of a negative time solution.
GIVEN: Position of a particle as a function of time.
FORMULATE EQUATIONS:
x(t) = 20 m + (180 m/s)t − (30 m/s3 )t3
SOLVE: First we need to solve for the times at which the speed is zero:
v(t) =
d
x(t) = (180 m/s) − (90 m/s3 )t2 = 0
dt
√
t2 = 2 s 2 ⇒ t = ± 2 s
The plot shows how speed varies with time. If the process could be started at some time before
the first zero crossing then the particle would encounter two instants at which its speed was zero.
If, however, a real system is being modeled for which the given t = 0 really corresponds to the
physical time at which the process begins, then the first solution is unattainable, occurring as it
does before the system is set into motion.
65
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.27
GOAL: Find a rock’s impact velocity when dropped from 100 feet.
GIVEN: Downward acceleration is g − cd ṡ2
DRAW:
ASSUME: The rock experiences accelerations due only to that of gravity and drag.
FORMULATE EQUATIONS:
Measuring from the clifftop we have
s(0) = 0 and ṡ(0) = 0
Because of the air resistance our ball’s acceleration is
s̈(t) = 32.2 ft/s2 − (0.01 ft−1 )ṡ2
Using s̈ds = ṡdṡ or
ds =
SOLVE:
Integrating gives
ṡdṡ
32.2 m/s − (0.01 ft−1 )ṡ2
2
ṡ
Z100 m
ds =
impact
Z
0
0
ṡdṡ
32.2 ft/s − (0.01 ft−1 )ṡ2
2
ṡ
1m
100 m = −
ln(32.2 ft/s2 − (0.01 m−1 )ṡ2 )
0.02
impact
0

−2 = ln 
2
−1
32.2 ft/s − (0.01 m
32.2 ft/s2
ṡimpact = 52.8 ft/s
)ṡ2impact


66
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
2.1.28
GOAL: Find the position of the probe at a prescribed value of the speed.
GIVEN: Acceleration as a function of the speed for a rectilinear motion.
DRAW:
FORMULATE EQUATIONS: For simplicity we’ll introduce z as the coordinate which is zero
at the release point and points down from there.
To get an equation for the probe’s position z after release, use the general relationship
The particle acceleration is given by:
ż dż = z̈ dz
(1)
z̈ = g − c ż 2
(2)
where g = 7.5 m/s2 and c = 1.2×10−4 m−1 . Thus there is a gravitational acceleration pulling the
probe down and a deceleration that depends upon the probe’s speed ż.
SOLVE: The terminal speed, żterm is found when the probe’s acceleration is equal to zero:
żterm = 250 m/s
(2) ⇒
Thus we see that, given enough time, the probe will move down at a speed of 245 m/s.
We want to find the distance travelled when ż = 0.98żterm = 0.98(250 m/s) = 245 m/s.
To find z, we integrate (1):
ż dż = (g − c ż 2 )dz
ż dż
= dz
g − c ż 2
−2cż dż
= −2cdz
g − c ż 2
d(g − c ż 2 )
= −2cdz
g − c ż 2
ż
Z2
ż =0
1
ln
g − cż22
g
!
d(g − c ż 2 )
= −2c
g − c ż 2
z
Z2
dz
z =0
1
7.5 m/s2 − (1.2×10−4 m−1 )(245 m/s)2
= −2cz2 ⇒ ln
7.5 m/s2
!
= −2(1.2×10−4 m−1 )z2
z2 = 1.35×104 m = 13.5 km
67
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2.1. STRAIGHT-LINE MOTION
CHAPTER 2. KINEMATICS OF PARTICLES
2.1.29
GOAL: Investigate how fast a particle reaches s = 8 when its velocity/displacement follows the
graph given in the homework statement.
GIVEN: A plot of speed vs. position.
DRAW:
FORMULATE EQUATIONS: For this problem we don’t have speed as a function of time but
rather as a function of position. Thus our governing equation is a bit different than usual.
From the graph, we see that
5
ds
= − s + 10
dt
4
Separating variables yields
ds
= dt
5
− 4 s + 10
SOLVE: Integrating gives us
Z s
0
ds
=
5
− 4 s + 10
which provides the equation
dt
0
s
4
5
−
ln 10 − s
5
4
Z t
t
=t
0
0
Analyzing over the bounds gives
−
Thus
5
4
ln 10 − s
5
4
+
4
(ln 10) = t
5
10 − 45 s
4
t = − ln
5
10
!
(1)
• To find the values of velocity that the particle will obtain, plug s = 8 into equation (1). The
resulting time is t = ∞. Thus the particle will have velocities 0 < ṡ ≤ 10 .
68
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CHAPTER 2. KINEMATICS OF PARTICLES
2.1. STRAIGHT-LINE MOTION
• Using 7.99 in (1) gives t = 5.348
Using 7.999 into (1) gives t = 7.190
Using 7.9999 into (1) gives t = 9.032
Thus one can calculate (by subtraction) that ∆t = 1.842 s .
• To see where this constant jump is coming from, introduce a coordinate ǫ that runs to the
left from s = 8:
s=8−ǫ
Substituting this into our expression for t gives
10 − 45 (8 − ǫ)
4
t = − ln
5
10
=
=
=
=
!
5
ǫ
4
− ln 4
5
10
4
5ǫ
− ln
5
40
5
4
ln
+ ln (ǫ)
−
5
40
4
5
4
− ln(ǫ)
− ln
5
40
5
!
In our problem ǫ was 0.01, 0.001, etc. So let’s let ǫ be equal to powers of 0.1:
ǫ = 0.1n
n = 1, 2, 3, · · ·
Then
5
4
−
t = − ln
5
40
5
4
−
= − ln
5
40
4
ln(0.1n )
5
4n
ln(0.1)
5
The only thing changing is the second term as we move closer and closer to s = 8. And the
change is equal to - 45 ln(.1), which, when evaluated gives us a time change of 1.842 , just as
we saw numerically.
69
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2.2. CARTESIAN COORDINATES
2.2
CHAPTER 2. KINEMATICS OF PARTICLES
Cartesian Coordinates
70
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.1
⇀
⇀
GOAL: Determine the coordinate transformation between ⇀
c 1, ⇀
c 2 and b 1 , b 2 .
GIVEN: Structural configuration of the connected links.
DRAW:
SOLVE: In the illustration φ was made “small” so that the relationship between the unit vectors
⇀
⇀
is more immediately apparent. Clearly b 1 and ⇀
c 1 point in essentially opposite directions, as do b 2
and ⇀
c 2 . Hence the entry in the coordinate transformation array will be − cos φ. More specifically,
⇀
⇀
we see that to get to the tip of ⇀
c 1 we need to go opposite from b 1 and then a bit in the b 2 direction.
Thus we have
⇀
⇀
⇀
c 1 = − cos φ b 1 + sin φ b 2
⇀
⇀
Similarly, to get to the tip of ⇀
c 2 means going opposite to b 2 and then a little bit in the − b 1 direction:
⇀
⇀
⇀
c 2 = − cos φ b 2 − sin φ b 1
Putting these into a transformation array gives us
⇀
⇀
b1
b2
⇀
⇀
c1
c2
− cos φ − sin φ
sin φ − cos φ
71
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2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.2.2
⇀
GOAL: Determine k r⇀(t)k, kv⇀(t)k, and ka(t)k
at time t = 2 s.
GIVEN: The position of particle P in the horizontal plane is given by r⇀(t) = 5t2 ⇀
ı + 2te0.1t ⇀
 m.
DRAW:
FORMULATE EQUATIONS:
The velocity and acceleration vectors for particle P are defined as, respectively,
v⇀(t) =
d r⇀(t)
dt
(1)
⇀
a(t)
=
dv⇀(t)
dt
(2)
SOLVE:
The distance from the origin to the particle at t = 2 s is given by
⇀
k r (t)k =
q
⇀
⇀
⇀
r (t) · r (t) =
k r (2)k =
q
q
(5t2 )2 + (2te0.1t )2 m
[5(2)2 ]2 + 2(2)e0.1(2)
2
m
k r⇀(2)k = 20.6 m
The particle’s speed at the given instant is
 m/s
 + 0.2te0.1t ⇀
v⇀(t) = 10t ⇀
ı + 2e0.1t ⇀
(1) ⇒
v⇀(t) = 10t ⇀
ı + 2e0.1t + 0.2te0.1t ⇀
 m/s
kv⇀(t)k =
q
v⇀(t) · v⇀(t) =
kv⇀(2)k =
q
q
(10t)2 + (2e0.1t + 0.2te0.1t )2 m/s
[10(2)]2 + 2e0.1(2) + 0.2(2)e0.1(2)
72
2
m/s
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
kv⇀(2)k = 20.2 m/s
The acceleration that the particle experiences at t = 2 s is
⇀
 m/s2
 + 0.02te0.1t ⇀
 + 0.2e0.1t ⇀
a(t)
= 10 ⇀
ı + 0.2e0.1t ⇀
(2) ⇒
⇀
 m/s2
a(t)
= 10 ⇀
ı + 0.4e0.1t + 0.02te0.1t ⇀
⇀
ka(t)k =
q
⇀
⇀
a(t) · a(t) =
⇀
ka(2)k
=
q
q
100 + (0.4e0.1t + 0.02te0.1t )2 m/s2
100 + 0.4e0.1(2) + 0.02(2)e0.1(2)
2
m/s2
⇀
ka(2)k
= 10.0 m/s2
73
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2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.2.3
⇀
⇀
GOAL: Determine the coordinate transformation between b 1 , b 2 and ⇀
c 1, ⇀
c 2 and between ⇀
ı, ⇀

⇀
⇀
and c 1 , c 2 .
GIVEN: Angular configuration of the dial and indicator needle.
DRAW: All unit vector pairings are shown. The inclination of the ⇀
c i are made up of the rotation
of the dial plus the indicator needle’s rotation with respect to the dial.
⇀ ⇀
SOLVE: Our middle pairing allows us to form the transformation array between b 1 , b 2 and ⇀
c 1, ⇀
c 2:
Putting these into a transformation array gives us
⇀
⇀
c1
c2
⇀
⇀
b1
b2
− sin φ cos φ
− cos φ − sin φ
Using the rightmost pairing gives us our final transformation array:
⇀
⇀
c1
c2
⇀
⇀
ı

cos(θ + φ) sin(θ + φ)
− sin(θ + φ) cos(θ + φ)
74
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.4
GOAL: Determine the horizontal location before the island at which to drop the supplies.
GIVEN: System geometry and initial speed: H = 150 ft, v = 200 mph.
DRAW:
FORMULATE EQUATIONS:
The x position of the airlifted supplies is given by
x = vt
(1)
1
y = H − gt2
2
(2)
The y position of the supplies is given by
SOLVE:
The total horizontal distance traveled is s, so (1) becomes
(1) ⇒
s = vt
(3)
To find the time when the supplies land, set y = 0 in (2), so
(2) ⇒
1
0 = H − gt2
2
(4)
(4) ⇒
s
(5)
t=
2H
g
Eliminate time in (3) to find the distance:
(5) → (3) ⇒
s=v
hr
s = (200 mph)
3600 s
s
2H
g
5280 ft
mi
s
2(150 ft)
32.2 ft/s2
s = 895.4 ft
75
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2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.2.5
⇀
GOAL: Determine k r⇀(t)k, kv⇀(t)k, and ka(t)k
at time t = 1 s.
GIVEN: The position of particle P in the horizontal plane is described by r⇀(t) = 0.3t3 ⇀
ı +
0.4t2 sin 2t ⇀
 ft.
DRAW:
FORMULATE EQUATIONS:
The velocity and acceleration vectors for particle P are defined as, respectively,
v⇀(t) =
d r⇀(t)
dt
(1)
⇀
a(t)
=
dv⇀(t)
dt
(2)
SOLVE:
The particle’s distance from the origin at t = 1 s is
k r⇀(t)k =
q
r⇀(t) · r⇀(t) =
⇀
k r (1)k =
q
q
(0.3t3 )2 + (0.4t2 sin 2t)2 ft
[0.3(1)3 ]2 + [0.4(1)2 sin(2 · 1)]2 ft
k r⇀(1)k = 0.471 ft
The speed of the particle at the given instant is
ı + 0.8t sin 2t ⇀
 + 0.8t2 cos 2t ⇀
 ft/s
v⇀(t) = 0.9t2 ⇀
(1) ⇒
v⇀(t) = 0.9t2 ⇀
ı + 0.8t sin 2t + 0.8t2 cos 2t ⇀
 ft/s
⇀
kv (t)k =
q
kv⇀(1)k =
⇀
⇀
v (t) · v (t) =
q
q
(0.9t2 )2 + (0.8t sin 2t + 0.8t2 cos 2t)2 ft/s
[0.9(1)2 )]2 + [0.8(1) sin(2 · 1) + 0.8(1)2 cos(2 · 1)]2 ft/s
76
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
kv⇀(1)k = 0.983 ft/s
The particle’s acceleration at t = 1 s is given by
(2) ⇒
⇀
a(t)
= 1.8t ⇀
ı + 0.8 sin 2t ⇀
 + 1.6t cos 2t ⇀
 + 1.6t cos 2t ⇀
 − 1.6t2 sin 2t ⇀
 ft/s2
⇀
a(t)
= 1.8t ⇀
ı + 0.8 sin 2t + 3.2t cos 2t − 1.6t2 sin 2t ⇀
 ft/s2
⇀
ka(t)k =
⇀
ka(1)k
=
q
q
⇀
⇀
a(t) · a(t) =
q
(1.8t)2 + (0.8 sin 2t + 3.2t cos 2t − 1.6t2 sin 2t)2 ft/s2
[1.8(1)]2 + [0.8 sin(2 · 1) + 3.2(1) cos(2 · 1) − 1.6(1)2 sin(2 · 1)]2 ft/s2
⇀
ka(1)k
= 2.73 ft/s2
77
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2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.2.6
⇀
⇀
GOAL: Re-express a vector in terms of the b 1 , b 2 frame.
GIVEN: Structural configuration of the connected links.
DRAW:
The transformation array is shown below.
⇀
⇀
b1
b2
⇀
⇀
ı

cos θ sin θ
− sin θ cos θ
SOLVE:
p⇀ = 4 ⇀
ı − 8⇀

⇀
⇀
⇀
⇀
= 4(cos θ b 1 − sin θ b 2 ) − 8(sin θ b 1 + cos θ b 2 )
⇀
⇀
= (4 cos θ − 8 sin θ) b 1 + (−4 sin θ − 8 cos θ) b 2
Evaluating at θ = 130 degrees yields
⇀
⇀
p⇀ = −8.70 b 1 + 2.08 b 2
78
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.7
GOAL: Re-express a given vector in a new set of unit vectors.
GIVEN: Orientation of two sets of unit vectors on a movable link.
DRAW:
⇀
⇀
b1
b2
⇀
⇀
ı

− cos θ − sin θ
sin θ − cos θ
FORMULATE EQUATIONS:
⇀
⇀
SOLVE: Express p⇀ in terms of b 1 and b 2 :
p⇀ = 3 ⇀
ı +4⇀

(1)
⇀
⇀
⇀
⇀
= 3 (− cos θ b 1 + sin θ b 2 ) + 4 (− sin θ b 1 − cos θ b 2 )
⇀
⇀
= b 1 (−3 cos θ − 4 sin θ) + b 2 (3 sin θ − 4 cos θ)
(2)
(3)
For θ = 53◦ , we get
⇀
⇀
p⇀ = −5.00 b 1 − 0.01 b 2
79
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(4)
2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.2.8
GOAL: Find whether a magnet thrown off the end of a car will attach to the car, or fall to the
ground.
√ ⇀
GIVEN: Initial velocity of the magnet is v⇀ = 5 i . The magnitude/car interaction as an acceler⇀
ation is equal to −am i where am = 10 m/s2 . Also, the sheet metal extends a length, L, equal to
1 m.
DRAW:
ASSUME: Magnet follows a parabolic trajectory once it is thrown from the car.
SOLVE:
t2
x = vt − am
2
y = −g
t2
2
Solve for t for x(t) = 0
a t∗
am t2
⇒ t∗ (v − m ) = 0
2
2
∗
t = 0 (initial condition)
0 = vt∗ −
t∗ =
y(t∗ ) = −
g
2
2v
am
2
2v
(return time)
am
=−
2(9.81 m/s2 )(5 m2 / s2 )
2gv 2
=
−
a2m
(10 m/s2 )2
y(t∗ ) = −0.98 m
So, just barely, the magnet hits the car (since missing would imply |y(t∗ )| > 1 m)
80
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.9
GOAL: Determine the minimum launch angle and corresponding time of flight.
GIVEN: System geometry and initial speed: d = 200 ft, h = 3 ft, v0 = 100 ft/s.
DRAW:
FORMULATE EQUATIONS:
The x position of the pumpkin is given by
x = (v0 cos θ)t
(1)
The y position of the pumpkin is given by
1
y = h + (v0 sin θ)t − gt2
2
(2)
SOLVE:
The total horizontal distance traveled is d, so (1) becomes
d = (v0 cos θ)t
d
t=
(v0 cos θ)
(1) ⇒
(3) ⇒
(3)
(4)
To find the launch angle, first set y = 0 in (2), so
1
0 = h + (v0 sin θ)t − gt2
2
(2) ⇒
(5)
Then eliminate the time t:
(4) → (5) ⇒
sin θ
d
1
0=h+
d− g
cos θ
2
v0 cos θ
0 = h cos2 θ + d sin θ cos θ −
2
gd2
2v02
1
gd2
0 = h cos2 θ + d sin(2θ) − 2
2
2v0
0 = 2h cos2 θ + d sin(2θ) −
gd2
v02
gd2
v02
= 2h cos2 θ + d sin(2θ)
(32.2 ft/s2 )(200 ft)2
= 2(3 ft) cos2 θ + (200 ft) sin(2θ)
(100 ft/s)2
128.8 ft = 6 cos2 θ ft + 200 sin(2θ) ft
81
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(6)
2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
❢
Using MATLAB R to solve for the minimum angle that satisfies (6),
θmin = 0.333 rad = 19.06◦
(7) → (4) ⇒
t=
200 ft
(100 ft/s)(cos(19.06◦ ))
t = 2.116 s
82
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(7)
CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.10
GOAL: Determine the time of flight, the horizontal distance traveled, and the maximum height
achieved.
GIVEN: System geometry and initial speed: h = 20 ft, θ = 45◦ , v0 = 25 ft/s.
DRAW:
FORMULATE EQUATIONS:
The x position of the craft is given by
x = (v0 cos θ)t
(1)
1
y = h + (v0 sin θ)t − gt2
2
(2)
The y position of the craft is given by
SOLVE:
Find the time of flight by setting y = 0 in (2) and solving for t:
1
0 = h + (v0 sin θ)t − gt2
(2) ⇒
2
p
−v0 sin θ ± (v0 sin θ)2 + 2gh
(3) ⇒
t=
−g
t =
−(25 ft/s) sin(45◦ ) ±
t = −0.693 s, 1.791 s
q
(3)
((25 ft/s) sin(45◦ ))2 + 2(32.2 ft/s2 )(20 ft)
−32.2 ft/s2
t = 1.791 s
(4)
Use the time of flight to solve for the distance traveled:
(4) → (1) ⇒
x = (25 ft/s) cos(45◦ )(1.791 s)
x = 31.67 ft
Determine the maximum height achieved by first differentiating (2) and setting it equal to 0:
dy
= v0 sin θ − gt = 0
(5)
Differentiate (2) = 0 ⇒
dt
83
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2.2. CARTESIAN COORDINATES
v0 sin θ
g
2
(v0 sin θ)
(v0 sin θ)2
=h+
−
g
2g
tpeak =
(5) ⇒
(6) → (2) ⇒
CHAPTER 2. KINEMATICS OF PARTICLES
ymax
(v0 sin θ)2
2g
((25 ft/s) sin(45◦ ))2
= 20 ft +
2(32.2 ft/s2 )
ymax = h +
ymax
ymax = 24.85 ft
84
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(6)
CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.11
GOAL: Determine the separation distance and the constant running speed needed to catch the
ball.
GIVEN: System geometry and initial speed: H = 30 ft, h = 10 ft, d = 20 ft, θ = 30◦ , v = 20 ft/s.
DRAW:
FORMULATE EQUATIONS:
The x position of the ball is given by
x = (v cos θ)t
(1)
The y position of the ball is given by
1
y = H + (v sin θ)t − gt2
2
The x position of the catcher is given by
x∗ = x∗0 + v ∗ t
(2)
(3)
SOLVE:
The separation distance is s, so (1) becomes
s = (v cos θ)t
(1) ⇒
We need the time of flight to solve for s, so set y = h in (2) and solve for t:
1
h = H + (v sin θ)t − gt2
(2) ⇒
2
1
0 = − gt2 + (v sin θ)t + (H − h)
2
p
−v sin θ ± (v sin θ)2 + 2g(H − h)
(5) ⇒
t=
−g
t =
−(20 ft/s) sin(30◦ ) ±
t = −0.847 s, 1.468 s
q
(4)
(5)
((20 ft/s) sin(30◦ ))2 + 2(32.2 ft/s2 )(30 ft − 10 ft)
−32.2 ft/s2
t = 1.468 s
85
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(6)
2.2. CARTESIAN COORDINATES
(6) → (4) ⇒
CHAPTER 2. KINEMATICS OF PARTICLES
s = (20 ft/s) cos(30◦ )(1.468 s)
s = 25.42 ft
To find the running speed,
(3) ⇒
x∗ − x∗0 = v ∗ t
−d = v ∗ t
d
v∗ = −
t
20 ft
v∗ = −
1.468 s
v ∗ = −13.63 ft/s (running to the left)
86
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.12
GOAL: Determine the ball’s launch speed and how long it is airborne.
GIVEN: The launch ramp is angled at θ = 45◦ with respect to the ground, and its end is located
h = 2 ft above the base of the backboard, which is inclined at β = 20◦ to the horizontal. The hole
is L = 3 ft along the backboard from its base.
DRAW:
FORMULATE EQUATIONS:
The x position of the ball is given by
x = x0 + ẋ0 t
(1)
1
y = y0 + ẏ0 t − gt2
2
(2)
The y position of the ball is given by
SOLVE:
We’ll need to eliminate the time t to first find the launch speed v0 , and so from (1) we get that
(1) ⇒
L cos β = v0 cos θ t
t=
L cos β
v0 cos θ
(3)
By eliminating the time t in (2), we find that the ball’s launch speed v0 must be
(3) → (2) ⇒
L sin β = h + v0 sin θ
L cos β
v0 cos θ
!
1
− g
2
(L sin β − h) cos2 θ = L cos β (sin θ cos θ) −
L cos β
v0 cos θ
!2
gL2 cos2 β
2v02
gL2 cos2 β
1
= (h − L sin β) cos2 θ + L cos β sin 2θ
2
2v0
2
87
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2.2. CARTESIAN COORDINATES
v0 =
v0 =
s
s
CHAPTER 2. KINEMATICS OF PARTICLES
gL2 cos2 β
2 (h − L sin β) cos2 θ + L cos β sin 2θ
(32.2 ft/s2 )(3 ft)2 cos2 (20◦ )
2 [2 ft − (3 ft) sin(20◦ )] cos2 (45◦ ) + (3 ft) cos(20◦ ) sin(2 · 45◦ )
v0 = 8.21 ft/s
Therefore, the ball’s time of flight is
(3) ⇒
t=
(3 ft) cos(20◦ )
(8.21 ft/s) cos(45◦ )
t = 0.485 s
88
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.13
GOAL: Determine when the football player should jump to catch the ball and how far away he is.
GIVEN: System geometry, the football’s initial speed, and the catcher’s jump speed: H = 8 ft,
h = 1 ft, θ = 50◦ , v0 = 40 ft/s, vj = 9 ft/s.
DRAW:
FORMULATE EQUATIONS:
The x position of the football is given by
x = (v0 cos θ)t
(1)
The y position of the football is given by
1
y = (v0 sin θ)t − gt2
2
The y position of the catcher is given by
1
yj = y0 + vj tj − gt2j
2
(2)
(3)
SOLVE:
First find the total time taken for the football to reach the catching height H by setting y = H in
(2) and solving for t:
1
H = (v0 sin θ)t − gt2
2
1
0 = − gt2 + (v0 sin θ)t − H
2
(2) ⇒
(4) ⇒
t=
t =
−v0 sin θ ±
−(40 ft/s) sin(50◦ ) ±
t = 0.312 s, 1.591 s
q
p
(v0 sin θ)2 − 2gH
−g
((40 ft/s) sin(50◦ ))2 − 2(32.2 ft/s2 )(8 ft)
−32.2 ft/s2
t = 1.591 s when the ball reaches H on the way down
Next find the time it takes the player to jump and catch the football:
89
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(4)
2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
1
yj − y0 = vj tj − gt2j
2
(3) ⇒
1
h = vj tj − gt2j
2
1 2
0 = − gtj + vj tj − h
2
(5) ⇒
−vj ±
tj =
tj
tj
=
q
vj2 − 2gh
−g
−9 ft/s ±
q
(9 ft/s)2 − 2(32.2 ft/s2 )(1 ft)
−32.2 ft/s2
= 0.153 s, 0.406 s
tj = 0.153 s when the catcher reaches H on the way up
The catcher needs to jump at
t∗ = t − tj = 1.591 s − 0.153 s
t∗ = 1.438 s
The distance the ball travels before being caught is d, so
(1) ⇒
d = (v0 cos θ)t
d = (40 ft/s)(cos(50◦ ))(1.591 s)
d = 40.9 ft
90
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(5)
CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.14
GOAL: Determine the ball’s launch speed v0 and time of flight.
GIVEN: The ball is launched toward point A, which is d = 5 m up an incline angled at θ = 15◦
with respect to the horizontal. The ball is projected at an angle β = 50◦ to ground at a distance
s = 4 m from the base of the incline.
DRAW:
FORMULATE EQUATIONS:
The x position of the ball is given by
x = x0 + ẋ0 t
(1)
1
y = y0 + ẏ0 t − gt2
2
(2)
The y position of the ball is given by
SOLVE:
Solving for the launch speed v0 first requires us to eliminate the time t, which we can express as
s + d cos θ = v0 cos β t
(1) ⇒
t=
s + d cos θ
v0 cos β
(3)
Eliminating the time t in (2) gives us the launch speed v0 needed so that the ball lands at A:
(3) → (2) ⇒
d sin θ = v0 sin β
s + d cos θ
v0 cos β
!
1
− g
2
d sin θ cos2 β = (s + d cos θ) (sin β cos β) −
s + d cos θ
v0 cos β
!2
g (s + d cos θ)2
2v02
91
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2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
g (s + d cos θ)2
1
= (s + d cos θ) sin 2β − d sin θ cos2 β
2
2v0
2
v0 =
v0 =
s
s
g (s + d cos θ)2
(s + d cos θ) sin 2β − 2d sin θ cos2 β
(9.81 m/s2 ) [4 m + (5 m) cos(15◦ )]2
[4 m + (5 m) cos(15◦ )] sin(2 · 50◦ ) − 2(5 m) sin(15◦ ) cos2 (50◦ )
v0 = 10.0 m/s
Thus, the ball is airborne for
(3) ⇒
t=
4 m + (5 m) cos(15◦ )
(10.0 m/s) cos(50◦ )
t = 1.37 s
92
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.15
GOAL: Show that the transformation array going from a set A to B to C can be expressed in
terms of the first and last vector set along with the total angle change.
GIVEN: Physical configuration of the connected links.
DRAW:
The two transformation arrays are shown below
⇀
⇀
b1
b2
⇀
⇀
ı

cos θ sin θ
− sin θ cos θ
⇀
⇀
c1
c2
⇀
FORMULATE EQUATIONS:
⇀
ı

cos φ sin φ
− sin φ cos φ
⇀
b 1 = cos θ ⇀
ı + sin θ ⇀

⇀
⇀
(1)
⇀
b 2 = − sin θ ı + cos θ 
⇀
ı = cos φ ⇀
c 1 − sin φ ⇀
c2
⇀
⇀
⇀
 = sin φ c 1 + cos φ c 2
(2)
(3)
(4)
SOLVE:
We can use (3) and (4) to reexpress (1) and (2) in terms of ⇀
c 1 and ⇀
c 2:
⇀
b 1 = cos θ(cos φ ⇀
c 1 − sin φ ⇀
c 2 ) + sin θ(sin φ ⇀
c 1 + cos φ ⇀
c 2)
⇀
b 2 = − sin θ(cos φ ⇀
c 1 − sin φ ⇀
c 2 ) + cos θ(sin φ ⇀
c 1 + cos φ ⇀
c 2)
Next we can regroup terms and reexpress them in terms of a more compact trigonometric form:
⇀
b 1 = (cos θ cos φ + sin θ sin φ) ⇀
c 1 + (sin θ cos φ − cos θ sin φ) ⇀
c2
⇀
b 2 = −(sin θ cos φ − cos θ sin φ) ⇀
c 1 + (cos θ cos φ + sin θ sin φ) ⇀
c2
⇀
b 1 = cos(θ − φ) ⇀
c 1 + sin(θ − φ) ⇀
c2
⇀
b 2 = − sin(θ − φ) ⇀
c 1 + cos(θ − φ) ⇀
c2
93
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2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
The final array transformation looks just like the original ones except it’s in terms of the difference
of two angles rather than in terms of a single angle. If I make superimpose the two unit vector sets
⇀ ⇀
b 1 , b 2 and ⇀
c 1, ⇀
c 2 , as done below, you can see that the same transformation is obtained.
⇀
c1
c2
cos(θ − φ) sin(θ − φ)
− sin(θ − φ) cos(θ − φ)
⇀
⇀
b1
b2
⇀
94
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.16
GOAL: Determine launch angle
GIVEN: Dimensions of range and barrier.
DRAW:
ASSUME:
r⇀(t1 ) = (67.5 ⇀
ı + 17.58 ⇀
 )m
(1)
⇀
(2)
⇀
r (t2 ) = 79.5 ı m
⇀
⇀
⇀
v (0) = v(cos θ ı + sin θ  )
(3)
1
 + v⇀(0)t + r⇀(0)
r⇀(t) = − gt2 ⇀
2
(4)
FORMULATE EQUATIONS:
SOLVE:
(1)→(4)⇒
⇀
ı:
⇀
:
(2)→(4)⇒
⇀
ı:
g
 + v(cos θ ⇀
ı + sin θ ⇀
 )t1
(67.5 ı + 17.58  ) m = − t21 ⇀
2
⇀
⇀
v0 cos θt1 = 67.5 m
67.5 m
t1 =
v0 cos θ
g
v0 sin θt1 − t21 = 17.58 m
2
g 67.5 m 2
(67.5 m) tan θ −
= 17.58 m
2 v0 cos θ
(5)
(6)
g
 + v(cos θ ⇀
ı + sin θ ⇀
 )t2
79.5 ı m = − t22 ⇀
2
⇀
v0 cos θt2 = 79.5 m
79.5 m
t2 =
v0 cos θ
95
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(7)
2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
g
v0 sin θt2 − t22 = 0
2
g 79.5 m 2
= 0
(79.5 m) tan θ −
2 v0 cos θ
⇀
:
(6)×79.52 -(8)×67.52 ⇒
tan θ 67.5(79.5)2 − 79.5(67.5)2
= 17.58(79.5)2
tan θ =
θ =
17.58(79.5)2
67.5(79.5)2 − 79.5(67.5)2
59.9 degrees
96
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(8)
CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.17
GOAL: Find the position of impact for a pea projected at an angle onto a slope.
GIVEN: System configuration.
DRAW:
FORMULATE EQUATIONS:
The x position of the dried pea is given by the equation
xp = v0 t cos η
The y position of the pea is given by the equation
1
yp = v0 t sin η − gt2
2
The x and y positions of the intercept on the slope are
√
3
xs =
h
2
and
1
ys = h
2
SOLVE:
Setting xp = xs and yp = ys yields a set of simultaneous equations that can be solved for h. The
equations are
√
3
h
(1)
v0 t cos η =
2
and
1
1
v0 t sin η − gt2 = h
(2)
2
2
Solving (1) for t gives
t=
(3) ⇒ (2) gives
√
3h
2v0 cos η
√
3v0 h sin η
3gh2
1
− 2
= h
2
2v0 cos η
8v0 cos η
2
97
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(3)
2.2. CARTESIAN COORDINATES
Solving for h gives
h=
CHAPTER 2. KINEMATICS OF PARTICLES
√
3 sin(η) − cos(η)
2 cos η
!
8v02 cos2 η
3g
Simplification yields
4v 2 √
h = 3g0
3 sin η cos η − cos2 η
98
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.18
GOAL: Find where a cricket lands on an inclined slope
GIVEN: Angle of slope.
DRAW:
FORMULATE EQUATIONS:
x = v0 t cos η
1
y = v0 t sin η − gt2
2
√
3
h
at A : x =
h, y =
2
2
SOLVE:
√
3
h = v0 t∗ cos η
2
h
1
= v0 t∗ sin η − g(t∗ )2
2
2
t=
(1) ⇒
h
=
2
(3) → (2) ⇒
√
Let f (η) =
√
3 sin η cos η − cos2 η −
3 hg
4 v2
0
√
(1)
(2)
√
3h
2v0 cos η
(3)
3h2
3h sin η 1
− g 2
2 cos η
2 4v0 cos2 η
3 sin η cos η − (cos η)2 −
3 hg
=0
4 v02
and find the value(s) of η for which f (η) = 0
Using the Newton-Raphson procedure to solve for the roots gives:
√
df (η) √
= 3 cos(2η) + 2 cos η sin η = 3 cos(2η) + sin(2η)
dη
ηi+1 = ηi −
f
df
dη
√
= ηi −
3
2
sin(2η) − cos2 η − 0.3660
√
3 cos(2η) + sin(2η)
(4)
Initial guess: η0 = 0.7. Using (4) yields η1 = 0.7762. Using (4) again yields η2 = 0.7852. One last
iteration gives η3 = 0.7854.
η = 0.7854 rad = 45◦
99
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2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.2.19
GOAL: Determine initial launch speed for a launched watermelon to impact a slope at a specified
position.
GIVEN: Angle of slope and angle of launch.
DRAW:
FORMULATE EQUATIONS:
1
r (t) = − gt2 ⇀
 + v⇀(0)t + r⇀(0)
2
⇀
ASSUME:
SOLVE:
(2)→(1)⇒
⇀
ı:
⇀
:
r⇀(tf ) = 10(cos(35◦ ) ⇀
ı + sin(35◦ ) ⇀
 ) ft
⇀
⇀
◦
v (0) = v0 (cos(45 ) ı + sin(45◦ ) ⇀
 ) ft/s
1
10(cos(35 ) ı + sin(35 )  ) ft = − gt2f
2
◦
◦
⇀
⇀
 + v0 (cos(45◦ ) ⇀
ı + sin(45◦ ) ⇀
 )tf
v
u
2
◦
u
u (− 1 )(32.2 ft/s2 ) 10 cos(35 ) ft
u
2
cos(45◦ )
= t
◦
◦
◦
v0 =
(2)
⇀
10 cos(35◦ ) ft = v0 cos(45◦ )tf
10 cos(35◦ )
tf =
sec
v0 cos(45◦ )
1
10 sin(35◦ ) ft = v0 sin(45◦ )tf − gt2f
2
2
◦)
10
cos(35
10 cos(35◦ )
1
10 sin(35◦ ) ft = v0 sin(45◦ )
−
sec
g
sec
v0 cos(45◦ )
2
v0 cos(45◦ )
v0
(1)
10(sin(35 ) − tan(45 ) cos(35 )) ft
29.7 ft/s
100
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(3)
CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.20
GOAL: Find how far along an inclined surface a tennis ball will impact when it rebounds at a
given velocity.
GIVEN: Tennis ball’s initial speed and slope’s inclination.
DRAW:
FORMULATE EQUATIONS:
SOLVE:
x = |v1 |(sin β)t
(1)
y = |v1 |(cos β)t − 4.905t2
(2)
y=
(1) → (2) ⇒
−4.905x2
x
2
2 + tan β
|v1 | (sin β)
(3)
A point on the sloped surface will satisfy
y = −(tan β)x
−(tan β)x =
(3), (4) ⇒
"
(4)
−4.905x2
x
2 + tan β
⇀ 2
|v 1 | (sin β)
1
4.905x
x ⇀ 2
2 − (tan β) + tan β
|v 1 | (sin β)
#
=0
x(.03139x − 2.3094) = 0
x = 73.57 m
m
R = 73.57
cos β = 84.95 m
101
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2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.2.21
GOAL: Calculate required acceleration of hand to catch glass and speed of hand on contact with
glass.
GIVEN: Positions at which glass falls and at which hand is initially.
DRAW:
FORMULATE EQUATIONS: Let r⇀G be the position vector of the glass and r⇀H be the position
vector of the hand during the fall.
1
r⇀G (t) = − gt2 ⇀
 + v⇀G (0)t + r⇀G (0)
2
1
(t − 0.25 s)2 a⇀H + r⇀H (0)
r⇀H (t) =
2
(1)
(2)
ASSUME:
SOLVE:
(1),(2)⇒
(3),(4),(5),(6)→(8)⇒
⇀
ı:
⇀
:
r⇀G (0) = (2 ⇀
ı + 3⇀
 ) ft
(3)
⇀
r H (0) = 2  ft
(4)
r⇀G (tf ) = r⇀H (tf ) = (2 ⇀
ı + 0.5 ⇀
 ) ft
⇀
⇀
2
ı
−
1.5

a⇀H = a √
22 + 1.52
(5)
⇀
(6)
1
1
 + v⇀G (0)t + r⇀G (0) = (t − 0.25 s)2 a⇀H + r⇀H (0)
− gt2 ⇀
2
2
1
1
2⇀
ı − 1.5 ⇀

− gt2 ⇀
 + (2 ⇀
ı + 3⇀
 ) ft = (t − 0.25 s)2 a √
2
2
22 + 1.52
+ (2 ⇀
 ft)
a(t − 0.25 s)2
√
22 + 1.52
!
1 2
1.5 a(t − 0.25 s)2
√
− gt + 3 ft = −
+ 2 ft
2
2
22 + 1.52
2 ft =
102
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(7)
(8)
(9)
(10)
CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
g
− t2 + 3 ft = (−1.5 + 2) ft
2
s
2(−2.5 ft)
= 0.394 s
t =
−32.2 ft/s2
(9)→(10)⇒
a =
(11)→(9)⇒
a =
v⇀H
√
(2 ft) 22 + 1.52
(0.394 s − .25 s)2
241 ft/s2
2⇀
ı − 1.5 ⇀

(0.394 s − 0.25 s)
= a⇀H t + v⇀H (0) = (241 ft/s2 ) √
22 + 1.52
|vH | = (241 ft/s2 )[(0.394 − 0.25) s] = 34.7 ft/s2
103
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(11)
2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.2.22
GOAL: Calculate minimum speed v to complete stunt.
GIVEN: Positions at which the car leaves the ramp, angle of the ramp and target location.
DRAW:
FORMULATE EQUATIONS:
ASSUME:
SOLVE:
(2)→(1)⇒
(1)
v⇀(0) = v(cos(20◦ ) ⇀
ı + sin(20◦ ) ⇀

⇀
⇀
r (0) = 10  ft
r⇀(tf ) = (79 ⇀
ı + 21 ⇀
 ) ft
(2)
1
 + v(cos(20◦ ) ⇀
ı + sin(20◦ ) ⇀
 )t + 10 ⇀
 m = 79 ⇀
ı m + 21 ⇀
 m
−g t2 ⇀
2
v cos(20◦ )t = 79 ft
79 ft
t =
v cos(20◦ )
⇀
ı:
21 ft = −
⇀
:
1
r⇀(t) = t2 a⇀ + v⇀(0)t + r⇀(0)
2
v =
g
2
79 ft
v cos(20◦ )
79 ft
cos(20◦ )
s
2
+ v sin(20◦ )
79 ft
v cos(20◦ )
(3)
+ 10 ft
32.2 ft/s2
2((79 ft) tan(20◦ ) − 11 ft)
v = 80.1 ft/s = 55 mph
104
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.23
GOAL: Determine the time it takes the ball to hit the water’s surface, where it hits, the impact
speed, and the angle of impact.
GIVEN: The ball starts from rest at H = 70 ft. The wind speed is vw = 80 ft/s at h = 50 ft.
DRAW:
FORMULATE EQUATIONS:
The y position of the ball is given by
1
y = H − gt2
2
The time it takes for the ball to fall from where the wind pushes it to the water’s surface is
(1)
t∗ = t − tA
(2)
s = vw t∗
(3)
The horizontal distance traveled due to the wind is
Since it starts from rest, the ball’s final vertical speed is given by
ẏ 2 = 2gH
f
SOLVE:
Find the time to impact by setting y = 0 in (1):
(1) ⇒
1
0 = H − gt2
2
t =
s
2H
g
t =
s
2(70 ft)
32.2 ft/s2
t = 2.085 s
The time it takes to fall to where the wind pushes the ball (point A) is
1
h = H − gt2A
(1) ⇒
2
105
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(4)
2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
s
tA =
s
2(70 ft − 50 ft)
32.2 ft/s2
= 1.115 s
tA =
tA
2(H − h)
g
From point A to the water’s surface,
t∗ = 2.085 s − 1.115 s
(2) ⇒
t∗ = 0.971 s
The distance traveled due to the wind is then
s = (80 ft/s)(0.971 s)
(3) ⇒
s = 77.6 ft
The impact speed is
q
vf =
ẋ2 + ẏ 2
f
(5)
f
The final horizontal speed is just the wind speed, so
(4) → (5) ⇒
vf =
vf =
q
2 + 2gH
vw
q
(80 ft/s)2 + 2(32.2 ft/s2 )(70 ft)
|vf | = 104.4 ft/s
The impact angle is
−1
θ = tan
(4) → (6) ⇒
−1
θ = tan
q
θ = tan−1 
√
ẏf
ẋf
!
2gH
vw
(6)
!
2(32.2 ft/s2 )(70 ft)
80 ft/s
|θ| = 40.0◦


106
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.24
GOAL: Calculate the acceleration and jerk of the wheel’s center in the vertical direction.
GIVEN: Size of wheel and bump.
DRAW
ASSUME: ẋ = v0 , ẍ = 0, ṙ = 0
FORMULATE EQUATIONS: The relation between x, y, and r is:
x2 + y 2 = r2
(1)
y 2 = r2 − x2
(2)
Rearranging (1) gives us
1
(3)
2y ẏ = −2xẋ
(4)
y=
r2 − x2
/2
SOLVE: Differentiating (2) with respect to time yields:
ẏ =
−xẋ
y
(5)
ẏ =
(3) → (5) ⇒
(r2
−xv0
− x2 )1/2
ẏ 2 + y ÿ = −ẋ2 − xẍ
Differentiating (4) ⇒
(6)
(7)
Because ẍ = 0 and ẋ = v0 , (7) can be solved for ÿ to yield:
ÿ =
(3), (6) → (8) ⇒
−v0 2 − ẏ 2
y
ÿ =
ÿ =
x2 v0 2
−v0 2 r2
r2 − x2 =
(r2 − x2 )3/2
(r2 − x2 )1/2
−v0 2 −
−v0 2 r2
(r2 − x2 )3/2
107
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(8)
(9)
2.2. CARTESIAN COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2ẏ ÿ + ẏ ÿ + yy (3) = 0
Differentiating (7) ⇒
(10)
where y (3) represents the third derivative of y with respect to time.
Solving (10) for y (3) ⇒
y
(3)
−xv0
2
(r − x2 )1/2
3ẏ ÿ
1
= −3 2
=−
y
(r − x2 )1/2
y (3) =
!
−v0 2
(r2 − x2 )3/2
!
(11)
−3v0 3 xr2
(r2 − x2 )5/2
108
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.25
GOAL: Find an expression for the necessary launch angle θ in terms of φ, d, h, L, g and vC .
GIVEN: System geometry.
DRAW:
FORMULATE EQUATIONS:
1
 + v⇀B (0)t + r⇀B (0)
r B (t) =
− gt2 ⇀
2
r⇀C (t) = v⇀C (0)t + r⇀C (0)
⇀
(1)
(2)
ASSUME:
r⇀B (0)
v⇀B (0)
r⇀C (0)
v⇀C (0)
r⇀B (tf )
=
=
=
=
=
0
v0 (cos θ ⇀
ı + sin θ ⇀
)
⇀
⇀
d ı +h
vC (cos φ ⇀
ı + sin φ ⇀
)
⇀
r C (tf ) = (d + L cos φ) ⇀
ı + (h + L sin φ) ⇀

(3)
SOLVE:
v⇀C (0)tf + r⇀C (0) = = r⇀C (0) + L(cos φ ⇀
ı + sin φ ⇀
)
L
vC tf = L ⇒ tf =
vC
(3)=(2)⇒
(3)=(1)⇒
1
− gt2f
2
1
− gt2f
2
(4)
⇀
 + v⇀B (0)tf + r⇀B (0) = (d + L cos φ) ⇀
ı + (h + L sin φ) ⇀

⇀
 + v0 (cos θ ⇀
ı + sin θ ⇀
 )tf
= (d + L cos φ) ⇀
ı + (h + L sin φ) ⇀

(5)
v0 tf cos θ = d + L cos φ
⇀
ı:
⇀
:
v0 =
v (d + L cos φ)
d + L cos φ
= C
tf cos θ
L cos θ
g
− t2f + v0 tf sin θ = h + L sin φ
2
(4),(6)→(7)⇒
109
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(6)
(7)
2.2. CARTESIAN COORDINATES
g
−
2
L
vC
!2
+
CHAPTER 2. KINEMATICS OF PARTICLES
vC (d + L cos φ)
L cos θ
!
L
vC
!
sin θ = h + L sin φ
tan θ(d + L cos φ) = h + L sin φ +
θ =
gL2
2
2vC
gL2
 h + L sin φ + 2 
2vC 


tan−1 


d + L cos φ




110
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CHAPTER 2. KINEMATICS OF PARTICLES
2.2. CARTESIAN COORDINATES
2.2.26
GOAL: Find launch speed of mass so it strikes middle of bat when θ = π2 .
GIVEN: Launch angle φ is fixed. Launching tube is a distance d from the bat.
DRAW:
FORMULATE EQUATIONS:
θ̈ = αt
θ̇ =
αt2
+ θ̇(0)
2
αt3
+ θ̇(0)t + θ(0)
6
(1)
g
 + v⇀(0)t + r⇀(0)
r⇀(t) = − t2 ⇀
2
(2)
θ(0) = π
θ̇(0) = 0
θ(t∗ ) = π2
(3)
r⇀(0) = d ⇀
ı
v⇀(0) = v0 (− cos φ ⇀
ı + sin φ ⇀
)
L⇀
⇀ ∗
r (t ) = 2 
(4)
θ=
ASSUME:
SOLVE:
αt∗ 3
+ π = π/2
6
(3)→(1)⇒
3π
t = −
α
∗
(4)→(2)⇒
⇀
ı:
1
3
(5)
L
g
 + v0 (− cos φ ⇀
ı + sin φ ⇀
 )t∗ + d ⇀
ı = ⇀

− t∗ 2 ⇀
2
2
−v0 t∗ cos φ + d = 0
d
cos φ =
v0 t∗
⇒
v
u
u
sin φ = t1 −
d
v0 t∗
!2
111
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(6)
2.2. CARTESIAN COORDINATES
⇀
:
(6),(5)→(7)⇒
CHAPTER 2. KINEMATICS OF PARTICLES
g
L
− t∗ 2 + v0 t∗ sin φ =
2
2
v
u
u
∗t
v0 t 1 −
d
v0 t∗
v0 t
!2
∗
2
v02
=
L + gt∗ 2
2
2
−d =
=
(7)
L + gt∗ 2
2
L+gt∗2 2
2
t∗ 2
!2
+ d2
v
u
2 2
u
u L + g − 3π 3
+ 4d2
α
u
v0 = u
2
t
3
4 − 3π
α
112
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3
Polar and Cylindrical Coordinates
113
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.1
GOAL: Determine the absolute speed and acceleration of a running boy on a turntable.
GIVEN: The turntable is rotating at 0.4 rad/s and its radius is 10 m. Andy begins running from
rest and accelerates at a constant rate of 3 m/s2 .
DRAW
FORMULATE EQUATIONS:
We’ll use our expressions for angular velocity and acceleration:
v⇀A = ṙ e⇀r + rθ̇ e⇀θ
a⇀A = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
as well as our formula for constant acceleration for a particle starting from rest (and r = 0):
1
r̈ r = ṙ2
2
ASSUME: Andy runs straight toward Ben.
SOLVE: After moving 10 m Andy is moving at q
ṙ =
(1) ⇒
||v⇀A || =
q
(rθ̇)2 + (ṙ)2 =
q
(1)
2(3 m/s2 )(10 m) = 7.75 m/s
[(10 m)(0.4 rad/s)]2 + (7.75 m/s)2
||v⇀A || = 8.72 m/s
||a⇀A || =
q
(r̈ − rθ̇2 )2 + (rθ̈ + 2ṙθ̇)2 =
q
[3 m/s2 − (10 m)(0.4 rad/s)2 ]2 + [2(7.75 m/s)(0.4 rad/s)]2
||a⇀A || = 6.35 m/s2
114
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.2
GOAL: Determine the absolute speed and acceleration of a person running across a carousel when
he reaches the edge.
GIVEN: The carousel is rotating at 0.6 rad/s (counter-clockwise) and its radius is 40 ft. When
Bill reaches the center O he has a speed of 5 ft/s and an acceleration of 2 ft/s2 (t = 0 is referenced to
the point at which he reaches the center). At t = 0 the carousel begins to decelerate at a constant
rate of 0.1 rad/s2 .
DRAW
FORMULATE EQUATIONS:
We’ll use our expressions for angular velocity and acceleration
v⇀A = ṙ e⇀r + rθ̇ e⇀θ
(1)
a⇀A = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
(2)
and our position, speed, constant acceleration relationships
a(t) = a0 ,
v(t) = v0 + a0 t,
x(t) = x0 + v0 t + 0.5a0 t2
SOLVE: First we solve for the elapsed time for Bill to go from O to B (radial motion):
40 ft = (5 ft/s)t + 0.5(2 ft/s2 )t2
which has solution t = 4.30 s. Evaluating the radial speed corresponding to this time gives us
ṙ = 5 ft/s + (2 ft/s2 )(4.30 s) = 13.6 ft/s. At t = 4.30 s the carousel is rotating at 0.6 rad/s −
(0.1 rad/s2 )(4.30 s) = 0.170 rad/s.
(1) ⇒
v⇀Bill = [13.6 e⇀r + (40 ft)(0.170 rad/s) e⇀θ ] ft/s
v⇀Bill = (13.6 e⇀r + 6.80 e⇀θ ) ft/s
(2) ⇒
a⇀Bill = [(2 ft/s2 − (40 ft)(0.170 rad/s)2 ) e⇀r + ((40 ft)(−0.1 rad/s2 ) + 2(13.6 ft/s)(0.170 rad/s)) e⇀θ ] ft/s2
a⇀Bill = (0.845 e⇀r + 0.6225 e⇀θ ) ft/s2
115
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.3
GOAL: Find the acceleration of a fly on a fan blade.
GIVEN: Fan dimensions and rotation rate.
DRAW:
FORMULATE EQUATIONS:
a⇀F = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
(1)
ASSUME: The fly isn’t moving outward and therefore ṙ = r̈ = 0. The rotation rate is constant
and therefore θ̈ = 0.
SOLVE: We’re told that the fan blade rotates 2π radians in 1.3 seconds. Thus
θ̇ =
2π
rad/s
1.3
a⇀F = −rθ̇2 e⇀r = −(1 m)
2
2π
rad/s
1.3
e⇀r
a⇀F = −23.4 e⇀r m/s2
116
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.4
GOAL: Find an example for which
d| r⇀A |
=0
dt
and
FORMULATE EQUATIONS:
|v⇀A | = 1
v⇀A = ṙA e⇀r + rA θ̇ e⇀θ
(1)
⇀
d| r |
SOLVE: We must have dtA = 0 and |v⇀A | = 1. The first relation says rA is constant (ṙA = 0).
This means A must be moving in a circle about the origin and eliminates the ṙA of (1), leaving us
with
v⇀A = rA θ̇ e⇀θ
|v⇀A | = 1 implies the speed of rotation is constant (|rθ̇| = 1).
117
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.5
GOAL: Find the velocity and acceleration of the tip of a fire ladder.
GIVEN: System dimensions and rates.
DRAW:
⇀
⇀
er
e⇀θ
⇀
ı

cos θ sin θ
− sin θ cos θ
FORMULATE EQUATIONS:
SOLVE:
(1) ⇒
v⇀B = ṙ e⇀r + rθ̇ e⇀θ
(1)
a⇀B = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
(2)
v⇀B = (0.4 m/s) e⇀r + (8 m)(0.05 rad/s) e⇀θ
v⇀B = (0.4 e⇀r + 0.4 e⇀θ ) m/s
a⇀B = [0.1 m/s2 −(8 m)(0.05 rad/s)2 ] e⇀r +[(8 m)(0.04 rad/s2 )+2(0.4 m/s)(0.05 rad/s)] e⇀θ
(2) ⇒
Using θ =
π
6
a⇀B = (0.08 e⇀r + 0.36 e⇀θ ) m/s2
rad and the coordinate transformation array gives us
v⇀B = (0.146 ⇀
ı + 0.546 ⇀
 ) m/s
a⇀B = (−0.111 ⇀
ı + 0.352 ⇀
 ) m/s2
118
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.6
GOAL: Find |vB | and |aB |.
GIVEN: System geometry and rates
DRAW
FORMULATE EQUATIONS:
SOLVE:
v⇀B = ṙ e⇀r + rθ̇ e⇀θ
(1)
a⇀B = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
(2)
v⇀B = (−0.0549 m/s) e⇀r + (0.3 m)(−0.683 rad/s) e⇀θ
|vB | =
√
0.05492 + 0.2052 m/s = 0.212 m/s
a⇀B = [0.0375 m/s2 − (0.3 m)(−0.683 rad/s)2 ] e⇀r
+[(0.3 m)(−0.1585 rad/s2 ) + 2(−0.0549)(−0.683 rad/s)] e⇀θ
= [−0.102 e⇀r + 0.0274 e⇀θ ]m/s2
|aB | = 0.106 m/s2
119
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.7
GOAL:
A particle’s velocity is given. Express the velocity in polar coordinates and determine θ̇.
GIVEN: System geometry and aircraft’s speed.
DRAW:
e⇀r
e⇀θ
⇀
⇀

√ı
3/2 √1/2
−1/2
3/2
FORMULATE EQUATIONS:
v⇀plane = −1200 ⇀
ı km/hr = −333.3̄ ⇀
ı m/s
!
√
3⇀
1
e r − e⇀θ m/s
= −333.3̄
2
2
= (−288.7 e⇀r + 166.6̄ e⇀θ ) m/s
(1)
v⇀plane = ṙ e⇀r + rθ̇ e⇀θ
(2)
v⇀plane is also expressable as
SOLVE:
(1), (2) ⇒
ṙ e⇀r + rθ̇ e⇀θ = (−288.7 e⇀r + 166.6̄ e⇀θ ) m/s
ṙ e⇀r + (3000 m)θ̇ e⇀θ = (−288.7 e⇀r + 166.6̄ e⇀θ ) m/s
e⇀r :
ṙ = −288.7 m/s
(3)
e⇀θ :
(3000 m)θ̇ = 166.6̄ m/s
(4)
(4) ⇒
θ̇ =
166.6̄ m/s
3000 m
= .05̄ rad/s
120
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.8
GOAL: Find the radial speed of car at B relative to point A in terms of v.
GIVEN: Dimensions to point A, dimensions to point B, ∠AOB = 45◦
DRAW:
⇀
⇀
er
e⇀θ
⇀
ı

◦
cos 63.4
sin 63.4◦
− sin 63.4◦ cos 63.4◦
FORMULATE EQUATIONS:
v⇀B = ṙ e⇀r + rθ̇ e⇀θ
SOLVE:
θ = arctan
e⇀r :
0.2 km
= 63.4◦
0.1 km
v⇀B ( expressed in e⇀r , e⇀θ ) = v⇀B ( expressed in ⇀
ı,⇀
)
√
√
2⇀
2⇀
ı −v

ṙ e⇀r + rθ̇ e⇀θ = −v
2
2
√
√
2
2
◦⇀
◦⇀
⇀
⇀
(cos 63.4 e r − sin 63.4 e θ ) − v
(sin 63.4◦ e⇀r + cos 63.4◦ e⇀θ )
ṙ e r + rθ̇ e θ = −v
2
2
√
√
2
2
◦
ṙ = −v
(cos 63.4 ) − v
(sin 63.4◦ ) = −.949v
2
2
|ṙB/ | = |ṙ| = .949v
A
|ṙB/ | = .949v
A
121
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.9
⇀
d r (t)
d ⇀
GOAL: Why is || dt || generally not equal to dt
|| r (t)||?
GIVEN: Two expressions involving a time derivative of r⇀.
FORMULATE EQUATIONS:
r⇀(t) = r⇀ = r e⇀r
v⇀p = ṙ e⇀r + rθ̇ e⇀θ
polar velocity
SOLVE:
note that all variables represent a function of time.
d ⇀
consider dt
|| r (t)|| :
|| r⇀|| = ||r e⇀r || = r
d
d ⇀
|| r (t)|| = r = ṙ
dt
dt
d r⇀(t)
= v⇀p = ṙ e⇀r + rθ̇ e⇀θ
dt
⇀
now consider ||
d r (t)
dt ||
:
d r⇀(t)
||
|| = ||v⇀p || =
dt
ṙ 6=
q
q
ṙ2 + (rθ̇)2
ṙ2 + (rθ̇)2
⇀
d r (t)
|| dt || represents the magnitude of the polar velocity function, or the actual speed of a particle
described over time. This speed compensates for both the speed at which a particle moves away
d ⇀
from the origin and the speed at which a particle moves around the origin.
dt || r (t)|| is ṙ or the
radial speed describing the speed at which a particle moves away from the origin, but not around
it.
122
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.10
d ⇀
d ⇀
GOAL: Determine when dt
| r (t)| is equal to | dt
( r (t))|.
⇀
GIVEN: Two expressions involving r .
FORMULATE EQUATIONS:
v⇀ = ṙ e⇀r + rθ̇ e⇀θ
SOLVE:
| r⇀| is the magnitude of r⇀, namely, r.
dr
dt is the speed at which r changes, namely, ṙ.
⇀
dr
⇀
dt is the velocity of the vector r , including ṙ and r θ̇ terms:
d r⇀
=
dt
q
ṙ2 + (rθ̇)2
q
ṙ is equal to ṙ2 + (rθ̇)2 only when θ̇ = 0. Thus, the particle must be moving outward but not
around the origin. ṙ has to be positive (moving away from the origin) for the signs to match.
123
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.11
d ⇀
GOAL: Determine if dt
( r ) is zero.
GIVEN: ṙ = 3 ft/s and rθ̇ = −3 ft/s.
FORMULATE EQUATIONS:
We’ll use v⇀ = ṙ e⇀r + rθ̇ e⇀θ .
SOLVE:
d ⇀
Although ṙ and rθ̇ are the magnitudes of the components of dt
( r ), these components are in orthog√
d ⇀
⇀
⇀
onal directions. dt r would equal (3 e r − 3 e θ ) ft/s, with a resultant magnitude of 3 2 ft/s. Thus the
final answer to the question is “no.”
124
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.12
GOAL: Compare the effect on acceleration of a particle moving horizontally as opposed to one
moving in a circular manner, both with a sinusoidal variation of path.
GIVEN: In case (a) y = r0 − r1 cos(2ωt) and in case (b) θ = 2ωt, r = r0 − r1 cos(θ).
DRAW:
FORMULATE EQUATIONS: Acceleration in Cartesian and polar frames are given by
a⇀A = ÿ ⇀

a⇀A = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
SOLVE:
Case (a): We can differentiate our expression for y twice, yielding
ÿ = 4ω 2 r1 cos(2ωt)
Case (b): We can use our expression for θ in our expression for r, giving us
r = r0 − r1 cos(2ωt)
Now we need to consider the component of acceleration in the e⇀r direction:
ar = r̈ − rθ̇2
Differentiating our expression for r twice yields
r̈ = 4ω 2 r1 cos(2ωt)
and differentiating θ gives us θ̇ = 2ω.
Put these results into our expression for ar and we have
ar = 4ω 2 r1 cos(2ωt) − 4ω 2 [r0 − r1 cos(2ωt)]
ar = −4r0 ω 2 + 8ω 2 r1 cos(2ωt)
We see two basic things. First, the fact that the particle in Case (b) is, on average, moving around
a circle, there is a constant centripetal component −4r0 ω 2 . Interestingly, in addition to this, there
is a further increase in the time varying acceleration as compared to the horizontal scenario (Case
(a)). Rather than an acceleration magnitude of 4ω 2 r1 , the rotational case has a magnitude that’s
100% larger: 8ω 2 r1 .
125
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.13
GOAL: Determine the angular speed of rotation needed to keep the image of a bird centered in a
camera’s viewfinder and calculate the total acceleration of the far end of the lens.
GIVEN: Distance to the end of the lens from the point of rotation is 17 inches. The viewfinder
displays a 40 inch wide target when the target is 80 feet away. The swallow moves at a constant
40 mph along a straight line. The closest approach of the swallow to the camera is h = 80 ft.
DRAW:
FORMULATE EQUATIONS: The equations for velocity and acceleration in polar coordinates
are
v⇀S = ṙ e⇀r + rθ̇ e⇀θ
(1)
and
a⇀S = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ
(2)
SOLVE: 40 mph is equivalent to 58.7 ft/s. At θ = 90◦ e⇀r = ⇀
 and e⇀θ = − ⇀
ı . Using the provided
parameters in (1) gives us
h
v⇀S = −58.7 ⇀
ı ft/s = ṙ e⇀r + (80 ft)θ̇ e⇀θ
ṙ = 0, θ̇ =
58.7 ft/s
= 0.733 rad/s
80 ft
i
θ̇ = 0.733 rad/s
Using this data along with the provided parameters in (2) yields
h
i
h
i
a⇀S = 0 = r̈ − (80 ft)(0.733 rad/s)2 e⇀r + 2(0)(0.733 rad/s) + (80 ft)θ̈ e⇀θ
r̈ = 43.0 ft/s2 , θ̈ = 0
Our equation for the acceleration of end of the lens is a bit different than the swallow’s acceleration
because the swallow is moving in a straight line whereas the lens is rotating in a circle. Hence for
the lens both ṙ and r̈ are zero. The acceleration is found from
126
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
a⇀A = (r̈ − rθ̇2 ) e⇀r + (2ṙθ̇ + rθ̈) e⇀θ = −
17 in
(0.733 rad/s)2 + 0 = −0.762 ⇀
 ft/s2
12 in/ ft
a⇀A = −0.762 ⇀
 ft/s2
127
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.14
GOAL: Determine the constant angular velocity of the camera that will cause the image of a
swallow to disappear from the viewfinder in 0.5 s. Compare this angular speed to the angular
speed needed to perfectly track the bird and determine the percent variation from the ideal case.
GIVEN: The camera’s telephoto lens captures 40 inches of target image when the object is 80 feet
away. The swallow is 8 inches long and initially centered. The bird is at θ = 90◦ and traveling at
a constant 40 mph with a constant radius of 80 ft from the camera.
DRAW:
FORMULATE EQUATIONS: The equation for velocity in polar coordinates is
v⇀m = ṙ e⇀r + rθ̇ e⇀θ
(1)
SOLVE: Using the provided parameters in (1) gives us a nominal rotation rate:
40 mph
e⇀ = 58.7 e⇀θ ft/s = (80 ft)θ̇ e⇀θ ⇒ θ̇nominal = 0.733 rad/s
v⇀S =
60 mph/(88 ft/s) θ
If the camera is being rotated too slowly, the bird’s image will move to the left in the viewfinder.
It’s initially centered and so in order to completely disappear from the viewfinder it needs to move
16 in + 8 in = 24 in = 2 ft. The time for this to occur is given as 0.5 s. This implies a speed of
2 ft/(0.5 s) = 4 ft/s. The corresponding angular speed is
4 ft/s
= 0.05 rad/s
θ̇ =
80 ft
Thus the angular rate of the camera would be the nominal rate ± this variation. Assuming a
camera pan that’s too slow (the usual problem) we have θ̇ = 0.733 rad/s − 0.05 rad/s = 0.683 rad/s.
θ̇ = 0.683 rad/s
0.05 :
The percent variation from nominal is given by 100 0.733
percent variation = 6.8%
The implication is that it’s not easy to keep a quickly moving object in the viewfinder, an observation
that’s very much supported by experience in the field.
128
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.15
GOAL: Calculate at what angle θ the camera’s angular acceleration is a maximum and determine
its angular acceleration at θ = 0 and θ = 90◦ .
GIVEN: The swallow moves at a constant 40 mph along a straight line. The closest approach of
the swallow to the camera is h = 80 ft.
DRAW:
⇀
⇀
er
e⇀θ
⇀
ı

cos θ sin θ
− sin θ cos θ
FORMULATE EQUATIONS: The equations for velocity and acceleration in polar coordinates
are
v⇀S = ṙ e⇀r + rθ̇ e⇀θ
(1)
and
SOLVE:
(1) ⇒
a⇀S = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ
−v ⇀
ı = −v[cos θ e⇀r − sin θ e⇀θ ] = ṙ e⇀r + rθ̇ e⇀θ
e⇀r :
ṙ = −v cos θ
θ̇ =
e⇀θ :
v sin θ
r
(3)
(4)
0 = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ
(2) ⇒
e⇀r :
r̈ − rθ̇2 = 0
e⇀θ :
2ṙθ̇ + rθ̈ = 0
θ̈ = −
(3), (4), (6) ⇒
From geometry we see that r =
(2)
h
sin θ
(5)
(6)
2v 2 sin θ cos θ
2ṙθ̇
=
r
r2
and hence we have
2v 2 sin3 θ cos θ
θ̈ =
h2
129
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(7)
2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
d θ̈ = 0:
We find where θ̈ is maximized (or minimized) by setting d
θ
d θ̈
6 sin2 θ cos2 θ − 2 sin4 θ
=0
=
dθ
h2
3 cos2 θ = sin2 θ ⇒ tan2 θ = 3 ⇒ tan θ =
√
3
θ = 60◦
Evaluating (7) at θ = 0 gives us the same answer as for θ = 90◦ :
θ̈ = 0
130
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.16
GOAL: Determine r, ṙ, r̈, θ̇, θ̈.
GIVEN: System geometry and battleship velocity: d = 300 m, θ = 50◦ , v⇀ = 25 ⇀
 m/s.
DRAW:
FORMULATE EQUATIONS:
The coordinate transformation array is
⇀
⇀
er
e⇀θ
⇀
ı

cos θ sin θ
− sin θ cos θ
The battleship velocity in terms of polar coordinates is
v⇀ = ṙ e⇀r + rθ̇ e⇀θ
(1)
The battleship acceleration in terms of polar coordinates is
SOLVE:
From the diagram we see that
a⇀ = (r̈ − rθ̇2 ) e⇀r + (2ṙ θ̇ + rθ̈) e⇀θ
(2)
d = r cos θ
d
r =
cos θ
300 m
r =
cos(50◦ )
r = 466.7 m
To find ṙ and θ̇,
(1) ⇒
v⇀
 = ṙ e⇀r + rθ̇ e⇀θ
131
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(3)
2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
v sin θ e⇀r + v cos θ e⇀θ = ṙ e⇀r + rθ̇ e⇀θ
ṙ = v sin θ
e⇀r :
ṙ = (25 m/s) sin(50◦ )
ṙ = 19.15 m/s
e⇀θ :
(4)
rθ̇ = v cos θ
v cos θ
r
(25 m/s) cos(50◦ )
θ̇ =
466.7 m
θ̇ =
(3) → (5) ⇒
θ̇ = 3.44 × 10−2 rad/s
(5)
(6)
To find r̈ and θ̈,
(2) ⇒
e⇀r :
⇀
0 = (r̈ − rθ̇2 ) e⇀r + (2ṙ θ̇ + rθ̈) e⇀θ
r̈ − rθ̇2 = 0
r̈ = rθ̇2
(3), (6) → (7) ⇒
(7)
r̈ = (466.7 m)(3.44 × 10−2 rad/s)2
r̈ = 0.553 m/s2
e⇀θ :
2ṙ θ̇ + rθ̈ = 0
2ṙ θ̇
r
2(19.15 m/s)(3.44 × 10−2 rad/s)
θ̈ = −
(466.7 m)
θ̈ = −
(3), (4), (6) → (8) ⇒
θ̈ = −2.83 × 10−3 rad/s2
132
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(8)
CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.17
GOAL: Determine ṙ, r̈, θ̇, θ̈.
GIVEN: System geometry, and the exerciser’s foot velocity and acceleration: r = 2 ft, θ = 15◦ ,
v⇀ = 4.33 ⇀
ı + 2.5 ⇀
 ft/s, a⇀ = 0.87 ⇀
ı + 0.5 ⇀
 ft/s2 .
DRAW:
FORMULATE EQUATIONS:
The coordinate transformation array is
⇀
⇀
er
e⇀θ
⇀
ı

cos θ sin θ
− sin θ cos θ
The velocity of the exerciser’s foot in terms of polar coordinates is
v⇀ = ṙ e⇀r + rθ̇ e⇀θ
(1)
The acceleration of the exerciser’s foot in terms of polar coordinates is
a⇀ = (r̈ − rθ̇2 ) e⇀r + (2ṙ θ̇ + rθ̈) e⇀θ
(2)
SOLVE:
To find ṙ and θ̇,
(1) ⇒
 = ṙ e⇀r + rθ̇ e⇀θ
vx ⇀
ı + vy ⇀
vx (cos θ e⇀r − sin θ e⇀θ ) + vy (sin θ e⇀r + cos θ e⇀θ ) = ṙ e⇀r + rθ̇ e⇀θ
e⇀r :
ṙ = vx cos θ + vy sin θ
ṙ = (4.33 ft/s) cos(15◦ ) + (2.5 ft/s) sin(15◦ )
ṙ = 4.830 ft/s
e⇀θ :
rθ̇ = −vx sin θ + vy cos θ
133
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(3)
2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
1
θ̇ = [−vx sin θ + vy cos θ]
r
θ̇ =
1
[−(4.33 ft/s) sin(15◦ ) + (2.5 ft/s) cos(15◦ )]
2 ft
θ̇ = 0.647 rad/s
(4)
To find r̈ and θ̈,
ax ⇀
ı + ay ⇀
 = (r̈ − rθ̇2 ) e⇀r + (2ṙ θ̇ + rθ̈) e⇀θ
(2) ⇒
ax (cos θ e⇀r − sin θ e⇀θ ) + ay (sin θ e⇀r + cos θ e⇀θ ) = (r̈ − rθ̇2 ) e⇀r + (2ṙ θ̇ + rθ̈) e⇀θ
e⇀r :
r̈ − rθ̇2 = ax cos θ + ay sin θ
r̈ = rθ̇2 + ax cos θ + ay sin θ
(4) → (5) ⇒
(5)
r̈ = (2 ft)(0.647 rad/s)2 + (0.87 ft/s2 ) cos(15◦ ) + (0.5 ft/s2 ) sin(15◦ )
r̈ = 1.807 ft/s2
e⇀θ :
(3), (4) → (6) ⇒
2ṙ θ̇ + rθ̈ = −ax sin θ + ay cos θ
1
θ̈ = [−2ṙ θ̇ − ax sin θ + ay cos θ]
r
1
θ̈ =
[−2(4.830 ft/s)(0.647 rad/s) − (0.87 ft/s2 ) sin(15◦ )
2 ft
+ (0.5 ft/s2 ) cos(15◦ )]
θ̈ = −2.996 rad/s2
134
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(6)
CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.18
⇀
GOAL: Determine kv⇀k and kak.
GIVEN: The radial position of the robotic arm’s end effector is governed by r(θ) = 0.5 + 0.25[1 −
cos(2θ)] m. At the given instant, θ = 45◦ , θ̇ = 0.5 rad/s, θ̈ = 0.
DRAW:
FORMULATE EQUATIONS:
The velocity of the end effector in terms of polar coordinates is
v⇀ = ṙ e⇀r + rθ̇ e⇀θ
(1)
The acceleration of the end effector in terms of polar coordinates is
SOLVE:
a⇀ = (r̈ − rθ̇2 ) e⇀r + (2ṙ θ̇ + rθ̈) e⇀θ
(2)
r(45◦ ) = 0.5 + 0.25[1 − cos(2 · 45◦ )] m
r(45◦ ) = 0.75 m
Differentiate r(θ) ⇒
(3)
ṙ(θ) = 0.5 sin(2θ)θ̇ m/s
ṙ(45◦ ) = 0.5 sin(2 · 45◦ )(0.5) m/s
ṙ(45◦ ) = 0.25 m/s
Differentiate ṙ(θ) ⇒
(4)
r̈(θ) = cos(2θ)θ̇2 + 0.5 sin(2θ)θ̈ m/s2
r̈(45◦ ) = cos(2 · 45◦ )(0.5)2 + 0 m/s2
r̈(45◦ ) = 0 m/s2
(1) ⇒
(3), (4) → (6) ⇒
kv⇀k =
kv⇀k =
q
q
v⇀ · v⇀ =
q
ṙ2 + r2 θ̇2
(0.25 m/s)2 + (0.75 m)2 (0.5 rad/s)2
kv⇀k = 0.451 m/s
135
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(5)
(6)
2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
(2) ⇒
(3), (4), (5) → (7) ⇒
⇀
kak =
⇀
kak
=
q
q
⇀
⇀
a·a =
q
(r̈ − rθ̇2 )2 + (2ṙ θ̇ + rθ̈)2
{0 − (0.75 m)(0.5 rad/s)2 }2 + {2(0.25 m/s)(0.5 rad/s) + 0}2
⇀
kak
= 0.313 m/s2
136
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(7)
CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.19
GOAL: Determine r, ṙ, r̈, θ̇, θ̈.
GIVEN: System geometry and the pole’s constant angular speed: L = 20 ft, θ = 40◦ , ω =
0.25 rad/s.
DRAW:
ASSUME: Neglect the mass of the pole, and analyze only the lumped mass on the end.
FORMULATE EQUATIONS:
The coordinate transformation array is
⇀
⇀
er
e⇀θ
⇀
ı

cos θ sin θ
− sin θ cos θ
With respect to the pole, the lumped mass’s velocity and acceleration are, respectively,
v⇀ = Lω e⇀t = −Lω ⇀
ı
⇀
a=
Lω 2 e⇀n
2⇀
= −Lω 
(1)
(2)
The velocity and acceleration of the lumped mass in terms of polar coordinates are, respectively,
SOLVE:
From the diagram we see that
v⇀ = ṙ e⇀r + rθ̇ e⇀θ
(3)
a⇀ = (r̈ − rθ̇2 ) e⇀r + (2ṙ θ̇ + rθ̈) e⇀θ
(4)
L = r sin θ
L
r =
sin θ
20 ft
r =
sin(40◦ )
r = 31.11 ft
To find ṙ and θ̇,
137
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(5)
2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
(1) = (3) ⇒
−Lω ⇀
ı = ṙ e⇀r + rθ̇ e⇀θ
−Lω(cos θ e⇀r − sin θ e⇀θ ) = ṙ e⇀r + rθ̇ e⇀θ
ṙ = −Lω cos θ
e⇀r :
ṙ = −(20 ft)(0.25 rad/s) cos(40◦ )
ṙ = −3.830 ft/s
e⇀θ :
(6)
rθ̇ = Lω sin θ
Lω
sin θ
r
(20 ft)(0.25 rad/s)
θ̇ =
sin(40◦ )
31.11 ft
(7)
θ̇ =
(5) → (7) ⇒
θ̇ = 0.103 rad/s
(8)
To find r̈ and θ̈,
 = (r̈ − rθ̇2 ) e⇀r + (2ṙ θ̇ + rθ̈) e⇀θ
−Lω 2 ⇀
(2) = (4) ⇒
−Lω 2 (sin θ e⇀r + cos θ e⇀θ ) = (r̈ − rθ̇2 ) e⇀r + (2ṙ θ̇ + rθ̈) e⇀θ
e⇀r :
r̈ − rθ̇2 = −Lω 2 sin θ
r̈ = rθ̇2 − Lω 2 sin θ
(5), (8) → (9) ⇒
(9)
r̈ = (31.11 ft)(0.103 rad/s)2 − (20 ft)(0.25 rad/s)2 sin(40◦ )
r̈ = −0.472 ft/s2
e⇀θ :
(5), (6), (8) → (10) ⇒
2ṙ θ̇ + rθ̈ = −Lω 2 cos θ
1
θ̈ = − [2ṙ θ̇ + Lω 2 cos θ]
r
1
θ̈ = −
[2(−3.830 ft/s)(0.103 rad/s)
31.11 ft
+ (20 ft)(0.25 rad/s)2 cos(40◦ )]
(10)
θ̈ = −5.34 × 10−3 rad/s2
138
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.20
GOAL: Determine kv⇀A k and ka⇀A k.
GIVEN: The merry-go-round has radius R = 5 ft and is spun to a constant θ̇ = 3 rad/s. The RC
toy car is accelerated from rest at the center straight out to point A at a rate r̈ = 3 ft/s2 .
DRAW:
FORMULATE EQUATIONS:
With respect to the rotating merry-go-round, the RC toy car’s velocity at point A is given by
2
= 2r̈R
ṙA
(1)
The velocity of the RC car at A in terms of polar coordinates is
v⇀A = ṙA e⇀r + Rθ̇ e⇀θ
(2)
The acceleration of the RC car at A in terms of polar coordinates is
a⇀A = (r̈ − Rθ̇2 ) e⇀r + (2ṙA θ̇) e⇀θ
SOLVE:
kv⇀A k =
(2) ⇒
q
v⇀A · v⇀A =
kv⇀A k =
(1) → (4) ⇒
⇀
kv A k =
q
q
q
2 + R2 θ̇ 2
ṙA
(3)
(4)
2r̈R + R2 θ̇2
2(3 ft/s2 )(5 ft) + (5 ft)2 (3 rad/s)2
kv⇀A k = 15.97 ft/s
ka⇀A k =
(3) ⇒
q
a⇀A · a⇀A =
ka⇀A k =
(1) → (5) ⇒
⇀
kaA k =
r
q
q
2 θ̇ 2
(r̈ − Rθ̇2 )2 + 4ṙA
(r̈ − Rθ̇2 )2 + 8r̈Rθ̇2
(3 ft/s2 ) − (5 ft)(3 rad/s)2
2
+ 8(3 ft/s2 )(5 ft)(3 rad/s)2
ka⇀A k = 53.33 ft/s2
139
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(5)
2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.21
GOAL: Find ṙ, r̈, θ̇, θ̈ of the dolphin.
GIVEN: r = 300 m, vr = 10 m/s, vθ = 17.3 m/s, ar = 0 m/s2 , aθ = 0 m/s2
DRAW:
FORMULATE EQUATIONS:
v⇀p = ṙ e⇀r + rθ̇ e⇀θ
polar velocity
v⇀p = vr e⇀r + vθ e⇀θ
a⇀p = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
polar acceleration
a⇀p = ar e⇀r + aθ e⇀θ
SOLVE:
vr :
vr = ṙ = 10 m/s
vθ :
vθ = rθ̇
θ̇ =
vθ
17.3 m/s
=
= 0.058 rad/s
r
300 m
ar = r̈ − rθ̇2 = 0 m/s2
ar :
r̈ = rθ̇2 = (300 m)(0.058 rad/s)2 = 0.998 m/s2
aθ = rθ̈ + 2ṙθ̇ = 0 m/s2
aθ :
θ̈ =
(−2)(10 m/s)(0.058 rad/s)
−2ṙθ̇
=
= −3.84×10−3 rad/s2
r
300 m
ṙ = 10 m/s
θ̇ = 0.058 rad/s
r̈ = 0.998 m/s2
θ̈ = −3.84×10−3 rad/s2
140
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.22
GOAL: Find the gecko’s speed.
GIVEN: System geometry and rate the thread exits the spinometer.
DRAW:
⇀
⇀
ı

cos θ sin θ
− sin θ cos θ
e⇀r
e⇀θ
ASSUME: The gecko moves vertically and therefore
v⇀G = vG ⇀
ı
(1)
v⇀G = ṙ e⇀r + rθ̇ e⇀θ
(2)
FORMULATE EQUATIONS:
−1
θ = tan
0.5
2.2
= 12.8◦
1
r = [.052 + 2.22 ] 2 m = 2.26 m
SOLVE: (1),(4) and ṙ = 0.78 m/s give us:
v⇀G = (0.78 m/s) e⇀r + (2.26 m)θ̇ e⇀θ = vG ⇀

(0.78 m/s)(cos(12.8◦ ) ⇀
ı + sin(12.8◦ ) ⇀
 ) + (2.26 m)θ̇(− sin(12.8◦ ) ⇀
ı + cos(12.8◦ ) ⇀
 ) = vG ⇀

(0.761 m/s − 0.5θ̇ m) ⇀
ı + (0.173 m/s + 2.2θ̇ m) ⇀
 = vG ⇀

0.761 m/s − 0.5θ̇ m = 0 ⇒ θ̇ = 1.52 rad/s
0.173 m/s + (2.2 m)θ̇ = vG
vG = [0.173 m/s + (2.2 m)(1.52 rad/s)] = 3.52 m/s
141
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(3)
(4)
2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.23
GOAL: Find magnitude of particle’s velocity and acceleration.
GIVEN: Velocity of point A.
DRAW:
FORMULATE EQUATIONS:
v⇀ = ṙ e⇀r + rθ̇
Velocity:
(1)
a⇀ = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ
Acceleration:
(2)
Differentiating equation for r:
ṙ = a − 3bt2
Differentiating equation for ṙ:
r̈ = −6bt
(4)
Differentiating equation for θ:
θ̇ = −ce−t
(5)
Differentiating equation for θ̇:
SOLVE:
θ̈ = ce−t
(6)
kv⇀k =
(3), (5) → (1) ⇒
q
(3)
(a − 3bt2 )2 + (at − bt3 )2 (−ce−t )2
(3) − (6) → (2) ⇒
⇀
kak =
rh
−6bt − (at − bt3 ) (−ce−t )2
i2
+ [2 (a − 3bt2 ) (−ce−t ) + (at − bt3 ) (ce−t )]2
142
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.24
GOAL: Determine r̈ for a plane in a circular loop.
GIVEN: Path of plane, position of plane and observer and speed/acceleration information.
DRAW
FORMULATE EQUATIONS: Define the velocity and acceleration in terms of the e⇀l , e⇀β unit
vectors, referenced to the center of the loop C:
v⇀ = l˙e⇀ + lβ̇ e⇀
P
l
β
a⇀P = (¨l − lθ̇2 ) e⇀l + (lβ̈ + 2l˙β̇) e⇀β
(1)
as well as the e⇀r , e⇀θ unit vectors, referenced to the ground O:
v⇀P = ṙ e⇀r + rθ̇ e⇀θ
a⇀P = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
ASSUME: We’re given the fact that the path is circular and so can deduce that
l˙ = ¨l = 0
SOLVE:
(1), (2) ⇒
(r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ = (¨l − lθ̇2 ) e⇀l + (lβ̈ + 2l˙β̇) e⇀β
(3), (4) ⇒
(r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ = −lθ̇2 e⇀l + lβ̈ e⇀β
(2)
(3)
(4)
Realizing that for the configuration under consideration e⇀θ = − e⇀β and e⇀r = − e⇀l gives us
(r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ = lθ̇2 e⇀r − lβ̈ e⇀θ
We’re already given the fact that the acceleration is equal to 193.6 ft/s2 in the e⇀r direction and
thus have
r̈ − rθ̇2 = 193.6 ft/s2
r̈ = rθ̇2 + 193.6 ft/s2
Both r and θ̇ are clearly non-zero and thus r̈ 6= 193.6 ft/s2
143
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.25
GOAL: Find the velocity and acceleration of A which can only move vertically
GIVEN: r = 2 m, θ = 25◦ , ṙ = −2 m/s, r̈ = 0 m/s2
DRAW:
⇀
e⇀r
e⇀θ
⇀
ı

◦
cos 295
sin 295◦
◦
− sin 295 cos 295◦
FORMULATE EQUATIONS:
v⇀p = ṙ e⇀r + rθ̇ e⇀θ
polar velocity
a⇀p = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
polar acceleration
SOLVE:
polar velocity
 ) = vA ⇀
 +0 ⇀
ı
ı +cos 295◦ ⇀
 )+rθ̇(− sin 295◦ ⇀
ı +sin 295◦ ⇀
v⇀p = ṙ(cos 295◦ ⇀
ṙ cos 295◦ − rθ̇ sin 295◦ = 0 m/s
⇀
ı:
θ̇ =
⇀
:
(−2 m/s) cos 295◦
ṙ cos 295◦
=
= 0.466 rad/s
◦
r sin 295
(2 m) sin 295◦
ṙ sin 295◦ + rθ̇ cos 295◦ = vA
vA = (−2 m/s) sin 295◦ + (2 m)(0.466 rad/s) cos 295◦ = 2.207 m/s
polar accel
⇀
ı:
 ) = aA ⇀
 +0 ⇀
ı
ı +cos 295◦ ⇀
 )+(rθ̈+2ṙθ̇)(− sin 295◦ ⇀
ı +sin 295◦ ⇀
a⇀p = (0 m/s2 −rθ̇2 )(cos 295◦ ⇀
−rθ̇2 cos 295◦ − rθ̈ sin 295◦ − 2ṙθ̇ sin 295◦ = 0 m/s2
θ̈ =
rθ̇2 cos 295◦ + 2ṙθ̇ sin 295◦
−r sin 295◦
144
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
θ̈ =
⇀
:
(2 m)(0.466 rad/s)2 cos 295◦ + (2)(−2 m/s)(0.466 rad/s) sin 295◦
= 1.03 rad/s2
−(2 m) sin 295◦
−rθ̇2 sin 295◦ + (rθ̈ + 2ṙθ̇) cos 295◦ = aA
aA = −(2 m)(0.466 rad/s)2 sin 295◦ +((2 m)(1.03 rad/s2 )+(2)(−2 m/s)(0.466 rad/s)) cos 295◦ = 0.480 m/s2
v⇀A = (2.207 ⇀
 ) m/s
a⇀A = (0.480 ⇀
 ) m/s2
145
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.26
GOAL: Express the acceleration equation for the end of a robotic arm in terms of ⇀
ı and ⇀
 , then
calculate the angular velocity and extensional velocity of the arm.
GIVEN: Tip acceleration and assorted kinematic data.
DRAW:
⇀
⇀
ı

cos θ sin θ
− sin θ cos θ
e⇀r
e⇀θ
SOLVE:
The acceleration of the end of the robotic arm is
a⇀ = (4 e⇀r + 5 e⇀θ ) m/s2
Converting to the coordinates ⇀
ı and ⇀
 (θ = 45◦ ) gives
√ !
√
√ !
√
2⇀
2⇀
2⇀
2⇀
2
2
⇀
a = (4 m/s )
ı +
 + (5 m/s ) −
ı +

2
2
2
2
Combining terms gives
√
a⇀ = −
2⇀
2 ı
+
√
9 2⇀

m/s2
2
SOLVE:
To determine θ̇ and ṙ, start with the equation for acceleration in polar coordinates,
a⇀ = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ
Set that equal to the acceleration of the end of the arm
r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ = (4 e⇀r + 5 e⇀θ ) m/s2
Then, equate the coefficients. So, for e⇀r
r̈ − rθ̇2 = 4 m/s2 ⇒ 10 m/s2 − (1.5 m)θ̇2 = 4 m/s2
θ̇ = ±2 rad/s
146
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
Because there are two possible values of θ̇, there are two possible solutions for ṙ. Using the
acceleration in the e⇀θ direction gives us
rθ̈ + 2ṙθ̇ = 5 m/s2 ⇒ (1.5 m)(−0.5 rad/s2 ) + 2ṙ(±2 rad/s) = 5 m/s2
ṙ = ±1.44 m/s
147
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.27
GOAL: Find ac and L̈.
GIVEN: vt , v̇t and car’s position.
DRAW:
⇀
e⇀r
e⇀θ
⇀
ı

−1 0
0 −1
FORMULATE EQUATIONS: The acceleration vector is
a⇀c = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
rθ̇ = v
⇒
θ̇ =
88
(50 mph)( 60
)
v
=
= 0.24 rad/s
r
300 ft
and
rθ̈ = 0.3(32.2 ft/s2 )
⇒
θ̈ =
0.3(32.2)
= 0.0322 rad/s2
300
also
ṙ = r̈ = 0
SOLVE: Calculate the acceleration
a⇀c = −(300 ft)(0.24 rad/s)2 e⇀r + (300 ft)(0.0322 rad/s2 ) e⇀θ
a⇀c = (−17.9 e⇀r + 9.66 e⇀θ ) ft/s2 = (17.9 ⇀
ı − 9.66 ⇀
 ) ft/s2
⇀
er
e⇀θ
⇀
ı
⇀
√1
2
− √12
√1
2
√1
2

Next we’ll find L̈. The velocity vector v⇀c is
1
1
v⇀c = L̇ e⇀L + Lβ̇ e⇀β = −v ⇀
 = −v( √ e⇀L + √ e⇀β )
2
2
148
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
73.3 ft/s
v
= −51.9 ft/s
L̇ = − √ = − √
2
2
e⇀L :
β̇ = − √
e⇀β :
73.3 ft/s
v
= −√ √
= −0.122 rad/s
2(L)
2( 2(300 ft))
The acceleration vector a⇀c is
a⇀c = (L̈ − Lβ̇ 2 ) e⇀L + (Lβ̈ + 2L̇β̇) e⇀β = (17.9 ⇀
ı − 9.66 ⇀
 ) ft/s2
⇀
9.66 17.9
L̈ − Lβ̇ 2 = − √ + √
2
2
⇀
9.66 17.9
Lβ̈ + 2L̇β̇ = − √ − √
2
2
eL :
eβ :
ft/s2
ft/s2
SOLVE: Solve for L̈
L̈ =
√
2(300 ft)(0.122 rad/s)2 −
√1 (9.66
2
− 17.9) ft/s2 = 12.2 m/s2
149
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.28
GOAL: Find velocity and acceleration functions of time that describe the motion of point A
GIVEN: r = aθ,
a = 10 ft/rad θ̇ = 10 rad/s
DRAW:
ASSUME: θ̈ = 0
FORMULATE EQUATIONS:
v⇀A = ṙ e⇀r + rθ̇ e⇀θ
polar velocity:
a⇀A = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
polar acceleration:
SOLVE:
θ=
Z
θ̇dt = 10 rad/s
Z
dt = (10 rad/s)t
r = aθ = (10 ft/rad)[(10 rad/s)t] = (100 ft/s)t
ṙ = 100 ft/s
polar velocity
r̈ = 0
v⇀A = ṙ e⇀r + rθ̇ e⇀θ = (100 ft/s) e⇀r + (100 ft/s)t(10 rad/s) e⇀θ
v⇀A = (100 ft/s) e⇀r + (1000 ft/s2 )t e⇀θ
polar accel
a⇀A = (0 − rθ̇2 ) e⇀r + (0 + 2ṙθ̇) e⇀θ
a⇀A = [−(100 ft/s)t(10 rad/s)2 ] e⇀r + (2)(100 ft/s)(10 rad/s) e⇀θ
a⇀A = −(10, 000 ft/s3 )t e⇀r + (2000 ft/s2 ) e⇀θ
150
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.29
GOAL: (case 1) Find acceleration of A for a constant 5000 rpm. (case 2) Suppose that the e⇀θ
acceleration component was equal in magnitude to the previous e⇀r acceleration component. Find
θ̈ that yields this condition and the time required for drill to spin to rest with this acceleration.
GIVEN: r = 0.003 m, θ̇ = 5000 rpm, θ̈ = 0 (for case 1)
DRAW:
FORMULATE EQUATIONS:
a⇀A = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
polar acceleration
ASSUME: Because the point A is fixed on the periphery of the drill we have ṙ = r̈ = 0.
SOLVE:
θ̇ = 5000 rpm = 524 rad/s
case 1
a⇀A = (0−rθ̇2 ) e⇀r +(0+0) e⇀θ = (−(0.003 m)(524 rad/s)2 ) e⇀r = −822 e⇀r m/s2
case 1:
case 2
a⇀A = −822 e⇀r m/s2
aθ = (rθ̈ + 2ṙθ̇) = (rθ̈ + 0) = −822 m/s2
θ̈ =
−822 m/s2
= −2.74×105 rad/s2
0.003 m
θ̇(t) = θ̈t + θ̇0 = 0
t=
case 2:
−θ̇0
θ̈
=
−524 rad/s
= 1.9×10−3 s
−2.74×105 rad/s2
θ̈ = −2.74×105 rad/s2 ,
t = 1.9×10−3 s
151
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.30
GOAL: Find velocity and acceleration of point A when θ = π2 rad.
GIVEN: rA = L(1 + sin θ), θ̇ = (2.0 rad/s2 )t, θ = 0 when t = 0.
DRAW:
FORMULATE EQUATIONS:
v⇀p = ṙ e⇀r + rθ̇ e⇀θ
polar velocity
a⇀p = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
polar acceleration
SOLVE:
θ=
(2.0 rad/s2 )t dt = (1.0 rad/s2 )t2 + θ0
θ = 0 when t = 0,
θ0 = 0,
s
s
θ
=
1.0 rad/s2
θ = (1.0 rad/s2 )t2
π
2
rad
= 1.25 s
1.0 rad/s2
d
d
θ̇ = (2.0 rad/s2 )t = 2.0 rad/s2
dt
dt
π
rA = L(1 + sin θ) = L(1 + sin ) = 2L
2
θ̈ =
r:
r̈:
Z
θ̇ = (2.0 rad/s2 )t = (2.0 rad/s2 )(1.25 s) = 2.51 rad/s
θ̈:
ṙ:
θ̇dt =
t=
t:
θ̇:
Z
ṙ =
π
d
L(1 + sin θ) = θ̇L cos θ = (2.51 rad/s)(L)(cos ) = 0
dt
2
d
r̈ = θ̇L cos θ = θ̈L cos θ − θ̇2 L sin θ
dt
π
π
r̈ = (2.0 rad/s2 )(L)(cos ) − (2.51 rad/s)2 (L)(sin ) = −6.28L s−2
2
2
v⇀A :
v⇀A = 0 + rθ̇ e⇀θ = (2L)(2.51 rad/s) e⇀θ = 5.01L s−1 e⇀θ
a⇀A :
a⇀A = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 0) e⇀θ
152
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
a⇀A = [(−6.28L s−2 )−(2L)(2.51 rad/s)2 ] e⇀r +[(2L)(2.0 rad/s2 )] e⇀θ = −18.85L s−2 e⇀r +4L s−2 e⇀θ
v⇀A = 5.01L s−1 e⇀θ ,
a⇀A = −18.85L s−2 e⇀r + 4L s−2 e⇀θ
153
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.31
GOAL: Find gecko’s speed and acceleration.
GIVEN: Thread feed rate and acceleration.
DRAW:
FORMULATE EQUATIONS:
v⇀ = ṙ e⇀r + rθ̇ e⇀θ
Velocity
(1)
a⇀ = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ
Acceleration
The gecko is contrained to stay on the wall:
(2)
r cos θ = 2.2 m
Differentiating this constraint with respect to time yields:
ṙ cos θ − rθ̇ sin θ = 0 ⇒ θ̇ =
ṙ cos θ
r sin θ
(3)
Differentiating again yields:
r̈ cos θ − ṙθ̇ sin θ − ṙθ̇ sin θ − rθ̈ sin θ − r(θ̇)2 cos θ = 0
⇒ θ̈ =
r̈ cos θ−2ṙ θ̇ sin θ−rθ̇2 cos θ
r sin θ
(4)
At the given point in time:
r=
tan θ =
p
0.52 + 2.22 m
sin θ
(2 − 1.5) m
0.5
=
=
cos θ
2.2 m
2.2
SOLVE:
2.2 ⇀
v = ṙ e r +
e
0.5 θ
⇀
(6),(3)→(1)⇒
r
⇀
kv k = (0.78 m/s) 1 +
(3),(4),(5),(6)→ (2)⇒
h
(5)
i
2.2 2
0.5
⇀
(6)
= 3.52 m/s
a⇀ = −1.2 m/s2 − (2.26 m)(1.52 rad/s)2 e⇀r + 2(0.78 m/s)(1.52 rad/s) + (2.26 m)(−13.6 rad/s2 ) e⇀θ
a⇀ = − (6.42 e⇀r + 28.25 e⇀θ ) m/s2
154
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.32
GOAL: Find Xiyalian transport’s velocity and acceleration as functions of time.
GIVEN: Radial position and tangential velocity.
DRAW:
FORMULATE EQUATIONS:
v⇀ = ṙ e⇀r + rθ̇ e⇀θ
Velocity
SOLVE:
Differentiating the equation governing radial position,
ṙ =
d
a(t0 − t) = −a
dt
(1)⇒
(2)
(3)
r̈ = 0
(4)
v⇀ · e⇀θ = b ⇒ rθ̇ = b
(5)
Differentiating again:
v⇀ = −a e⇀r + b e⇀θ
(5),(3)→(1)⇒
(4), (5),(6) →(2) ⇒
(1)
a⇀ = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ
Acceleration
Differentiating (5)⇒
d rθ̇ = ṙθ̇ + rθ̈ = 0 ⇒ rθ̈ = −ṙθ̇
dt
a⇀ = −
b2
e⇀ − b e⇀
a(t0 − t) r t0 − t θ
155
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(6)
2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.33
⇀
⇀
⇀
GOAL: Find time for B to reach π
2 rad and evaluate r B , v B aB
GIVEN: r and θ as functions of time.
DRAW
⇀
e⇀r
e⇀θ
FORMULATE EQUATIONS:
⇀
ı

cos θ sin θ
− sin θ cos θ
r⇀B/ = bt2 e⇀r
(1)
O
v⇀B = 2bt e⇀r + bt2
a⇀B = 2b e⇀r + 2bt
d e⇀r
= 2bt e⇀r + bt2 θ̇ e⇀θ
dt
(2)
e⇀
e⇀r
+ 2btθ̇ e⇀θ + bt2 θ̈ e⇀θ + bt2 θ̇ r
dt
dt
= 2b e⇀r + 4btθ̇ e⇀θ + bt2 θ̈ e⇀θ − bt2 θ̇2 e⇀r
= (2b − bt2 θ̇2 ) e⇀r + (4btθ̇ + bt2 θ̈) e⇀θ
(3)
θ = at2
(4)
θ̇ = 2at
(5)
θ̈ = 2a
(6)
v⇀B = 2bt e⇀r + 2abt3 e⇀θ
(7)
a⇀B = (2b − 4ba2 t4 ) e⇀r + 10abt2 e⇀θ
(8)
(2), (5) ⇒
(3), (5), (6) ⇒
SOLVE:
Find t such that θ = π
2:
(1) ⇒
q
π
π
= at2 ⇒ t = 2a
2
⇀
⇀
r⇀B = bt2 (cos θ i + sin θ j )
156
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(9)
CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
At t =
q
π
π
2a and θ = 2 we have
(5), (7) ⇒
Evaluating at t =
⇀
r⇀B = bπ
2a j
⇀
q
⇀
v⇀B = (2bt cos θ − 2abt3 sin θ) i + (2bt sin θ + 2abt3 cos θ) j
π
π
2a , θ = 2 yields
v⇀B = −πb
q
π ⇀
2a ı + 2b
q
π ⇀
2a j
⇀
⇀
(3), (5), (6) ⇒
a⇀B = [(2b − 4ba2 t4 ) cos θ − 10abt2 sin θ] i + [(2b − 4ba2 t4 ) sin θ + 10abt2 cos θ] j
q
π , θ = π we have
At t = 2a
2
⇀
⇀
a⇀B/ = −5πb i + b(2 − π 2 ) j
A
157
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.34
GOAL: Determine kinematic data of a satellite from ground information.
GIVEN: Position and assorted kinematic data.
DRAW:
e⇀r2
e⇀β
e⇀r1
e⇀θ
cos(θ − β) − sin(θ − β)
sin(θ − β) cos(θ − β)
FORMULATE EQUATIONS:
We’ll be using the general formulas for velocity and acceleration in a polar frame, expressed in
terms of the two sets of unit vectors:
v⇀ = ṙ e⇀r + rθ̇ e⇀θ
a⇀ = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ
SOLVE:
Using the provided values gives
a⇀ = r̈2 − (5280 ft/mi)(100 mi)(−4.88×10−2 rad/s)2 e⇀r2 + (5280 ft/mi)(100 mi)β̈ e⇀β
a⇀ = r̈2 − 1257 ft/s2 e⇀r2 + 5.28×105 ft β̈ e⇀β
The acceleration of the satellite is
so
a⇀ = −30.7 e⇀r2 ft/s2
−30.7 e⇀r2 ft/s2 = a⇀ = r̈2 − 1257 ft/s2 e⇀r2 + 5.28×105 ft β̈ e⇀β
Equating coefficients gives for e⇀r2
r̈2 − 1257 ft/s2 = −30.7 ft/s2
r̈2 = 1227 ft/s2
158
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
and for e⇀β
5.28×105 β̈ = 0
β̈ = 0
The first step to solve for the values with respect to Station A is to find the velocity of the satellite
with respect to Station B. The velocity is
v⇀B = ṙ2 e⇀r2 + r2 θ̇ e⇀β
v⇀B = 0 e⇀r2 + (5.28×105 ft)(−4.88×10−2 rad/s) e⇀β
v⇀B = −2.58×104 e⇀β ft/s
Applying a coordinate transform with θ = 45◦ , β = 90◦ gives
!
√
√
2⇀
2⇀
4
⇀
v A = (−2.58×10 ft/s) −
e +
e
2 r1
2 θ
So
v⇀A = [1.82×104 e⇀r1 − 1.82×104 e⇀θ ] ft/s
Now set this equation equal to the general equation for velocity in polar coordinates
ṙ1 e⇀r1 + r1 θ̇ e⇀θ = [1.82×104 e⇀r1 − 1.82×104 e⇀θ ] ft/s
Use the station’s geometry to solve for r1
r1 =
100 mi
sin45o
So
r1 = 7.47×105 ft
Equating coefficients in (1) gives for e⇀r1
ṙ1 = 1.82×104 ft/s
and for e⇀θ
r1 θ̇ = −1.82×104 ft/s
So
θ̇ = −2.44×10−2 rad/s
Now, to solve for the rest of the variables, write the acceleration equations
a⇀B = −30.7 e⇀r2 ft/s2
Transforming to station A’s coordinates
a⇀A = (−21.7 e⇀r1 − 21.7 e⇀θ ) ft/s2
159
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(1)
2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
Writing the general equation
a⇀A = r̈1 − r1 θ̇2 e⇀r1 + 2ṙ1 θ̇ + r1 θ̈ e⇀θ
and equating coefficients for e⇀r1 gives
−21.7 ft/s2 = r̈1 − (7.47×105 ft)(−2.44×10−2 rad/s)2
r̈1 = 423 ft/s2
Equating for e⇀θ gives
−21.7 ft/s2 = 2(1.82×104 ft/s)(−2.44×10−2 rad/s) + (7.47×105 ft)θ̈
θ̈ = 1.16×10−3 rad/s2
160
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CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.35
GOAL: Find acceleration of probe tip.
GIVEN: radial, angular, and vertical positions
DRAW:
FORMULATE EQUATIONS:
Acceleration
⇀
a⇀ = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ + ḧk
Differentiating the equation governing radial position,
(1)
d
a(1 − cos(ω1 t)) = θ̇ = aω1 sin(ω1 t)
dt
(2)
Differentiating again,
d
aω1 sin(ω1 t) = θ̈ = aω12 cos(ω1 t)
dt
Differentiating the equation governing height twice,
(3)
ḧ = bω22 cos(ω2 t)
(4)
SOLVE:
(2),(3),(4)→ (1)⇒
⇀
a⇀ = −ra2 ω12 sin2 (ω1 t) e⇀r + raω12 cos(ω1 t) e⇀θ + bω22 cos(ω2 t)k
161
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.36
GOAL: Find the velocity and acceleration of a horse on a carousel.
GIVEN: System dimensions, position relation of height to rotation angle, and rotational speed as
a function of time.
DRAW:
FORMULATE EQUATIONS:
v⇀h = ṙh e⇀r + rh θ̇ e⇀θ + ż e⇀z
(1)
a⇀h = (r̈h − rh θ̇2 ) e⇀r + (rh θ̈ + 2ṙh θ̇) e⇀θ + z̈ e⇀z
(2)
ASSUME: The horse can’t move radially and thus
ṙh = r̈h = 0
(3)
z = z0 + z1 sin (ω2 θ)
(4)
SOLVE:
ω1 −bt
e
b
(5)
ω1 ω2 −bt
e )
b
(6)
−ω1 ω2 −bt
e )
b
(7)
θ̇ = ω1 e−bt ⇒ θ = −
(4),( 5) ⇒
z = z0 + z1 sin (−
(6) ⇒
(7) ⇒
ż = z1 ω1 ω2 e−bt cos (
z̈ = −z1 bω1 ω2 e−bt cos (
−ω1 ω2 −bt
−ω1 ω2 −bt
e ) + z1 ω12 ω22 e−2bt sin (
e )
b
b
(3),(5),(7) → (1) ⇒
v⇀h = rh ω1 e−bt e⇀θ + z1 ω1 ω2 e−bt cos (
−ω ω −bt ⇀
1 2e
)ez
b
(3),(5),(8) → (2) ⇒
162
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(8)
CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
a⇀h
= −rh ω12 e−2bt e⇀r − rh bω1 e−bt e⇀θ −
−ω1 ω2 −bt
−ω1 ω2 −bt ⇀
e ) + z1 ω12 ω22 e−2bt sin (
e ) ez
z1 bω1 ω2 e−bt cos (
b
b
163
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2.3. POLAR AND CYLINDRICAL COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.3.37
GOAL: Find magnitude and direction of thief’s velocity and acceleration.
GIVEN: Magnitude of thief’s speed and dimensions of the museum floor.
DRAW:
FORMULATE EQUATIONS:
v⇀ = ṙ e⇀r + rθ̇ + ḣ e⇀z
Velocity
Acceleration
(1)
a⇀ = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ + ḧ e⇀z
The thief is constrained to remain on the floor,
(2)
12
θ ft
(3)
2π
where, h0 is the initial height (in ft) and θ is the thief’s angular position specified in radians.
Differentiating (3) twice,
12
ḣ = − θ̇ ft/s
(4)
2π
h = h0 −
Constraint
ḧ = −
SOLVE:
(4) → (1) ⇒
Speed is 12 mph ⇒
12
θ̈ ft/s2
2π
(5)
12 ⇀
θ̇ e ft/s
2π z
q
144
v⇀ · v⇀ = r2 θ̇2 + 2 θ̇2 = 12 mph
4π
v⇀ = rθ̇ e⇀θ −
θ̇ = 0.4395 rad/s
(7) ⇒
v⇀ = 17.58 e⇀θ − 0.84 e⇀z ft/s
direction of velocity = tan−1
0.84
17.58
= 2.7◦
a⇀ = −40θ̇2 e⇀r ft/s2
(8), (5) → (2) ⇒
⇀
kak
= −7.73 ft/s2 , a⇀ is in radial direction
164
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(6)
(7)
(8)
CHAPTER 2. KINEMATICS OF PARTICLES
2.3. POLAR AND CYLINDRICAL COORDINATES
2.3.38
GOAL: Find a) revolutions of wingnut in given time, b) percentage increase in speed due to moving
down.
GIVEN: vertical speed, constant angular velocity, and total acceleration of fixed point on wingnut.
DRAW:
FORMULATE EQUATIONS:
Because wingnut has constant angular velocity, it must also have constant vertical velocity due to
constant thread pitch:
0.9 in
ḣ =
(1)
Vertical velocity
1.2 s
Acceleration
SOLVE:
(2) ⇒
a⇀A = r̈ − rθ̇2 e⇀r + 2ṙθ̇ + rθ̈ e⇀θ + ḧ e⇀z
(2)
ka⇀A k = k − rθ̇2 e⇀r k = 2.193 × 103 in/s2
(3)
θ̇ = 52.357 rad/s
(4)
(3) ⇒
Z 1.2
Integrating (4) ⇒
θ̇dt = 52.357(1.2) rad
0
θ = 62.83 rad = 10 revolutions
kv⇀1 k = rθ̇
Speed at constant height ⇒
Speed at variable height ⇒
kv⇀2 k =
% increase in speed ⇒
100 ×
(5), (6) → (7) ⇒
r
q
(5)
(rθ̇)2 + ḣ2
kv⇀2 k − kv⇀1 k
kv⇀1 k
[(0.8 in)(52.357 rad/s)]2 +
0.9 in
1.2 s
2
− (0.8 in)(52.36 rad/s)
(0.8 in)(52.357 rad/s)
% increase in speed = 0.016 %
165
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(6)
(7)
2.4. PATH COORDINATES
2.4
CHAPTER 2. KINEMATICS OF PARTICLES
Path Coordinates
166
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.1
GOAL: Find the normal acceleration felt by the airplane if the pilot pulls into an upward loop.
GIVEN: The radius of the loop is 1300 ft.
DRAW:
FORMULATE EQUATIONS: We’ll use the formula for acceleration in path coordinates:
v2
a⇀P = at e⇀t + an e⇀n = v̇ e⇀t + e⇀n
rc
SOLVE:
The radius of the loop is the radius of curvature for this problem:
(300 ft/s)2
v2
=
an =
r
1300 ft
an = 69.2 ft/s2
167
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.2
GOAL: Calculate the speed of a car.
GIVEN: a⇀P = (−7.0 e⇀t + 5.0 e⇀n ) m/s2 and r = 100 m.
FORMULATE EQUATIONS: The acceleration vector written in the ( e⇀t , e⇀n ) basis is
a⇀ = v̇ e⇀t +
v2 ⇀
e
rC n
where rC is the radius of curvature.
SOLVE: Given an = 5 m/s2 and rC = 100 m, we can solve for v :
v2
= 5 m/s2 ⇒ v = 22.4 m/s
100 m
v = 22.4 m/s
168
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.3
GOAL: Determine the total acceleration ka⇀A k at A and the radius of curvature rB at B.
GIVEN: At point A, the radius of curvature is rA = 2 m, and the marble’s speed is a constant
vA = 5 m/s. At point B, the marble is traveling at vB = 3 m/s with a total and tangential
acceleration of ka⇀B k = 10 m/s2 and aB,t = 2 m/s2 , respectively.
DRAW:
FORMULATE EQUATIONS:
The marble’s acceleration in terms of path coordinates is given by
vt2 ⇀
a = at e t + e n
r
⇀
⇀
(1)
SOLVE:
Since the marble is traveling at a constant speed at point A, its total and normal accelerations are
the same, and hence
2
vA
⇀
e⇀
a
=
(1) ⇒
A
rA n
ka⇀A k =
q
a⇀A · a⇀A =
ka⇀A k =
2
vA
rA
(5 m/s)2
2m
ka⇀A k = 12.5 m/s2
To find the radius of curvature at B, we’ll need to rearrange our expression for the marble’s
acceleration at that location:
v2
(1) ⇒
a⇀B = aB,t e⇀t + B e⇀n
rB
169
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2
ka⇀B k = a⇀B · a⇀B = aB,t
rB = r
2
+
2
vB
rB
!2
2
vB
ka⇀B k2 − aB,t
2
(3 m/s)2
rB = q
(10 m/s2 )2 − (2 m/s2 )2
rB = 0.919 m
170
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.4
GOAL: Calculate the speed of a race car.
GIVEN: a⇀P = (0 e⇀t + 0 e⇀n ) m/s2 and r = 90 m.
FORMULATE EQUATIONS: The acceleration vector written in the ( e⇀t , e⇀n ) basis is
a⇀ = v̇ e⇀t +
v2 ⇀
e
rC n
where rC is the radius of curvature.
SOLVE: Given an = 0 m/s2 and rC = 90 m, we can solve for v :
v2
= 0 m/s2 ⇒ v = 0 m/s
90 m
v=0
What’s interesting here is that normally one expects a knowledge of acceleration to give no information about velocity. For instance, if a particle is moving at a constant speed of (100 ⇀
ı + 50 ⇀
 ) m/s
it will have zero acceleration. Likewise it’ll have zero acceleration when moving at a constant speed
of (−4 ⇀
ı + 8⇀
 ) m/s. Hence knowing that the acceleration is zero doesn’t tell us anything about
the velocity. For path coordinates, though, we see that a non-zero speed and positive radius of
curvature implies a non-zero normal acceleration and the reverse holds as well: a zero normal acceleration implies zero speed (unless the radius of curvature is infinity, bringing us to pure rectilinear
motion).
171
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.5
GOAL: Determine the constant vmax before the motorcycle begins to slip at s = 20 ft.
GIVEN: The maximum allowable acceleration of the motorcycle before slip is amax = 30 ft/s2 .
The path’s radius of curvature is described by rc = 25−0.002s3 ft. The tangential speed is constant.
DRAW:
FORMULATE EQUATIONS:
The acceleration of the motorcycle in terms of path coordinates is
a⇀ =
v2 ⇀
e
rc n
SOLVE:
(1) ⇒
amax =
vmax
vmax
=
√
=
q
2
vmax
rc
amax rc
(30 ft/s2 ) {25 − 0.002(20)3 ft}
vmax = 16.43 ft/s
172
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(1)
CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.6
GOAL: Find the rotational speed of a Ferris wheel.
GIVEN: Magnitude of acceleration and dimensions.
DRAW:
FORMULATE EQUATIONS:
v⇀P = rC θ̇ e⇀t
Velocity of P
a⇀P = at e⇀t +
Acceleration of P
v2 ⇀
e
rC n
SOLVE:
a⇀P =
constant rotational speed⇒
a⇀P =
(1) → (3) ⇒
2 θ̇ 2
rC
rC
(1)
v2 ⇀
e
rC n
= 0.33 ft/s2
θ̇ = 0.105 rad/s = 1 rev/min
173
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(2)
(3)
2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.7
GOAL: Determine |at | for a car in a turn.
GIVEN: Radius of ramp is 180 m, v = 30 m/s, and the total acceleration magnitude is 7.07 m/s2 .
DRAW
FORMULATE EQUATIONS:
a⇀c = v̇ e⇀t +
v2 ⇀
e
rC n
SOLVE:
|ac | = 7.07 m/s2
a⇀c = at e⇀t +
(1), (2) ⇒
(30 m/s)2 ⇀
e
180 m n
(7.07 m/s2 )2 = a2t + (5 m/s2 )2
a2t = 25.0 (m/s2 )2
at = 5 m/s2
174
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(1)
(2)
CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.8
GOAL: Determine the marble’s constant speed vA at A and its tangential acceleration aB,t at B.
GIVEN: At point A, the radius of curvature is rA = 0.7 ft, and the marble’s total acceleration is
ka⇀A k = 14 ft/s2 . At point B, the marble is traveling at vB = 2.5 ft/s with a total acceleration of
ka⇀B k = 18 ft/s2 . The radius of curvature at B is rB = 0.4 ft.
DRAW:
FORMULATE EQUATIONS:
The marble’s acceleration in terms of path coordinates is given by
vt2 ⇀
a = at e t + e n
r
⇀
⇀
(1)
SOLVE:
We’re told that the marble’s speed at A is constant, and so its total acceleration is given by the
acceleration normal to the path:
2
vA
⇀
(1) ⇒
e⇀
aA =
rA n
⇀
kaA k =
q
vA =
vA =
q
⇀
⇀
aA · aA =
q
2
vA
rA
ka⇀A k rA
(14 ft/s2 )(0.7 ft)
vA = 3.13 ft/s
We can find the marble’s tangential acceleration at B by rearranging our expression for its total
acceleration at that point:
2
vB
⇀
⇀
e⇀
a
=
a
e
+
(1) ⇒
B
B,t t
rB n
175
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2
ka⇀B k = a⇀B · a⇀B = aB,t
v
u
u
aB,t = tka⇀B k2 −
aB,t =
s
(18 ft/s2 )2
2
+
2
vB
rB
2
vB
rB
!2
(2.5 ft/s)2
−
0.4 ft
!2
2
aB,t = 8.94 ft/s2
176
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.9
GOAL: Find rmin to support a maximum normal acceleration of 3.5g
GIVEN: Path and speed.
DRAW:
FORMULATE EQUATIONS:
a⇀ = v̇ e⇀t +
v2 ⇀
e
rC n
SOLVE:
60 mph = 88 ft/s
an =
v2
(88 ft/s)2
=
= (3.5)(32.2 ft/s2 )
rC
rC
rC = 68.7 ft
rmin = rC = 68.7 ft
177
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.10
GOAL: Find acceleration of point P at t = 3 s.
GIVEN: Tangential acceleration, initial conditions, and dimensions.
DRAW:
FORMULATE EQUATIONS:
at =
dv
dt
= ct ⇒
dv = c t dt ⇒
Z t
0
a⇀P = at e⇀t +
Acceleration of P
SOLVE:
v(3 s) =
(1) and given conditions⇒
(3) → (2) ⇒
dv =
"
Z t
0
τ dτ ⇒ v(t) − v(0) =
v(t)2 ⇀
v2 ⇀
e n = ct e⇀t +
e
rC
rC n
3
⇀
#
a⇀P (3 s) = (−1.20 e⇀t + 2.24 e⇀n ) ft/s2
178
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(1)
(2)
(−0.4 ft/s3 )(3 s)2
+ 10 ft/s = 8.2 ft/s
2
(8.2 ft/s)2 ⇀
en
aP (3 s) = (−0.4 ft/s )(3 s) e t +
30 ft
⇀
c t2
2
(3)
CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.11
GOAL: Determine which car (if any) will make it around the turn without sliding off the road and
how long it takes to do so.
GIVEN: Car A is attempting to make a semicircular turn of radius rA = 90 m at a constant speed
of vA = 24 m/s. Car B intends to travel at vB = 28 m/s through a path of radius rB = 95 m. The
cars will slide off the road when their normal accelerations exceed an,max = 0.75g.
DRAW:
FORMULATE EQUATIONS:
Since the speed of each car is constant through the turn, the acceleration can be expressed in terms
of path coordinates as
a⇀ =
vt2 ⇀
e
r n
(1)
The time it takes for each car to travel through the turn with a constant speed is governed by
v=
∆s
∆t
(2)
SOLVE:
We’re told that loss of traction will occur when the normal acceleration exceeds an,max = 0.75g =
7.36 m/s2 , so let’s start by checking if car A will make it through the turn:
2
vA
aA =
e⇀
rA n
⇀
(1) ⇒
⇀
kaA k =
q
⇀
⇀
aA · aA =
2
vA
rA
179
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
ka⇀A k = aA,n =
(24 m/s)2
90 m
aA,n = 6.40 m/s2 < an,max
Thus, we find that car A will maintain traction with the road, so it will make it through the turn
in
πr
(2) ⇒
tA = A
vA
π(90 m)
24 m/s
tA =
tA = 11.8 s
aA,n < an,max
⇒
Car A will make the turn in tA = 11.8 s.
Let’s now look at car B:
⇀
aB =
(1) ⇒
⇀
kaB k =
q
2
vB
rB
⇀
e⇀n
⇀
aB · aB =
ka⇀B k = aB,n =
2
vB
rB
(28 m/s)2
95 m
aB,n = 8.25 m/s2 > an,max
Therefore, we see that car B will exceed slip conditions, and so it won’t make it through the turn:
aB,n > an,max
⇒
Car B will slide off the road.
180
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.12
GOAL: Find the acceleration magnitude of a skier, S.
GIVEN: Skier’s path and speed.
DRAW
FORMULATE EQUATIONS:
a⇀ = v̇ e⇀t +
v2 ⇀
e
rC n
SOLVE: We need to find the skier’s speed at B. The acceleration is constant and so we have
v2
= a∆s = (28 ft/s2 )(200 feet) = 5, 600 (ft/s)2
2
v = 105.8 ft/s
a⇀S = 28 ft/s2 e⇀t +
(105.8 ft/s)2 ⇀
e n = [28 e⇀t + 50.9 e⇀n ] ft/s2
220 ft
|a⇀S | = 58.1 ft/s2 = 1.80 g
181
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.13
GOAL: Find direction and magnitude of velocity of point b immediately before and after reaching
point A.
GIVEN: Constraint of constant speed before reaching point A, acceleration after reaching point
A.
DRAW:
FORMULATE EQUATIONS:
⇀
v⇀before = v⇀b = vb i
Velocity of point b before reaching point A
v⇀after = v⇀b +
Velocity of point b after reaching point A
Z t
SOLVE:
v⇀after = v⇀b +
Immediately after passing point A, t → tA in (2)⇒
⇀
adt
t
A
Z t
A
t
A
⇀
adt
= v⇀b
⇀
v⇀before = v⇀after = vb i
182
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(1)
(2)
CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.14
GOAL: Find the time at which a sports car’s tangential acceleration is 10% of its normal acceleration.
GIVEN: The track has a radius of 100 m and the sports car’s speed is given by v = v0 1 − e−at
where v0 = 60 m/s and a = 0.3 s−1 .
DRAW:
FORMULATE EQUATIONS:
vt = v0 1 − e−at
at =
(1)
dvt
= av0 e−at
dt
(2)
v02 1 − e−at
(vt )2
=
an =
r
r
2
(3)
SOLVE:
We need to solve for when the tangential acceleration is 10% of the normal acceleration, hence
10at = an . Using (1)-(3) gives us
10av0 e−at =
Letting x =
e−at
v02 1 − e−at
2
r
and using the given parameter values yields the following quadratic equation
x2 − 7x + 1 = 0
This has only one physical solution: x = 0.146. Thus we have
−1
e−(0.3 s )t = 0.146
t = 6.42 s
183
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.15
GOAL: Determine the normal and tangential acceleration of the point P when it has the same
speed in both the ⇀
ı and ⇀
 directions.
GIVEN: ẏ = bx, with b = 12 s−1 .
DRAW:
⇀
e⇀t
en
⇀
⇀
ı

cos θ sin θ
− sin θ cos θ
FORMULATE EQUATIONS: We’ll determine the acceleration in terms of ⇀
ı ,⇀
 and then transform this into the path frame.
SOLVE:
Differentiating y = ax2 with respect to time gives us
ẏ = 2axẋ
(1)
ẋ and ẏ will be equal if 2ax = 1. Hence we have
x=
1
2a
=
1 ft
1
= 12
12 ft−1
as the position at which ẋ and ẏ are equal. Furthermore, we’re given
ẏ = bx
(2)
12 s−1
b
= 1.0 ft/s
(3)
=
2a
12 ft−1
If ẋ and ẏ are equal the e⇀t unit vector is oriented up by 45◦ (since it point along the velocity
vector).
1
(4)
ẏ = bx = (12 s−1 )( ft) = 1.0 ft/s
(2) ⇒
12
and
√
p
v = (1.0 ft/s)2 + (1.0 ft/s)2 = 2 ft/s
(1), (2) ⇒
2axẋ = bx ⇒ ẋ =
(1) ⇒
ÿ = 2aẋ2 + 2axẍ
(5)
(2) ⇒
ÿ = bẋ
(6)
(5), (6) ⇒
(6) ⇒
(12 ft−1 )(1.0 ft/s)2 + ẍ = (12 s−1 )(1.0 ft/s) ⇒ ẍ = 0
ÿ = (12 s−1 )(1.0 ft/s) = 12 ft/s2
184
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(7)
(8)
CHAPTER 2. KINEMATICS OF PARTICLES
(7), (8) ⇒
⇀
ı :
⇀
 :
at
1
1
√ ⇀
ı +√ ⇀

2
2
2.4. PATH COORDINATES
1
1
ft/s2 + an − √ ⇀
ı +√ ⇀

2
2
ft/s2 = 12 ⇀
 ft/s2
a
a
√t − √n = 0
2
2
a
a
√t + √n = 12 ft/s2
2
2
√
at = 6 2 ft/s2
(9)
(10)
(11)
√
an = 6 2 ft/s2
185
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.16
π.
GOAL: Determine the speed of a moving particle when it is at x = 2λ
1 ft−1 and a = 80 ft. When at x = π , a = (8.0 ⇀
GIVEN: y = a[1 − cos(λx)] with λ = 320
ı +
2λ P
2
⇀
2.0  ) ft/s .
DRAW:
⇀
e⇀t
en
⇀
⇀
ı

cos θ sin θ
− sin θ cos θ
FORMULATE EQUATIONS: We’ll determine the acceleration in terms of ⇀
ı ,⇀
 and then transform this into the path frame using
at e⇀t + an e⇀n = ax ⇀
ı + ay ⇀

dy at x = π :
SOLVE: To find the angle θ we can calculate dx
2λ
dy
π
1 −1
= aλ sin λ
ft ) = 0.25
= (80 ft)(
dx
2λ
320
dy
= 0.25 ⇒ θ = 14.0◦
dx
(8.0 ⇀
ı + 2.0 ⇀
 ) ft/s2 = at e⇀t + an e⇀n = at (0.970 ⇀
ı + 0.243 ⇀
 ) + an (−0.243 ⇀
ı + 0.970 ⇀
)
ı :
8.0 ft/s2 = 0.970at − 0.243an
⇀
2.0 ft/s2 = 0.243at + 0.970an
⇀
 :
2
Solving these two equations gives us an = 0. Since the normal acceleration is given by an = rv we
C
can have a zero value of this acceleration component through having a speed of zero or an infinite
radius of curvature. The radius of curvature at the give point is found from
rC =
h
dy 2
1 + ( dx
)
d2 y
dx2
i3
2
π gives us
Evaluating the second derivative of y at x = 2λ
π
d2 y
2
=
aλ
cos
λ
=0
2λ
dx2
Thus we see that our radius of curvature is infinite. We must therefore conclude that we can’t
determine v because any finite v will produce a zero normal acceleration component due to the
186
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
infinite radius of curvature.
v cannot be determined
187
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.17
GOAL: Determine the maximum constant acceleration in the straightaway from B to C so that the
car just makes it through the bend from C to D without losing traction, and find the corresponding
time to go from A to D.
GIVEN: From A to B, the car travels at a constant speed of vA = 25 mph through a 90◦ arc of
radius r1 = 70 ft. The straightaway from B to C is d = 40 ft long, and the bend from C to D is a
90◦ arc with a radius of r2 = 80 ft. The car’s speed is constant from C to D. The car will slide off
the road when its normal acceleration exceeds an,max = 0.75g.
DRAW:
FORMULATE EQUATIONS:
Since the car’s speed is constant through both bends, its acceleration during the turns can be
expressed in terms of path coordinates as
a⇀ =
vt2 ⇀
e
r n
(1)
The time it takes for the car to travel through each turn with a constant speed is governed by
v=
∆s
∆t
(2)
When the car is accelerating at a constant rate in the straightaway, we can say that
v 2 − v02 = 2a∆s
a=
∆v
∆t
(3)
(4)
SOLVE:
The car will lose traction when its normal acceleration exceeds an,max = 0.75g = 24.2 ft/s2 , so let’s
first verify that it will make it through the first turn from A to B:
188
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
a⇀AB =
(1) ⇒
ka⇀AB k =
ka⇀AB k = aAB,n =
q
h
2
vA
e⇀
r1 n
a⇀AB · a⇀AB =
(25 mph)
2
vA
r1
hr
3600 s
70 ft
5280 ft
mi
i2
2
aAB,n = 19.2 ft/s < an,max
We find that the car indeed makes it through the first bend, and thus the time it takes to go from
A to B is
πr1
tAB =
(2) ⇒
2vA
tAB =
π(70 ft)
2(25 mph)
hr
3600 s
tAB = 3.0 s
5280 ft
mi
We can find the maximum constant speed at which the car can travel through the bend from C to
D by letting ka⇀CD k = aCD,n = an,max :
an,max =
(1) ⇒
vC =
vC =
q
√
2
vC
r2
an,max r2
0.75(32.2 ft/s2 )(80 ft)
vC = 44.0 ft/s = 30.0 mph
Therefore, the time it takes to go from C to D is given by
πr2
(2) ⇒
tCD =
2vC
tCD =
π(80 ft)
2(44.0 ft/s)
tCD = 2.86 s
Now that we have the car’s speed at C (and we already know that the speeds at A and B are the
same), we can determine the constant acceleration in the straightaway:
(3) ⇒
aBC =
2 − v2
vC
B
2d
189
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2.4. PATH COORDINATES
aBC =
CHAPTER 2. KINEMATICS OF PARTICLES
h
(44.0 ft/s)2 − (25 mph)
hr
3600 s
2(40 ft)
5280 ft
mi
i2
aBC = 7.34 ft/s2
The corresponding time of travel is
(4) ⇒
tBC =
tBC =
vC − vB
aBC
44.0 ft/s − (25 mph)
hr
3600 s
2
7.34 ft/s
5280 ft
mi
tBC = 0.99 s
Lastly, the total time it takes to go from A to D is
ttotal = tAB + tBC + tCD
ttotal = 3.0 s + 2.86 s + 0.99 s
ttotal = 6.85 s
190
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.18
⇀
GOAL: Determine kak
at the given instant.
GIVEN: The snowball has a tangential speed vt = 10 m/s and tangential acceleration at = 2 m/s2
at x = 4 m. The hill profile is described by y = 0.25x2 m.
DRAW:
FORMULATE EQUATIONS:
The acceleration of the snowball in terms of path coordinates is
a⇀ = at e⇀t +
vt2 ⇀
e
rc n
(1)
The radius of curvature is
1+
rc =
dy 2
dx
d2 y
dx2
3
2
(2)
SOLVE:
Differentiate y(x) ⇒
Differentiate
dy
dx
⇒
dy
= 0.5x
dx
(3)
d2 y
= 0.5
dx2
(4)
3
(3), (4) → (2) ⇒
[1 + (0.5x)2 ] 2
m
rc (x) =
0.5
3
[1 + (0.5 · 4)2 ] 2
rc (4 m) =
m
0.5
rc (4 m) = 22.36 m
191
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(5)
2.4. PATH COORDINATES
(1) ⇒
(5) → (6) ⇒
CHAPTER 2. KINEMATICS OF PARTICLES
⇀
kak
=
⇀
kak
=
s
q
v
u
u
a⇀ · a⇀ = ta2t +
(2 m/s2 )2 +
vt2
rc
!2
(10 m/s)2
22.36 m
2
⇀
kak
= 4.90 m/s2
192
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(6)
CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.19
⇀
GOAL: Determine kak
in g’s, θ̇, θ̈.
GIVEN: The end of the test tube in the centrifuge has tangential velocity vt = 10 ft/s and
tangential acceleration at = 10 ft/s2 at the given instant. The radius of the path is R = 0.75 ft.
DRAW:
FORMULATE EQUATIONS:
The acceleration of the end of the test tube in terms of path coordinates is
a⇀ = at e⇀t +
vt2 ⇀
e
R n
(1)
SOLVE:
(1) ⇒
⇀
kak
=
⇀
kak =
s
q
v
u
u
a⇀ · a⇀ = ta2t +
(10 ft/s2 )2
+
vt2
R
!2
(10 ft/s)2
0.75 ft
2
⇀
kak
= 133.7 ft/s2 = 4.152g
(1) ⇒
vt2
R
θ̇ =
θ̇ =
= Rθ̇2
vt
R
10 ft/s
0.75 ft
θ̇ = 13.33 rad/s
(1) ⇒
at = Rθ̈
193
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
θ̈ =
θ̈ =
at
R
10 ft/s2
0.75 ft
θ̈ = 13.33 rad/s2
194
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.20
GOAL: Determine an and vt before the snowboarder becomes airborne. Also find the maximum
height achieved and the total time in the air.
⇀
GIVEN: Before becoming airborne, the snowboarder has kak
= 20 ft/s2 and at = 10 ft/s2 . The
halfpipe has a radius R = 20 ft.
DRAW:
FORMULATE EQUATIONS:
The acceleration of the snowboarder at the edge of the halfpipe in terms of path coordinates is
a⇀ = at e⇀t + an e⇀n = at e⇀t +
vt2 ⇀
e
R n
(1)
The maximum height achieved is governed by
vt2 = 2g∆y
(2)
The time to reach the maximum height is governed by
vt = g∆t
(3)
SOLVE:
⇀
kak
=
(1) ⇒
an
=
an
=
q
q
q
a⇀ · a⇀ =
q
a2t + a2n
⇀ 2
kak
− a2t
(20 ft/s2 )2 − (10 ft/s2 )2
an = 17.32 ft/s2
(1) ⇒
an =
vt =
q
(4)
vt2
R
an R
195
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(5)
2.4. PATH COORDINATES
(4) → (5) ⇒
CHAPTER 2. KINEMATICS OF PARTICLES
vt =
q
(17.32 ft/s2 )(20 ft)
vt = 18.61 ft/s
(2) ⇒
(6) → (7) ⇒
∆y =
∆y =
vt2
2g
(6)
(7)
(18.61 ft/s)2
2(32.2 ft/s2 )
∆y = 5.38 ft
∆t =
(3) ⇒
(6) → (8) ⇒
∆t =
vt
g
18.61 ft/s
32.2 ft/s2
∆t = 0.578 s
The total time in the air is simply twice the time to the maximum height, so
ttotal
= 2∆t
ttotal
= 2(0.578 s)
ttotal = 1.156 s
196
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(8)
CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.21
GOAL: Find constant tangential acceleration needed for car’s total acceleration to be 8 m/s2 at
point B.
GIVEN: Initial conditions and dimensions.
DRAW:
FORMULATE EQUATIONS:
a dx = v dv with constant acceleration⇒
a sB − s0 =
a⇀B = a e⇀t +
Acceleration at B
SOLVE:
h
(1) and initial conditions⇒
(2) ⇒
(3), (4) ⇒
v2 ⇀
e
rC n
i2
v(xB )
v(xB ) =
q
1
v(sB )2 − v(s0 )2
2
(1)
(2)
= 2a(100 m)
(200 m)a
v
u
u
v(xB )4
⇀
= 8 m/s
kaB k = ta2 +
r2
(3)
(4)
C
a2 = (8 m/s2 )2 −
(200 m)2 a2
(150 m)2
a = 4.8 m/s2
197
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(5)
2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.22
GOAL: Time before bicyclist B slips.
GIVEN: Initial conditions, dimensions, maximum sustainable acceleration, constraint of constant
speed.
DRAW:
FORMULATE EQUATIONS:
a⇀B = at e⇀t +
Acceleration of B
v2 ⇀
e
rC n
(1)
SOLVE:
20 mph = 29.3 ft/s
a⇀B =
(1) and constant speed condition⇒
(2) ⇒
(3) ⇒
rC =
v2 ⇀
e = 28 ft/s2 e⇀n
rC n
(29.3 ft/s)2
= 30.73 ft
28 ft/s2
rC = 50 − 0.0025s2 = 30.73 ft ⇒ s = 87.8 ft
t = 87.8 ft = 3 s
29.3 ft/s
198
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(2)
(3)
CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.23
GOAL: Find |vP | and |aP | of a point P on a ferris wheel.
GIVEN: Wheel’s radius, θ̇ as a function of time and elapsed time.
DRAW
FORMULATE EQUATIONS:
v⇀P = v e⇀t
a⇀P = v̇ e⇀t +
v2 ⇀
e
rC n
(1)
(2)
SOLVE:
θ̇ = ct ⇒ θ̈ = c
(3)
v = rC θ̇ = (35 ft)θ̇
(4)
Because P is moving in a circular fashion,
(3), (4) → (1) ⇒
v⇀P = (35 ft)[(0.05 rad/s2 )(5 s)] = 8.75 ft/s
v = |vp | = 8.75 ft/s
(3), (4) ⇒
(4), (5) → (2) ⇒
v̇ = rC θ̈ = (35 ft)(0.05 rad/s2 ) = 1.75 ft/s2
a⇀P = (1.75 ft/s2 ) e⇀t +
(8.75 ft/s)2 ⇀
e n = (1.75 e⇀t + 2.19 e⇀n ) ft/s2
35 ft
|a⇀P | = 2.80 ft/s2
199
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(5)
2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.24
GOAL: Find a track’s radius of curvature.
DRAW:
FORMULATE EQUATIONS:
To solve this problem, start with the equation ads = vdv. The acceleration is constant, a = 5 ft/s2 .
So, integrating the equation
Z
Z
sf
vf
ads =
s0
gives
vdv
v0
1
a(sf − s0 ) = (vf2 − v02 )
2
SOLVE:
Using the given values gives
1
(5 ft/s2 )(600 ft) = (vf2 − (80 ft/s)2 )
2
So the value for the final velocity is
vf = 111 ft/s
We know the magnitude of the car’s total acceleration is 14.6 ft/s2 at C. The equation for the
magnitude of the total acceleration is
⇀
||a||
=
q
a2n + a2t
2
Since an = rv we can compute the radius of curvature
C
v
!
u
u (111 ft/s)2 2
2
t
+ (5 ft/s2 )2
14.6 ft/s =
rC
and
rC = 899 ft
200
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.25
GOAL: Find the radius of curvature of a meteor’s path at the given instant.
GIVEN: Speed, heading and acceleration information.
DRAW:
⇀
e⇀t
en
⇀
⇀
ı

◦
− cos 3 − sin 3◦
sin 3◦ − cos 3◦
FORMULATE EQUATIONS:
We’ll be using the formula for acceleration as expressed in path coordinates:
a⇀ = v̇ e⇀t +
v2 ⇀
e
rC n
SOLVE:
The first step is to convert the gravitational acceleration into path coordinates of the meteor and
add that to the deceleration due to drag, so the total acceleration of the meteor is
a⇀m
= −9.5 ⇀
 m/s2 − 5.0 e⇀t m/s2
= −9.5(− cos 3◦ e⇀n − sin 3◦ e⇀t ) m/s2 − 5 e⇀t m/s2
= 9.49 e⇀n − 4.50 e⇀t
Converting the velocity from kph to m/s
v = (25, 000 kph)
1000 m/km
= 6944 m/s
3600 s/hr
Using our expression for the normal acceleration component an gives us
(6944 m/s)2
= 9.49 m/s2
rC
rC = 5.08×103 km
201
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(1)
2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.26
GOAL: Determine the acceleration of the rocket’s engine, the total acceleration in path coordinates, and the radius of curvature of the path.
GIVEN: Rocket’s acceleration and velocity.
DRAW:
⇀
⇀
et
e⇀n
⇀
ı

cos θ sin θ
sin θ − cos θ
FORMULATE EQUATIONS:
We’ll be using the formula for acceleration as expressed in path coordinates:
a⇀ = v̇ e⇀t +
v2 ⇀
e
rC n
SOLVE: The total acceleration of the rocket is
a⇀R = a⇀E + a⇀G
where a⇀E is the acceleration due to the engine and a⇀G is the acceleration due to gravity. So
(5.657 ⇀
ı − 3.843 ⇀
 ) m/s2 = (a1 ⇀
ı + a2 ⇀
 ) m/s2 − 9.5 ⇀
 m/s2
Equating coefficients gives a1 = 5.657 m/s2 a2 = 5.657 m/s2 The acceleration of the rocket engine
is
a⇀E = (5.657 ⇀
ı + 5.657 ⇀
 ) m/s2
To convert the rocket’s acceleration to path coordinates, you need to know the angle the rocket is
flying with reference to the vectors ⇀
ı and ⇀
 . The velocity vector,
v⇀R = (5, 000 ⇀
ı + 2, 000 ⇀
 ) m/s,
can be used to determine this angle.
θ = tan−1
2, 000
5, 000
= 21.8◦
Converting the acceleration to path coordinates
a⇀R = 5.657(cosθ e⇀t + sinθ e⇀n ) − 3.843(sinθ e⇀t − cosθ e⇀n )
202
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
Evaluating this equation gives
a⇀R = (3.823 e⇀t + 5.669 e⇀n )
To find the radius of curvature, use the equation
v2
r
C
= an with v =
5, 385 m/s and an = 5.669 m/s2 . Solving for rC gives
q
(50002 + 20002 ) (m/s)2 =
rC = 5115 km
203
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.27
GOAL: Determine the acceleration of a particle moving along the path xy = 36 m2 .
GIVEN: Horizontal speed of the particle.
FORMULATE EQUATIONS: We’ll use the facts that displacement differentiated with respect
to time gives us velocity and velocity differentiated with respect to time gives us acceleration.
m2 = 4 m.
SOLVE: xy = 36 m2 . If x = 9 m then y = 369 m
Differentiating xy = 36 m2 gives
y
ẋy + xẏ = 0 ⇒ ẏ = − ẋ
x
(1)
v⇀ = ẋ ⇀
ı + ẏ ⇀
 = ẋ ⇀
ı −
(1) ⇒
a = v˙
⇀
⇀
ẏ
y ẋ
y
 + ẋ − ⇀
 + 2⇀

= ẍ ı − ⇀
x
x
x
!
yẍ ẋẏ y ẋ2 ⇀
−
+ 2 
= ẍ ⇀
ı + −
x
x
x
⇀
y⇀

x
(3)
We’re given that v = 10 m/s. Using (2) gives us
v = ẋ 1 +
2 ! 12
y
x
= 10 m/s ⇒ ẋ = 9.138 m/s (using x = 9 m, y = 4 m)
From (1) we have ẏ = −4.061 m/s.
We’re also given that v is constant. Thus v̇ = 0. From (2) this gives

d 
v̇ =
ẋ 1 +
dt
ẍ 1 +
2 ! 12
y
x
+ ẋ 1 +
2 ! 21
y
x

=0
2 !− 12 y
y ẏ
x
x
y ẋ
− 2
x x
(2)
=0
Using the values for x, y, ẋ and ẏ gives ẍ = 3.06 m/s2 .
Using the values for x, y, ẋ, ẏ and ẍ in (3) gives
a⇀ = [3.06 ⇀
ı + 6.89 ⇀
 ] m/s2
204
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.28
GOAL: Find distance traveled and time elapsed when the magnitude of car’s acceleration is 0.9
m/s2 .
GIVEN: Constraint of constant tangential acceleration, initial conditions, and dimensions.
DRAW:
FORMULATE EQUATIONS:
v=
Speed
0
s=
Distance traveled
Z t
0
at dt = at t
(1)
1
vdt = at t2
2
(2)
a⇀ = at e⇀t +
Acceleration
SOLVE:
v2 ⇀
e
rC n
v
u
u
a4 t4
⇀
kak = ta2t + t 2 = 0.9 m/s2
(1) → (3) ⇒
(4) ⇒
Z t
rC
h
t4 = (0.9 m/s2 )2 − (0.75 m/s2 )2
i
(300 m)2
(0.75 m/s2 )4
t = 16.3 s
1
s = (0.75 m/s2 )(16.3 s)2
2
(6) → (2) ⇒
s = 99.5 m
205
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(3)
(4)
(5)
(6)
2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.29
GOAL: Calculate the radius of curvature for a race car.
GIVEN: Track dimensional data, car’s speed and car’s overall acceleration level.
FORMULATE EQUATIONS: The acceleration vector written in the ( e⇀t , e⇀n ) basis is
a⇀ = v̇ e⇀t +
v2 ⇀
e
rC n
where rC is the radius of curvature.
SOLVE: Given an = 0.5g and v = 200 km/hr = 55.5 m/s, we can solve for rC
rC =
v2
(55.5 m/s)2
= 629 m
=
.5g
0.5(9.81 m/s2 )
206
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
2.4.30
GOAL: Determine whether or not a car will lose traction during a turn.
GIVEN: The radius of curvature at entry and exit of the turn as well as the car’s speed.
DRAW:
FORMULATE EQUATIONS:
All we need is the path coordinate form for acceleration:
a⇀c = v̇ e⇀t +
SOLVE:
In this problem v̇ = 0 so
a⇀c =
v2 ⇀
e
ρc n
v2 ⇀
e
ρc n
Converting from mph to ft/s we have 100 mph=146.6̄ ft/s
At the start of the curve we have
v2
(146.6̄ ft/s)2
= 21.5 ft/s2
=
rc
1000 ft
At the end we have
v2
(146.6̄)
= 43 ft/s2
=
rc
500
43 ft/s2 exceeds the tire’s maximum sustainable acceleration of 35 ft/s2 and so the car slips before
exiting the turn.
207
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2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.31
GOAL: Determine vB , aBC , vD , and the time to complete one circuit.
GIVEN: The straightaways have a length of d = 50 m, and the rounds have radius r = 15 m. The
acceleration from A to B is a1 = 2 m/s2 , and the acceleration from C to D is a2 = 1 m/s2 . The
chariot has a constant velocity in the rounds.
DRAW:
FORMULATE EQUATIONS:
In the straightaways, the chariot’s displacement, velocity, acceleration, and travel time are related
by
v 2 − vi2 = 2as
(1)
vf − vi = at
(2)
f
In the rounds, the chariot’s displacement, velocity, acceleration, and travel time are related by
a⇀ =
v2 ⇀
e
r n
πr = vt
SOLVE:
From A to B in the first straightaway,
(1) ⇒
vB =
vB =
q
2a1 d
q
2(2 m/s2 )(50 m)
vB = 14.14 m/s
(2) ⇒
tAB =
vB
a1
208
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(3)
(4)
CHAPTER 2. KINEMATICS OF PARTICLES
tAB =
2.4. PATH COORDINATES
14.14 m/s
2 m/s2
tAB = 7.07 s
(5)
From B to C in the first round,
(3) ⇒
aBC =
aBC =
2
vB
r
(14.14 m/s)2
15 m
aBC = 13.33 m/s2
tBC =
(4) ⇒
tBC =
πr
vB
π(15 m)
14.14 m/s
tBC = 3.33 s
(6)
From C to D in the second straightaway,
(1) ⇒
vD =
vD =
q
q
2 + 2a d =
vC
2
q
2 + 2a d
vB
2
(14.14 m/s)2 + 2(1 m/s2 )(50 m)
vD = 17.32 m/s
(2) ⇒
tCD =
tCD =
v − vB
vD − vC
= D
a2
a2
17.32 m/s − 14.14 m/s
1 m/s2
tCD = 3.18 s
(7)
From D to A in the second round,
(4) ⇒
tDA =
tDA =
πr
vD
π(15 m)
17.32 m/s
tDA = 2.72 s
(5) + (6) + (7) + (8) ⇒
ttotal = 16.30 s
209
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(8)
2.4. PATH COORDINATES
CHAPTER 2. KINEMATICS OF PARTICLES
2.4.32
GOAL: Derive the formula for the radius of curvature.
GIVEN: Coordinate geometry.
DRAW
FORMULATE EQUATIONS:
Two positions are drawn, 1 and 2 and we have the relationship
∆s = rC ∆θ
(1)
between the two, i.e. the length of the curve traced out as θ changes slightly is equal to rC times
the angular change ∆θ. A further picture shows these two points along with their original and
altered x, y coordinates.
SOLVE:
Start with ∆θ :
dθ
∆x
(2)
∆θ =
dx
dy
, the usual definition of slope.
At 1 we have θ = arctan dx
dθ
To find dx
we need to differentiate:
tan θ =
dy
dx
d
d2 y
(tan θ) = 2
dx
dx
dθ
d2 y
d
(tan θ)
= 2
dθ
dx
dx
1 + tan2 θ
1+
dy
dx
dθ
dx
2 !
=
d2 y
dx2
dθ
d2 y
= 2
dx
dx
210
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CHAPTER 2. KINEMATICS OF PARTICLES
2.4. PATH COORDINATES
d2 y
dθ
= 2
dx
dx
∆s is the Euclidean distance between points 1 andq
2:
∆s =
dy
dx ∆x
in (4) yields
∆s =
(1), (2), (3), (5) ⇒
s
(∆x)2
+
dy
dx
dy
dx
2 !−1
(3)
(x + ∆x − x)2 + (y + ∆y − y)2
∆s =
Using ∆y =
1+
2
s
q
(∆x)2 + (∆y)2
(∆x)2 +
(∆x)2
= rC
dy
dx
2
d2 y
d2 x
(4)
(∆x)2
1+
(5)
dy
dx
2 !−1
∆x
Canceling ∆x and rearranging yields
rC =
"
1+
dy
dx
2 # 32
d2 y
d2 x
211
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5
Relative Motion and Constraints
212
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.1
GOAL: Determine v⇀A/B and a⇀A/B .
GIVEN: Jet A has velocity v⇀A = 660 ⇀
ı ft/s and acceleration a⇀t,A = 50 ⇀
ı ft/s2 , and it is at the
⇀
bottom of a loop with radius R = 100 ft. Jet B has velocity v B = −700 ⇀
ı ft/s and acceleration
a⇀B = −40 ⇀
ı ft/s2 .
DRAW:
FORMULATE EQUATIONS:
The velocity of jet A as seen by jet B is
v⇀A/B = v⇀A − v⇀B
(1)
The acceleration of jet A as seen by jet B is
⇀
aA/B
2
vA
⇀
= aA − aB = at,A ı +
 − a⇀B
R
⇀
⇀
⇀
SOLVE:
(1) ⇒
v⇀A/B = (660 ft/s) ⇀
ı − (−700 ft/s) ⇀
ı
v⇀A/B = 1360 ⇀
ı ft/s
(2) ⇒
a⇀A/B = (50 ft/s2 ) ⇀
ı +
(660 ft/s)2 ⇀
 − (−40 ft/s2 ) ⇀
ı
100 ft
a⇀A/B = 90 ⇀
ı + 4356 ⇀
 ft/s2
213
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(2)
2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.2
GOAL: Determine how far over from B the plane will be at the anticipated time of arrival (i.e.,
with no wind) and at what speed and angle the plane should fly at relative to the wind to actually
arrive at B in this time.
GIVEN: The plane initially has a velocity v⇀p = 200 ⇀
 mph and a strong wind blows at v⇀w =
⇀
80 ı mph when the plane is d = 100 mi from B.
DRAW:
FORMULATE EQUATIONS:
The anticipated time of arrival tB is based on when there is no wind, and so
tB =
d
vp
(1)
The plane, if it doesn’t course correct, will be pushed over a distance s to the right of B by the
wind according to
s = vw tB
(2)
v⇀p,a = v⇀w + v⇀p/w
(3)
The actual velocity of the plane is given by
SOLVE:
From (1), the plane’s anticipated time of arrival is in
(1) ⇒
tB =
100 mi
= 0.5 hr
200 mph
By (2), the plane will actually arrive to the right of B at a distance
(2) ⇒
s = (80 mph)(0.5 hr)
s = 40 mi
To arrive at B in the same amount of time as tB , the actual velocity of the plane including the
effects of the wind must be v⇀p,a = 200 ⇀
 mph, and so
v
tan θ = w
vp,a
⇒
−1
θ = tan
80 mph
200 mph
214
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
θ = 21.8◦
(3) ⇒
⇀
ı:
vp,a ⇀
 = vw ⇀
ı + vp/w (− sin θ ⇀
ı + cos θ ⇀
)
0 = vw − vp/w sin θ
⇒
vp/w =
80 mph
sin(21.8◦ )
vp/w = 215 mph
215
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.3
GOAL: Solve for the velocity vector of the bad guy with respect to the ground as he leaves the
car.
GIVEN: The car is moving to the right at 30 m/s and the ejector seat moves vertically at 10 m/s
with respect to the car.
DRAW:
FORMULATE EQUATIONS:
v⇀B = v⇀C + v⇀B/
C
SOLVE:
The bad guy (B) is traveling 30 m/s in the ⇀
ı direction when the ejector seat button is pushed.
Immediately after the button is pushed, he also has a 10 m/s velocity in the ⇀
 direction. The
overall velocity of the bad guy is therefore
v⇀B = (30 ⇀
ı + 10 ⇀
 ) m/s
216
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.4
GOAL: Determine the speed and acceleration of the disk’s top C.
GIVEN: The disk has a radius of R = 7 in. and spins at a constant angular speed of ω = 500 rpm
in the clockwise direction. The disk is released from rest far above the ground and allowed to fall
h = 3 ft.
DRAW:
FORMULATE EQUATIONS:
The absolute velocity and acceleration of the disk’s top C can be expressed as, respectively,
v⇀C = v⇀O + v⇀C/
(1)
a⇀C = a⇀O + a⇀C/
(2)
O
O
SOLVE:
Since the√disk is released from rest, the speed of its mass center O after falling a distance h is given
by vO = 2gh, and thus the speed of the disk’s top C is
p
v⇀C = − 2gh ⇀
 + Rω ⇀
ı
(1) ⇒
⇀
kv C k =
s
kv⇀C k =
7
ft
12
q
v⇀C · v⇀C =
rev
500
min
min
60 s
q
(Rω)2 + 2gh
2π rad
rev
2
+ 2(32.2 ft/s2 )(3 ft)
kv⇀C k = 33.6 ft/s
Noting that the disk is spinning at a constant angular speed and that its mass center is acted on
by gravity alone, we have that the acceleration of the disk’s top is

a⇀C = −g ⇀
 − Rω 2 ⇀
(2) ⇒
ka⇀C k =
ka⇀C k = 32.2 ft/s2 +
q
a⇀C · a⇀C = g + Rω 2
7
ft
12
500
rev
min
min
60 s
2π rad
rev
2
ka⇀C k = 1.63 × 103 ft/s2
217
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.5
GOAL: Solve for the velocity of the box.
GIVEN: The end of the rope is moving left at 2 m/s.
DRAW:
FORMULATE EQUATIONS:
The displacement of the worker is ∆x1 and the displacement of the box is ∆x2 . So,
2∆x2 + ∆x1 = 0
Now find the velocity relation by dividing by ∆t and taking the limit as ∆t goes to zero. This gives
you
ẋ1 = −2ẋ2
SOLVE:
1
1
ẋ2 = − ẋ1 = − (2 m/s) = −1 m/s
2
2
v⇀box = 1 ⇀
ı m/s
218
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.6
GOAL: Find the absolute velocity of B.
GIVEN: Pulley arrangement, ẋ and ẏ.
DRAW
ASSUME: An overall constraint for this problem is the fact that both A and B are affected in
the exact same manner by the central reel C’s motion.
FORMULATE EQUATIONS:
−∆y − ∆z + 2∆x = 0
SOLVE:
Differentiating gives us
−ẏ − ż + 2ẋ = 0
ż = 2ẋ − ẏ
v⇀B = (2ẋ − ẏ) ⇀

219
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.7
GOAL: Determine the velocity and acceleration of B as seen by A.
GIVEN: Two ball bearings (A and B) are constrained to move in the horizontal plane along
concentric circular paths of radii rA = 0.4 m and rB = 0.9 m. At the given instant, the velocity
of A is a constant v⇀A = −5 ⇀
ı m/s, whereas B is traveling at v⇀B = 8 ⇀
ı m/s with a tangential
acceleration of a⇀B,t = 0.7 ⇀
ı m/s2 .
DRAW:
FORMULATE EQUATIONS:
The velocity and acceleration of B as seen by A are given by, respectively,
v⇀B/ = v⇀B − v⇀A
(1)
a⇀B/ = a⇀B − a⇀A
(2)
A
A
SOLVE:
The velocity of B relative to A is
v⇀B/ = vB ⇀
ı − (−vA ) ⇀
ı
(1) ⇒
A
v⇀B/ = (8 m/s + 5 m/s) ⇀
ı
A
v⇀B/ = 13 ⇀
ı m/s
A
Since the velocity of A is constant at this instant, its acceleration has a normal component only,
and therefore the acceleration of B relative to A is
(2) ⇒
a⇀B/ = (aB,t ⇀
ı + aB,n ⇀
 ) − aA,n ⇀

A
⇀
B/
A
a
⇀
= aB,t ı +
2
vB
rB
−
2
vA
rA
!
⇀

220
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
⇀
B/
A
a
"
#
(8 m/s)2 (5 m)2 ⇀

−
= (0.7 m/s ) ı +
0.9 m
0.4 m
2
⇀
a⇀B/ = 0.7 ⇀
ı + 8.61 ⇀
 m/s2
A
221
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.8
GOAL: Determine the velocity of block B given the velocity of the free end A.
GIVEN: A is moving down at 2 m/s.
DRAW:
FORMULATE EQUATIONS:
Moving A (increasing x) will require us to add 2∆x units of rope. Moving B (increasing y) will
require us to add 2∆y units of rope. Applying conservation of rope tells us that the net change
must be zero:
2∆x + 2∆y = 0
Dividing by time ∆t and taking the limit as ∆t goes to zero gives
2ẋ + 2ẏ = 0 ⇒ ẏ = −ẋ
SOLVE:
We’re given ẋ = 2 m/s and thus ẏ = −2 m/s
v⇀B = 2 ⇀
 m/s
222
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.9
GOAL: Determine vC .
GIVEN: The motor at R reels in rope at vR = 3 m/s.
DRAW:
FORMULATE EQUATIONS:
By conservation of rope,
Rope 1 ⇒
3∆x + ∆xR = 0
(1)
Rope 2 ⇒
2∆xC − ∆x = 0
(2)
Differentiate (1) ⇒
3ẋ + vR = 0
(3)
Differentiate (2) ⇒
2vC − ẋ = 0
(4)
(4) → (3) ⇒
6vC + vR = 0
SOLVE:
1
vC = − vR
6
1
vC = − (3 m/s)
6
1
vC = − m/s
2
223
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.10
GOAL: Calculate the necessary constant acceleration for a raindrop to hit the floor of the train
car eight feet from the point A and calculate the angle made by its velocity vector with the floor.
GIVEN: The initial speed of the train is vA = 80 ft/s, the speed of the raindrop is vrd = 40 ft/s
and the height of the overhang is h = 8 ft.
DRAW
FORMULATE EQUATIONS:
Distance the point A travels in time t∗ :
d = vA t∗ +
Distance the raindrop falls in time t∗ :
aA t∗ 2
2
(1)
h = vrd t∗
(2)
Relative velocity of the raindrop:
v⇀rd = v⇀A + v⇀rd/
(3)
A
SOLVE: First we’ll determine the time at which the raindrop strikes the floor of the train car.
8 ft = (40 ft/s)t∗ ⇒ t∗ = 0.2 s
(2) ⇒
We want to ensure that in 0.2 s the point A moves forward by 8 ft:
8 ft = (80 ft/s)(0.2 s) +
(1) ⇒
aA (0.2 s)2
2
aA = −400 ft/s2
(Note that this deceleration is very nonrealistic for an actual train.)
v⇀rd = v⇀A + v⇀rd/
(3) ⇒
A
⇒ v⇀rd/ = v⇀rd − v⇀A
A
v⇀rd/ = −40 ⇀
 ft/s − 80 ⇀
ı ft/s + (400 ft/s2 )(0.2 s) ⇀
ı = −40 ⇀
 ft/s
A
The deceleration is such that at 0.2 s the horizontal speed of the raindrop with respect to the train
has gone to zero and the velocity is purely vertical.
angle with respect to the floor is 90◦
224
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.11
GOAL: Determine the approach speed kv⇀a k of the fuel nozzle with respect to the jet, and find
what speed the jet should slow down to if it wants to maintain the same kv⇀a k when the refueling
plane reduces its speed.
GIVEN: The refueling plane comes in at an angle β = 85◦ to the vertical with an initial speed
vp = 350 mph, while the jet flies at v⇀j = 360 ⇀
ı mph. The plane then reduces its speed to vp∗ =
340 mph.
DRAW:
FORMULATE EQUATIONS:
The approach velocity of the fuel nozzle as seen by the jet is
v⇀a = v⇀p − v⇀j
SOLVE:
v⇀a = vp (sin β ⇀
ı − cos β ⇀
 ) − vj ⇀
ı
(1) ⇒
kv⇀a k =
kv⇀a k =
q
q
v⇀a · v⇀a =
q
(vp sin β − vj )2 + (vp cos β)2
{(350 mph) sin(85◦ ) − 360 mph}2 + {(350 mph) cos(85◦ )}2
kv⇀a k = 32.5 mph
(2) ⇒
kv⇀a k = va =
q
(vp∗ sin β − vj∗ )2 + (vp∗ cos β)2
va2 = (vp∗ sin β − vj∗ )2 + (vp∗ cos β)2
0 = (vj∗ )2 − (2vp∗ sin β)vj∗ + (vp∗ sin β)2 + (vp∗ cos β)2 − va2
0 = (vj∗ )2 − (2vp∗ sin β)vj∗ + {(vp∗ )2 − va2 }
vj∗ =
vj∗
vj∗
(1)
2vp∗ sin β ±
q
(2vp∗ sin β)2 − 4{(vp∗ )2 − va2 }
2 p
2(340 mph) sin(85◦ ) ± {2(340 mph) sin(85◦ )}2 − 4{(340 mph)2 − (32.5 mph)2 }
=
2
= 325.3 mph, 352.2 mph
225
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(2)
2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
The slower speed corresponds to the jet and plane moving away from each other, so the jet speed
we want is
vj∗ = 352.2 mph
226
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.12
GOAL: Determine kv⇀A k and ka⇀A k at the given instant.
GIVEN: The car travels with v⇀c = 30 ⇀
ı mph and a⇀c = 5 ⇀
ı ft/s2 . The windmill has a radius of
r = 4 in. and spins at θ̇ = 200 rpm counterclockwise. At the given instant, e⇀t = − ⇀
ı and e⇀n = − ⇀

at A.
DRAW:
FORMULATE EQUATIONS:
The velocity of point A on the windmill is given by
v⇀A = v⇀c + v⇀A/c
(1)
a⇀A = a⇀c + a⇀A/c
(2)
The acceleration of A is given by
SOLVE:
(1) ⇒
v⇀A = vc ⇀
ı + rθ̇ e⇀t = (vc − rθ̇) ⇀
ı
kv⇀A k =
ft
kv A k = 30 mph − (4 in.)
12 in.
⇀
q
v⇀A · v⇀A = vc − rθ̇
mi
60 min
(200 rpm)
5280 ft
hr
2π rad
rev
kv⇀A k = 25.2 mph
(2) ⇒
a⇀A = ac ⇀
ı + rθ̇2 e⇀n = ac ⇀
ı − rθ̇2 ⇀

ka⇀A k =
q
a⇀A · a⇀A =
q
a2c + (rθ̇2 )2
v
"
u
2 # 2
u
200 · 2π
4
2 2
t
⇀
ft
rad/s
kaA k = (5 ft/s ) +
12
60
ka⇀A k = 146.3 ft/s2
227
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.13
GOAL: Find the agent’s acceleration so that he will be moving at zero velocity with respect to
the ground. Also, find the point on the ground where he will land.
GIVEN: Agent is 30 ft from the back of the bus and moving with a constant acceleration with
respect to the bus. Bus is moving at 10 mph with respect to ground.
DRAW:
FORMULATE EQUATIONS:
To move at zero velocity with respect to the ground the agent must be moving at −10 mph with
respect to the bus. You can solve for the acceleration with respect to the bus using
s̈ds = ṡ dṡ
(1)
Converting the speed from mph to ft/sec gives
10 mph = 14.6 ft/sec
SOLVE: Integrating (1) yields
Z 30
0
s̈ ds =
Z 14.6
ṡ dṡ
0
1
14.6 2 ⇒ s̈ = 3.59 ft/s2
2
When the agent leaves the bus, he is traveling with zero velocity with respect to the ground in the
⇀
ı direction and thus he’ll drop straight down from the back of the bus. To determine the location
of the bus when the agent jumps, we look at the equation for change of position under constant
acceleration
1
s = s̈t2 + ṡ0 t + s0 .
2
where ṡ0 and s0 are the speed and position of the agent with respect to the bus at t = 0. Since ṡ0
and s0 equal zero, the equation simplifies to
s̈(30) =
1
30 ft = (3.59 ft/s2 )t2
2
Solving for t gives t = 4.08 s. Using this fact and the fact that the bus is moving at a constant
speed gives us
x = ẋt
x = (14.6 ft/s)(4.08 s) = 60 ft
The position of the agent with respect to A is x = 60 ft
228
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.14
GOAL: Find the velocity and acceleration of car A relative to car B.
GIVEN: Absolute positions and velocities of cars A and B.
DRAW
FORMULATE EQUATIONS:
v⇀A/ = v⇀A − v⇀B
Velocity of car A relative to car B :
(1)
B
a⇀A/ = a⇀A − a⇀B
Acceleration of car A relative to car B :
(2)
B
v⇀A = ṙ e⇀r + rθ̇ e⇀θ
Absolute velocity of car A:
a⇀A = r̈ − rθ̇
Absolute acceleration of car A:
SOLVE:
⇀
A/
B
v
(1) ⇒
"
2
(4)
mph
(5)
e⇀r + 2ṙθ̇ + rθ̈ e⇀θ
√
3⇀
= 30  − 60 0.5 ı +

2
⇀
(3)
⇀
!#
v⇀A/ = (−30 ⇀
ı − 21.96 ⇀
 ) mph
B
Because car A is moving at a constant speed, in a circle of constant radius, at angular position
θ = 0 we have
v⇀A = rθ̇ e⇀θ = rθ̇ ⇀

(6)
(3) ⇒
2
2
(4) ⇒
a⇀A = −rθ̇ e⇀r = −rθ̇ ⇀
ı
We can get θ̇ from (6) ⇒
30 mph ⇀
 = (100 ft)θ̇ ⇀

θ̇ =
30 mph
100 ft
ft/s
1.4667
mph
(7)
= 0.44 rad/s
(8)
ı = −19.36 ft/s2 ⇀
a⇀A = −(100 ft)(0.44 rad/s)2 ⇀
ı
(8) → (7) ⇒
a⇀B = 0
Because car B is moving in a straight line:
(9)
(10)
a⇀A/ = −19.36 ft/s2 ⇀
ı
(9), (10) → (2) ⇒
B
a⇀A/ = −19.36 ft/s2 ⇀
ı
B
229
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.15
GOAL: Find how long it will take for a fisherman’s reel to fully unwind and the velocity and
acceleration of a swordfish with respect to the fisherman’s boat.
GIVEN: Relative positions of the swordfish and boat and both boat and swordfish’s velocity.
DRAW:
FORMULATE EQUATIONS:
r⇀S (t) = v⇀S (0)t + r⇀S (0)
(1)
⇀
r B (t) = v B (0)t + r B (0)
(2)
max| r⇀S (t) − r⇀B (t)| = 500 m
(3)
⇀
⇀
ASSUME:
r⇀S (0) = 50 ⇀
ı m
r B (0) = −40 ⇀
 m
⇀
v⇀S (0) = 10 ⇀
ı m/s
v B (0) = −3 ⇀
 m/s
(4)
⇀
SOLVE:
|((10 m/s)tf + 50 m) ⇀
ı + (−(3 m/s)tf − 40 m/s) ⇀
 | = 500 m
(1),(2),(4)→(3)⇒
q
((10 m/s)tf + (50 m))2 + ((3 m/s)tf + (40 m))2 = 500
(109 s−2 )t2f + (1240 s−1 )tf − 245900 = 0
tf
= −53.5 s, 42.1 s
Note that only the positive solution makes sense for this problem.
(1),(2)⇒
v⇀S/
= v⇀S − v⇀B = 10 ⇀
ı m/s − (−3 ⇀
 m/s)
v⇀S/
=
B
(10 ⇀
ı + 3⇀
 ) m/s
B
a⇀S/
=
B
d⇀
v
= 0
dt S/B
230
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(5)
CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.16
GOAL: Find the distance raindrops travel into the train, and the velocity of raindrops relative to
the train.
GIVEN: The speed of point A, vA , the speed of the raindrops, vrd , and the height of the overhang,
h.
DRAW
FORMULATE EQUATIONS:
Distance the train travels in time t∗ :
Distance the raindrops fall in time t∗ :
d = vA t∗
(1)
h = vrd t∗
(2)
Relative velocity of the raindrop:
v⇀rd = v⇀A + v⇀rd/
A
SOLVE:
A raindrop will fall by a height h in the same time as!the train moves forward by a distance d.
vA
30 mph
h=
(7 ft) = 8.4 ft
d=
(1), (2) ⇒
vrd
25 mph
d = 8.4 ft
Solving (3) for the velocity of the raindrop relative to point A gives us:
v⇀rd/ = v⇀rd − v⇀A = (−25 ⇀
 ) mph − (30 ⇀
ı ) mph
A
v⇀rd/ = (−30 ⇀
ı − 25 ⇀
 ) mph
A
231
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(3)
2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.17
GOAL: Find the distance you will travel in the ⇀
ı direction before landing on shore.
GIVEN: Absolute velocity of the river, v⇀r , velocity of the boat relative to the river, v⇀B/ , distance
r
to shore in the ⇀
 direction.
DRAW
FORMULATE EQUATIONS:
The position you will land on the shore relative to your starting position:
r⇀S/ = x ⇀
ı + 11 m ⇀

(1)
v⇀B = v⇀R + v⇀B/
(2)
O
The absolute velocity of the boat:
R
The velocity of the boat relative to the river:
 ) m/s
ı + 4 sin 45◦ ⇀
v⇀B/ = (4 cos 45◦ ⇀
(3)
R
The distance travelled in t∗ seconds:
r⇀S/ = v⇀B t∗
(4)
O
SOLVE:
(3) → (2) ⇒
(1),(5) → (4) ⇒
 m/s]
v⇀B = [(3 + 4 cos 45◦ ) ⇀
ı m/s + 4 sin 45◦ ⇀
x⇀
ı + 11 m ⇀
 = [(3 + 4 cos 45◦ ) ⇀
ı m/s + 4 sin 45◦ ) ⇀
 m/s] t∗
(5)
(6)
⇀
ı :
x = [(3 + 4 cos 45◦ ) m/s] t∗
(7)
⇀
11 m = [(4 cos 45◦ ) m/s] t∗
(8)
 :
(7), (8) ⇒
x=
4 cos 45◦ ) m/s
(3 +
(4 cos 45◦ ) m/s
(11 m) = 22.7 m
x = 22.7 m
232
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.18
GOAL: Find direction to paddle.
GIVEN: Absolute velocity of river, v⇀R , position vector of friend with respect to raft’s initial
position r⇀0 , time to reach friend, t∗ .
DRAW
ASSUME:
v⇀Ra/ = ẋ ⇀
ı + ẏ ⇀

Velocity of raft relative to the river:
R
FORMULATE EQUATIONS:
θ = tan
Direction to paddle:
−1
ẏ
ẋ
(1)
Absolute velocity of the raft:
v⇀Ra = v⇀R + v⇀Ra/ = (1 m/s + ẋ) ⇀
ı + ẏ ⇀

(2)
Distance travelled in time t∗ :
SOLVE:
(2) → (3) ⇒
r⇀0 = v⇀Ra t∗
(3)
48 m ⇀
 = [(1 m/s + ẋ) ⇀
ı + ẏ ⇀
 ] (120 s)
(4)
R
ẋ = −1 m/s
⇀
ı :
ẏ =
⇀
 :
Plugging these into (1) ⇒
θ = tan
48 m
= 0.375 m/s
120 s
−1
0.375 m/s
−1 m/s
= 159◦
233
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.19
GOAL: A rod is constrained to move within two independent, overlapping slots. Find the rod’s
velocity with respect to one of the constraining boards.
GIVEN: Speed of board C and orientation of slot.
DRAW:
FORMULATE EQUATIONS: We’ll use the relative velocity formula
v⇀B = v⇀A + v⇀B/
A
SOLVE:
v⇀Body
⇀
v Body
⇀
v B /Body
v⇀B /Body
A
C
A
= 0
(1)
⇀
= 1 ı m/s
= ẏ 
⇀
C
⇀
vB
= ṡ b 1 = ṡ
⇀
= v Body
=
v⇀B
(2)
⇀
⇀
ı
C
(3)
√
3⇀ 1⇀
ı + 
2
2
!
⇀
(4)
⇀
+ v B /BodyC = ı + ṡ
√
√
3⇀ 1⇀
ı + 
2
2
!
!
3
1
1 m/s +
ṡ + ṡ ⇀

2
2
= v⇀Body
A
+ v⇀B /Body
A
= ẏ ⇀

(5)
(6)
v⇀B must be the same in (5) and (6). Equating these two expressions gives
√ !
ṡ
3
⇀
ṡ + ⇀

= 0⇀
ı + ẏ ⇀

ı 1 m/s +
2
2
Equating Coefficients:
1 m/s +
⇀
ı :
√
3
ṡ = 0
2
ṡ
= ẏ
2
2
ṡ = − √ m/s
3
⇀
 :
(7) ⇒
234
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(7)
(8)
(9)
CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
(9) → (4) ⇒
v⇀B /BodyC = ṡ
√
3⇀
2 ı
+ 12 ⇀
 = −⇀
ı −
√1 ⇀

3
m/s
235
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.20
GOAL: Find velocity of block B.
GIVEN: Velocity of free end A.
DRAW:
ASSUME: Both ropes have a fixed length (conservation of rope).
FORMULATE EQUATIONS:
First consider the left pulley (Figure (a)):
2∆x2 + ∆x1 = 0
Increase x1 and x2 :
1
ẋ2 = − ẋ1
2
(1)
Next consider the right pulley (Figure (b)):
2∆x3 − ∆x2 = 0
Increase x2 and x3 :
1
ẋ3 = ẋ2
2
SOLVE:
ẋ3 =
(1), (2) ⇒
1
1
− ẋ1
2
2
1
= − ẋ1
4
We’re given ẋ1 = 3 m/s and so have
236
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(2)
CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
v⇀B = 43 ⇀
 m/s
237
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.21
GOAL: Find the absolute velocity of A.
GIVEN: Pulley arrangement, speed of B and rate at which reel is pulling in rope.
DRAW
ASSUME: An overall constraint for this problem is the fact that the speed with which the pulley
P drops is equal to the rate at which rope is fed from the reel.
FORMULATE EQUATIONS:
Ignoring motion of the pulley P we see by inspection that the velocities of A and B must be equal
in magnitude and opposite in sign. We also have the relative velocity relationships
v⇀A = v⇀P + v⇀A/
P
v⇀B = v⇀P + v⇀B/
P
SOLVE:
We’re given v⇀P = 14 ⇀
 in/s and v⇀B = −2 ⇀
 in/s. Using our relative velocity relationships gives us
v⇀B/ = v⇀B − v⇀P
P
v⇀B/ = −2 ⇀
 in/s − 14 ⇀
 in/s = −16 ⇀
 in/s
P
⇀
A/
P
We’ve already observed that v
= −v⇀B/ and so have
P
v⇀A/ = 16 ⇀
 in/s
P
We can now solve for v⇀A .
 in/s
v⇀A = v⇀P + v⇀A/ = 14 ⇀
 in/s + 16 ⇀
 in/s = 30 ⇀
P
238
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.22
GOAL: Find the absolute velocity of Block D.
GIVEN: Pulley arrangement and speed of the rope.
DRAW
FORMULATE EQUATIONS: Let’s use the “conservation-of-rope” principle. If we increase xA
by a small amount ∆xA , then we will need to add THREE sections of rope at pulley C, which is
3∆xC . But it is very important to note that when you increase xA , you’re actually taking rope
away, and not adding it to the overall length of the rope. Thus, our condition is
−∆ xA + 3∆ xC = 0
− vA + 3 vC = 0
Differentiating gives us
SOLVE:
0.4 m/s
1
vA = −
3
3
Because the velocity of Block D is the same as that of Block C, we get
vC =
Solving (2) for vC ⇒
v⇀D = 0.133 ⇀
 m/s
239
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(1)
(2)
2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.23
GOAL: Find the absolute velocity of A.
GIVEN: Pulley arrangement and speed of the rope.
DRAW
FORMULATE EQUATIONS: In addition to the constraint that there is a fixed amount of
rope, there is a constraint that the distance between pulley 1 and pulley 3 is fixed. This means
that if you change the position of one, the position of the other will change by the same amount.
Using the notation in the figure, this can be expressed as
∆ y 3 = ∆ y1
(1)
Using the “conservation-of-rope” principle, the other constraint is
∆ yB − 2∆ y3 + 2∆ y2 + 2∆ y1 = 0
SOLVE:
(1) → (2) ⇒
∆ yB + 2∆ y2 = 0
1
4 ft/s
v2 = − vB = −
= −2 ft/s
2
2
Differentiating (3) and rearranging ⇒
v⇀A = 2 ⇀
 ft/s
240
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(2)
(3)
CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.24
GOAL: Determine the velocity of a point on the rightmost rope of a pulley.
GIVEN: Pulley geometry and rate at which the reel is taking in pulley rope.
DRAW
ASSUME: To account for the reel we introduce a coordinate yR that shows the vertical position of
the leftmost pulley rope. A positive value for ẏR indicates that the rope is unreeling and a negative
value means the reel is retracting rope.
Conservation of rope gives us
−∆yR + 3∆yA = 0
(1)
1
10
ẏA = ẏR =
in/s
3
3
(2)
FORMULATE EQUATIONS:
(1) ⇒
SOLVE:
We can proceed simply using observation at this point. We’ve already seen that the pulley B is
moving up at 10
3 in/s. The rope that D’s a part of goes up over the top pulley (C) and then down
to pulley B. Hence, if B is rising at 10
3 in/s, the rope at D must be going down at the same rate.
⇀
v⇀D = − 10
3  in/s
241
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.25
GOAL: Find velocity of weight B.
GIVEN: Speed with which rope is drawn into reel (v0 ).
DRAW:
ASSUME: All three ropes have a constant length. Positive ∆yA indicates that the rope is being
unreeled.
FORMULATE EQUATIONS:
Conservation of rope gives us
−∆yA + 2∆yB = 0
(1)
−∆yB + 2∆yC = 0
(2)
−∆yC + 2∆yD = 0
(3)
SOLVE:
(1) ⇒
ẏA = 2ẏB
(4)
(2) ⇒
ẏB = 2ẏC
(5)
(3) ⇒
ẏC = 2ẏD
(6)
(4), (5), (6) ⇒
ẏA = 8ẏD
Realizing that the reel is taking in rope (ẏA = −v0 ) and that a positive ẏ implies that B is rising
(ẏD = −ẏ) we have
v

v⇀B = 80 ⇀
242
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.26
GOAL: Determine the acceleration of the mass given the acceleration of the free end of rope.
GIVEN: Free end of the rope is accelerating downward at 4 ft/s2 .
DRAW:
FORMULATE EQUATIONS:
The displacement of the free end is ∆y1 . The displacement of the mass is ∆y2 . So the displacement
relation is
3∆y2 + ∆y1 = 0
Taking the limit as ∆t goes to zero gives
1
ẏ2 = − ẏ1
3
Differentiating again to find acceleration gives
1
ÿ2 = − ÿ1
3
SOLVE:
 ft/s2 .
Letting ÿ1 = 4 ft/s2 , we get a⇀mass = 34 ⇀
243
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.27
GOAL: Determine the velocity of block B given the velocity of block A.
GIVEN: Block A is moving to the right at 10 m/s.
DRAW:
FORMULATE EQUATIONS:
Moving A (increasing x) will require us to lose 3∆x units of rope. Moving B (increasing y) will
require us to add 4∆y units of rope. Applying conservation of rope tells us that the net change
must be zero:
−3∆x + 4∆y = 0
Dividing by time ∆t and taking the limit as ∆t goes to zero gives
3
−3ẋ + 4ẏ = 0 ⇒ ẏ = ẋ
4
SOLVE:
We’re given ẋ = 10 m/s and thus
v⇀B = 43 (10 ⇀
ı m/s) = 7.5 ⇀
ı m/s
244
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.28
GOAL: Find Block B’s velocity.
GIVEN: Pulley geometry.
DRAW:
FORMULATE EQUATIONS:
2∆x + 2∆z + ∆y = 0
2ẋ + 2ż + ẏ = 0
SOLVE:
We’re given ẋ = 20 in./s and ẏ = 10 in./s. Using this information in (1) yields
ż =
1
−ẏ − 2ẋ
= − (10 in./s + 2(20 in./s)) = −25 in/s
2
2
v⇀B = 25 ⇀
 in./s
245
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(1)
2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.29
GOAL: Find v⇀A of a pulley system.
GIVEN: Pulley geometry.
DRAW:
ASSUME: The ropes connecting the pulleys are vertical.
FORMULATE EQUATIONS:
1
∆x + 4∆y = 0 ⇒ ẏ = − ẋ
4
SOLVE:
We’re given ẋ = 10 in/s. Thus we have
5
1
ẏ = − ẋ = − in/s
4
2
v⇀A = 25 ⇀
 in/s
246
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.30
GOAL: GOAL: Find the velocity of block B.
GIVEN: The velocity of the free end A.
DRAW:
ASSUME: Both ropes have a fixed length (conservation of rope).
FORMULATE EQUATIONS: We’ll examine the two ropes separately. Note that y and ∆y is
shown for both.
SOLVE:
From (a) we see that if the free end of the rope is being pulled down z decreases. Thus we have
ż = −vA .
Rope 1:
−∆z − ∆y = 0 ⇒ −ż = ẏ ⇒ ẏ = vA
Rope 2:
2
2∆y − 3∆x = 0 ⇒ ẋ = ẏ
3

v⇀B = ẋ ⇀
 = 32 vA ⇀
247
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.31
GOAL: Determine ẏ3 and ẏ4 .
GIVEN: The reels R1 and R2 are taking in rope at speeds v1 and v2 , respectively.
DRAW:
ASSUME: All three ropes have a fixed length (conservation of rope).
FORMULATE EQUATIONS: We’ll examine the three ropes separately. Note that the ∆yi
represent small changes to the position of the coordinates yi and hence small changes in the length
of rope.
SOLVE:
∆y1 − 3∆y4 + 2∆y3 = 0 ⇒ ẏ1 − 3ẏ4 + 2ẏ3 = 0
(1)
Rope 1:
Rope 2:
−∆y3 + 3∆y2 − 2∆y5 = 0 ⇒ −ẏ3 + 3ẏ2 − 2ẏ5 = 0
(2)
Rope 3:
∆y5 − ∆y4 = 0 ⇒ ẏ5 = ẏ4
(3)
(1), (2), (3) ⇒
ẏ1 − 3
−ẏ3 + 3ẏ2
2
+ 2ẏ3 = 0
ẏ3 = − 27 v1 + 79 v2
Using (2) and (3), along with our solution for ẏ3 gives us
9
1 2
v1 − v2 + 3v2
ẏ4 =
2 7
7
ẏ4 = 17 v1 + 67 v2
248
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.32
GOAL: Determine the velocity of the pulley rope’s free end.
GIVEN: Block A is observed to be dropping down at a steady 0.9 ft/s.
DRAW
FORMULATE EQUATIONS: We’ll use the “conservation of rope” principle. To lower Block
A we’ll have to lower the pulley directly above it. To allow the pulley to drop we have to “break”
the rope. If we increase x2 by a small amount (to x2 + ∆x2 ), then we’ll have two gaps (shown in
the figure at B) that will need to be filled in by additional rope. Similarly, bringing the free end of
the pulley rope down means adding rope (bringing it from x1 to x1 + ∆x1 ). Thus, our conservation
of rope condition is
∆ x1 + 2∆ x2 = 0
v1 + 2 v2 = 0
Differentiating gives us
SOLVE:
(2) ⇒
(1)
(2)
v1 = −2v2 = −2(0.9 ft/s) = −1.8 ft/s
x1 is oriented downward and thus the negative value for v1 indicates that the free end is moving
upward and has a velocity of
v⇀1 = 1.8 ⇀
 ft/s
249
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.33
GOAL: Derive an expression for vA .
GIVEN: A motor at R reels in rope at a rate of vR .
DRAW:
FORMULATE EQUATIONS:
If we define y as the diagonal length of rope between the pulley supporting block A and the pulley
above the motor at R, then conservation of rope gives us
∆xR + 2∆xA + ∆y = 0
(1)
The diagonal length of rope y is related to the block’s vertical position xA by
y=
SOLVE:
Differentiate (1) ⇒
Differentiate (2) ⇒
(4) → (3) ⇒
q
L2 + x2A
vR + 2vA + ẏ = 0
x v
ẏ = q A A
L2 + x2A
x v
=0
vR + 2vA + q A A
L2 + x2A
vA =
−vR
q
L2 + x2A
q
xA + 2 L2 + x2A
250
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(2)
(3)
(4)
CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.34
GOAL: Derive an expression for vB .
GIVEN: The rope is pulled down at A at a rate of vA .
DRAW:
FORMULATE EQUATIONS:
Let’s first define y as the diagonal length of rope between the pulley supporting block B and the
pulley above A. Likewise, we’ll let z be the diagonal length of rope between the pulley supporting
block B and the wall to its right. Thus, by conservation of rope, we have that
∆xA + ∆y + ∆z = 0
(1)
The diagonal lengths y and z are related to the block’s vertical position xB by, respectively,
q
s2 + x2B
(2)
d2 + x2B
(3)
Differentiate (1) ⇒
vA + ẏ + ż = 0
(4)
Differentiate (2) ⇒
ẏ = q
Differentiate (3) ⇒
ż = q
y=
z=
SOLVE:
q
xB vB
s2 + x2B
xB vB
d2 + x2B
251
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(5)
(6)
2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
(5), (6) → (4) ⇒

1
1

=0
vA + xB vB  q
+q
s2 + x2B
d2 + x2B
vB =

xB  q
−vA
1
s2 + x2B
1


+q
d2 + x2B
252
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.35
GOAL: Determine vA .
GIVEN: Rope is reeled in at R at the rate vR = 3 ft/s.
DRAW:
FORMULATE EQUATIONS:
By conservation of rope,
Rope 1 ⇒
∆xR + 2∆x1 = 0
(1)
Rope 2 ⇒
−∆x1 + 3∆xA = 0
(2)
Differentiate (1) ⇒
vR + 2ẋ1 = 0
(3)
Differentiate (2) ⇒
−ẋ1 + 3vA = 0
(4)
SOLVE:
vR + 6vA = 0
(4) → (3) ⇒
vA
vA
1
= − vR
6
1
= − (3 ft/s)
6
1
vA = − ft/s
2
253
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.36
GOAL: Determine vB .
GIVEN: The person pulls down at A with a speed vA = 2 ft/s.
DRAW:
FORMULATE EQUATIONS:
By conservation of rope,
Rope 1 ⇒
∆xA + 2∆x1 = 0
(1)
Rope 2 ⇒
−∆x1 + 3∆x2 = 0
(2)
Rope 3 ⇒
−∆x2 + 2∆x3 = 0
(3)
Rope 4 ⇒
−∆x3 + 2∆xB = 0
(4)
Differentiate (1) ⇒
vA + 2ẋ1 = 0
(5)
Differentiate (2) ⇒
−ẋ1 + 3ẋ2 = 0
(6)
Differentiate (3) ⇒
−ẋ2 + 2ẋ3 = 0
(7)
Differentiate (4) ⇒
−ẋ3 + 2vB = 0
(8)
SOLVE:
254
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
(6), (7), (8) → (5) ⇒
vB = −
vB = −
1
v
24 A
1
(2 ft/s)
24
vB = −
1
ft/s
12
255
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
2.5.37
GOAL: Determine the kayaker’s actual speed in the current, how far over the kayak is at C, the
angle the kayaker needs to travel at to get to B, and how much longer it takes than kayaking in
completely still water.
GIVEN: The kayaker paddles from A to B with a maximum velocity of v⇀k = 2 ⇀
 m/s. The water
has a current of v⇀w = 3 ⇀
ı m/s for d1 = 20 m into the pond. The water is still to the other side of
the pond, d2 = 50 m.
DRAW:
FORMULATE EQUATIONS:
The actual velocity of the kayaker is given by
v⇀a = v⇀k + v⇀w
(1)
The kayaker’s speed, distance traveled, and travel time are related by
d = vt
(2)
SOLVE:
The kayaker’s actual speed while in the current is
ı
v⇀a = vk ⇀
 + vw ⇀
(1) ⇒
va = kv⇀a k =
va =
q
q
v⇀A · v⇀A =
q
2
v 2 + vw
k
(2 m/s)2 + (3 m/s)2
va = 3.61 m/s
Using similar triangles to find how far the kayaker has moved over,
vk
d1
vw
s
v
s = d1 w
vk
=
256
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
s = (20 m)
3 m/s
2 m/s
s = 30 m
The actual distance traveled while in the current is
q
d21 + s2
d1,a
=
d1,a
=
d1,a
= 36.1 m
q
(20 m)2 + (30 m)2
The travel time in the current is
(2) ⇒
t1 =
t1
t1
d1,a
va
36.1 m
3.61 m/s
= 10.0 s
=
The angle the kayaker needs to paddle at to get to B is
θ = tan−1
θ = tan−1
d2
s
50 m
30 m
θ = 59.0◦
The actual distance traveled while in the still water portion is
d2,a
=
q
q
d22 + s2
(50 m)2 + (30 m)2
d2,a
=
d2,a
= 58.3 m
The travel time in the still water is
(2) ⇒
t2 =
d2,a
vk
257
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
58.3 m
2 m/s
= 29.15 s
t2
=
t2
The total travel time is
ttotal
ttotal
ttotal
= t 1 + t2
= 10.0 s + 29.15 s
= 39.15 s
When the entire pond is still water, the total travel time is
t∗ =
(2) ⇒
d1 + d2
vk
20 m + 50 m
2 m/s
= 35.0 s
t∗ =
t∗
The extra time it takes to paddle in the pond with a current is then
∆t = ttotal − t∗
∆t = 39.15 s − 35.0 s
∆t = 4.15 s
258
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.38
GOAL: Find the direction in which the student should paddle.
GIVEN: Distances to shore, velocity of the river, v⇀r , and speed of the canoe vc .
ASSUME: The absolute velocity vector of the canoe is given by
v⇀c = ẋc ⇀
ı + ẏc ⇀

(1)
The velocity vector of the canoe relative to the river is given by
ı + ẏc/ ⇀
v⇀c/ = ẋc/ ⇀

(2)
r
r
r
DRAW
FORMULATE EQUATIONS:
The direction the student should paddle, measured with respect to ⇀
ı , is
ẏc/
θ = tan−1
r
(3)
ẋc/
r
The absolute velocity of the canoe is
v⇀c = v⇀r + v⇀c/
(4)
r
Because the absolute velocity vector of the canoe should point directly at the point on the shore
where the student wishes to land, we get the relation
ẏc
10
=
ẋc
95
(5)
Lastly, the speed at which the student paddles is related to the two components of the relative
velocity by
q
vc = ẋ2c + ẏc2 = 3 m/s
(6)
/r
/r
SOLVE:

ı + ẏc/ ⇀
v⇀c = (4 + ẋc/ ) ⇀
(2) → (4) ⇒
r
r
(7)
By comparing (7) to (1), we can see that
ẋc = 4 + ẋc/
(8)
ẏc = ẏc/
(9)
r
r
259
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
ẏc/
r
(8), (9) → (5) ⇒
4 + ẋc/
r
ẋc/ =
Rearranging (6) ⇒
(11) → (10) ⇒
=
r
"
1+
10
95
2 #
2
ẏc
/r
q
2(10)(4 m/s)
−
ẏc/ +
r
95
10
95
(10)
vc2 − ẏc2
"
(11)
/r
10(4 m/s)
95
2
−
1.011ẏc2 − 0.842ẏc/ + 0.0776 = 0
r
/r
10
95
2
vc2
#
=0
(12)
Solving (12) yields two possible solutions ⇒
ẏc/ = 0.727 m/s, or 0.105 m/s
r
Plugging these into (11) ⇒
ẋc/ = 2.91 m/s, or − 3.00 m/s
r
Thus the two possible solution pairs are
ẋc/ = 2.911 m/s, ẏc/ = 0.727 m/s,
r
r
and
ẋc/ = −3.00 m/s, ẏc/ = 0.105 m/s,
r
r
Plugging these into (3) ⇒
−1
θ1 = tan
0.727 m/s
2.911 m/s
θ2 = π rad − tan−1
= 0.245 rad
0.105 m/s
3.00 m/s
= 3.11 rad
260
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.39
GOAL: Find the speed and acceleration of the pulley’s load as a function of x, ẋ and ẍ.
GIVEN: System configuration.
DRAW
⇀
e⇀r
e⇀θ
⇀
ı

sin θ − cos θ
cos θ sin θ
FORMULATE EQUATIONS:
Well use the general polar expressions for velocity and acceleration:
v⇀C = ṙ e⇀r + rθ̇ e⇀θ
a⇀C = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ
ASSUME: The motion of the point C is constrained such that v⇀C = ẋ ⇀
ı and a⇀C = ẍ ⇀
ı.
SOLVE:
Equating the two expressions for v⇀C gives us
ṙ e⇀r + rθ̇ e⇀θ = ẋ ⇀
ı = ẋ(sin θ e⇀r + cos θ e⇀θ )
e⇀r :
ṙ = ẋ sin θ
(1)
e⇀θ :
rθ̇ = ẋ cos θ
(2)
(1) ⇒
ṙ = xẋ(x2 + h2 )−0.5
(3)
(2) ⇒
θ̇ = ẋh(x2 + h2 )−1
(4)
We know from the kinematics of pulleys that the speed of the load will be one half the speed at
which rope pulls through the pulley. Hence we have
v⇀A = 21 xẋ(x2 + h2 )−0.5 ⇀

Moving on to acceleration, we have
a⇀C = (r̈ − rθ̇2 ) e⇀r + (rθ̈ + 2ṙθ̇) e⇀θ = ẍ ⇀
ı = ẍ(sin θ e⇀r + cos θ e⇀θ )
r̈ = rθ̇2 + ẍ sin θ
e⇀r :
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(5)
2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
(4), (5) ⇒
r̈ = ẋ2 h2 (x2 + h2 )−1.5 + ẍx(x2 + h2 )−0.5
We know the acceleration of the load is one half the acceleration with which rope pulls through the
pulley and thus have:
a⇀A =
1
2
2 2 2

ẋ h (x + h2 )−1.5 + ẍx(x2 + h2 )−0.5 ⇀
262
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
2.5.40
⇀
⇀
⇀
GOAL: Find time for B to reach π
2 rad and evaluate r B/A , v B/A , aB/A
GIVEN: Velocity of A and path of B
DRAW
⇀
e⇀r
e⇀θ
⇀
ı

cos θ sin θ
− sin θ cos θ
FORMULATE EQUATIONS:
r⇀B/ = bt2 e⇀r
(1)
O
v⇀B = 2bt e⇀r + bt2
a⇀B = 2b e⇀r + 2bt
d e⇀r
= 2bt e⇀r + bt2 θ̇ e⇀θ
dt
(2)
d e⇀
d e⇀r
+ 2btθ̇ e⇀θ + bt2 θ̈ e⇀θ + bt2 θ̇ θ
dt
dt
= 2b e⇀r + 4btθ̇ e⇀θ + bt2 θ̈ e⇀θ − bt2 θ̇2 e⇀r
(2), (5) ⇒
(3), (5), (6) ⇒
SOLVE:
(a) Find t such that θ = π
2:
= (2b − bt2 θ̇2 ) e⇀r + (4btθ̇ + bt2 θ̈) e⇀θ
(3)
θ = at2
(4)
θ̇ = 2at
(5)
θ̈ = 2a
(6)
v⇀B = 2bt e⇀r + 2abt3 e⇀θ
(7)
a⇀B = (2b − 4ba2 t4 ) e⇀r + 10abt2 e⇀θ
(8)
q
π
π
= at2 ⇒ t = 2a
2
263
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(9)
2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
(b)
⇀
⇀
v⇀A = vA i ⇒ r⇀A/ = vA t i and a⇀A = 0
(10)
O
r⇀B/ = r⇀B/ − r⇀A/
A
At t =
q
O
⇀
= bt2 e⇀r − vA t i
r⇀B/
(1), (10) ⇒
O
A
⇀
⇀
⇀
= bt2 (cos θ i + sin θ j ) − vA t i
⇀
⇀
= (bt2 cos θ − vA t) i + bt2 sin θ j
π
π
2a and θ = 2 we have
r⇀B/ = −vA
A
q
π ⇀ bπ ⇀
2a i + 2a j
v⇀B/ = v⇀B − v⇀A
A
⇀
B/
A
v
(5), (10) ⇒
⇀
3⇀
⇀
= 2bt e r + 2abt e θ − vA i
⇀
⇀
= (2bt cos θ − 2abt3 sin θ − vA ) i + (2bt sin θ + 2abt3 cos θ) j
Evaluating at t =
q
π
π
2a , θ = 2 yields
v⇀B/ = (−πb
A
q
⇀
π
2a − vA ) i + 2b
q
π ⇀
2a j
a⇀B/ = a⇀B − a⇀A
A
(5), (8), (10) ⇒
⇀
B/
A
a
= (2b −
4ba2 t4 ) e⇀r
+ 10abt2 e⇀θ
⇀
⇀
= [(2b − 4ba2 t4 ) cos θ − 10abt2 sin θ] i + [(2b − 4ba2 t4 ) sin θ + 10abt2 cos θ] j
At t =
q
π
π
2a , θ = 2 we have
⇀
⇀
a⇀B/ = −5πb i + b(2 − π 2 ) j
A
264
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
SYSTEM ANALYSIS: Chapter 2
265
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
SA2.1
GOAL:
Compare speed of a block being raised by a pulley for two different cases.
GIVEN: System geometry for the two configurations.
DRAW:
Question (a):
FORMULATE EQUATIONS:
(Case a): From Figure (a) we have
1
4∆y + ∆x = 0 ⇒ ∆y = − ∆x
4
SOLVE:
Block A rises at 41 ẋ.
1
1
ẏ = − ẋ = − (1.0 m/s) = −0.25 m/s
4
4
Block A is raised at 0.25 m/s
(Case b):
FORMULATE EQUATIONS:
From Figure (b) we see that the amount of rope used on the right pulley is more than in the left.
y is defined from the Block A’s initial position,
√ as shown in Figure (c).
The original length of hanging rope is 2b + 2 a2 + b2 and after the block moves up y the length of
rope is
q
2(b − y) + 2 (b − y)2 + a2
After the reel has retracted an amount x of rope we know the retracted rope and the rope currently
in the pulley is equal to the original amount
q of rope, giving us
p
x + 2(b − y) + 2 (b − y)2 + a2 = 2b + 2 a2 + b2
SOLVE:
p
x = 2y + 2
ẋ = 2ẏ − 2
a2
+
b2
q
− 2 (b − y)2 + a2
1
1
((b − y)2 + a2 )− 2 (−2bẏ + 2y ẏ)
2
1
((b − y)2 + a2 ) 2
ẏ = ẋ
2 ((b − y)2 + a2 ) 21 + b − y
266
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
To plot this we’ll need to numerically integrate from 0 to 3.5 seconds, a = 1 m, b = 5 m, using an
initial condition of y(0) = 0 The plot (obtained from a MATLAB integration of our equation for
ẏ) shows the elevation rate for the two cases (given in m/s). Note how the elevation rate is higher
from the start for the case of non-vertical supporting ropes and that the rate continues to increase
as the mass is lifted higher.
Question (b): Left as a design exercise.
267
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
SA2.2
GOAL: Analyze the behavior of an ejector system.
GIVEN: System configuration and vehicle launch kinematics.
DRAW:
Ejection phase:
FORMULATE EQUATIONS: Motion in the x and y directions are independent. Gravity
induces a negative vertical acceleration to the body and the ejection rocket provides an acceleration
directed along the long axis of the seat. The seat ejects at an angle of 20◦ and so we have x and y
components of the acceleration given by
ax = −[16 − (9 s−2 )t2 ](sin 20◦ )(32.2 ft/s2 )
ay = [(16 − (9 s−2 )t2 ) cos 20◦ − 1](32.2 ft/s2 )
SOLVE: The speed at launch is 600 KEAS= 600(1.15 mph) = 1012 ft/s.
The acceleration components can be evaluated to give
ax = −176 ft/s2 + (99.1 ft/s4 )t2
ay = 452 ft/s2 − (272 ft/s4 )t2
Integration gives us
vx
vy
x
y
= −(176 ft/s2 )t + (33.0 ft/s4 )t3 + 1012 ft/s
= (452 ft/s2 )t − (90.8 ft/s4 )t3
= −(88.1 ft/s2 )t2 + (8.26 ft/s4 )t4 + (1012 ft/s)t
= (226 ft/s2 )t2 − (22.7 ft/s4 )t4
Evaluating at t = 0.8 s yields
vx
vy
x
y
= 888 ft/s
= 315 ft/s
= 757 ft
= 135 ft
Drag phase
FORMULATE EQUATIONS:
We’re given that the acceleration follows
ax = −0.003vx2
We can determine the change in distance as a function of the initial and final speed through
268
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
evaluating
a dx = v dv
(1)
and find the elapsed time as a function of initial and final speed through
dv
=a
(2)
dt
SOLVE: The drag phase ends when the seat reaches 100 KEAS= 100(1.15) mph = 168.7 ft/s.
−(0.003 ft−1 )vx2 dsx = vx dvx
(1) ⇒
dv
−(0.003 ft−1 ) dsx
= vx
x
−(0.003 ft−1 )∆sx
= ln[vx ] − ln[vx ]
f
Using ln[vx ]
f
i
= 168.7 ft/s and ln[vx ] = 888 ft/s gives us
i
∆x = 554. ft
(2) ⇒
dvx
dt
= −(0.003 ft−1 )vx2
dvx
vx2
= −(0.003 ft−1 )dt
− v1
x
f
− v1
x
i
= −(0.003 ft−1 )∆t
Solving gives us
∆t = 1.60 s
Knowing how long the drag phase lasts and how far the seat travels due to the drag deceleration
lets us calculate the final horizontal and vertical positions of the seat.
x = 757 ft + (888 ft/s)(1.60 s) + 554 ft = 2732 ft
1
y = 135 ft + (315 ft/s)(1.60 s) − (32.2 ft/s2 )(1.60 s)2 = 598 ft
2
Calculating the range variation for c going from 0.001 ft−1 to 0.009 ft−1 yields the following plot,
which shows the x,y positions for increments of 0.001 ft−1 .
269
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
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CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS
SA2.3
GOAL:
Determine maximal acceleration levels of a carousel horse.
GIVEN: Position as a function of time for the horse.
DRAW:
FORMULATE EQUATIONS:
We’re given
z(t) = z0 + z1 cos(ωh t)
SOLVE:
Differentiating twice yields
z̈(t) = −z1 ω 2 cos(ωh t)
h
Parametric plots left as an open-ended exercise.
271
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2.5. RELATIVE MOTION AND CONSTRAINTS
CHAPTER 2. KINEMATICS OF PARTICLES
SA2.4
GOAL:
a) Construct coordinate system
b) Calculate time for ball (b) to reach the building
c) Calculate height of ball when it hits the building.
d) Calculate height of cover when ball hits the building.
e) Graphically find the permissible launch angles so the ball goes through the opening.
GIVEN:
 , v⇀B (0) = vB ⇀
 v = 40 m/s, vB =
r⇀b (0) = L3 ⇀
 , v⇀b (0) = v(cos θ ⇀
ı + sin θ ⇀
 ), r⇀B (0) = L4 ⇀
ı + L1 ⇀
6.0 m/s, L1 = 14 m, L2 = 25 m, L3 = 2 m, L4 = 100 m
DRAW:
FORMULATE EQUATIONS:
1
r b (t) =
− gt2 ⇀
 + v⇀b (0)t + r⇀b (0)
2
r⇀B (t) = v⇀B (0)t + r⇀B (0)
⇀
SOLVE:
(1)⇒
1
− gt2f
2
:
(2)
⇀
 + v⇀b (0)tf + r⇀b (0) = L4 ⇀
ı + h⇀

L4
tf = v cos
θ
⇀
(1)
(3)
h = − g2 t2f + v tf sin θ + L3
L4 2 + v sin θ
L4
= − g2 v cos
θ
v cos θ + L3
h = L4 tan θ −
gL24
+ L3
2v 2 cos2 θ
272
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(4)
CHAPTER 2. KINEMATICS OF PARTICLES
2.5. RELATIVE MOTION AND CONSTRAINTS

vB ⇀
 tf + L4 ⇀
ı + L1 ⇀
 = L4 ⇀
ı + hB ⇀
⇀
(3)→(2)⇒
:⇒
vB tf + L1 = hB
hB =
vB
L4
v cos θ
+ L1
Plotting the height of the cover, the height of the ball and the maximum height of the opening
produces the following graph, where (a) indicates the height of the ball, (b) indicates the height of
the moving cover and (c) indicates the height of the top edge of the opening. From the plot it can
be seen that the ball is above the cover at a launch angle of slightly less than 0.7 rad and moves
beyond the top of the opening at around 0.77 rad.
273
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