Phys
201
Fall
2009
Thursday,
September
17,
2009
&
Tuesday,
September
19,
2009
Chapter
3:
Mo?on
in
Two
and
Three
Dimensions
Displacement, Velocity and
Acceleration
Displacement describes the location of
a particle
Velocity is rate of change of
displacement
Acceleration is rate of change of
velocity
In more than one dimension, displacement,
velocity, and acceleration are all vectors.
3-D Kinematics
x = x(t)
y = y(t)
z = z(t)
dx
vx =
dt
dy
vy =
dt
dz
vz =
dt
dvx d 2 x
ax =
= 2
dt
dt
 
r = r (t )
d2y
ay =
= 2
dt
dt
dvy

 dr
v=
dt
dvz d 2 z
az =
= 2
dt
dt

 d r
a= 2
dt
2
3-D Kinematics
x = x(t)
dx
v=
dt
dv d 2 x
a=
= 2
dt dt
Displacement of a particle in two dimensions

r = r = x2 + y 2
θ
€
x = x cos θ
y = r sin θ
Change of displacement of a moving particle
average velocity is
€

 Δr
v =
Δt
Magnitude of velocity vector:
2
x
v = v +v
2
y
Direction of velocity vector described by θ:
€
vy
θ = tan
vx
−1
y
vy
θ
vx
€
x
Average velocity over time interval Δt:


Δr
v =v=
Δt
Instantaneous velocity:
€
€



lim Δ r d r
v (t) =
=
Δt → 0 Δt
dt
Relative velocity
•  Velocity is defined relative to a frame
of reference.
vx = 0
vx ≠ 0
Example 3-2: Flying plane in wind

Wind blows east (along x) with velocity v AG = 90 km/h
€
Pilot of plane that flies 200 km/h wishes to fly due north.
What direction should he point?
Example 3-2: Flying plane in wind

Wind blows east (along x) with velocity v AG = 90 km/h
€
Pilot of plane that flies 200 km/h wishes to fly due north.
What direction should he point?
Velocity of plane in x-direction = vAG-vpAsinθ.
To go due north, this component of the velocity
must be zero:
sinθ = vAG/vpA = (90 km/hr)/(200 km/hr)
sinθ = 0.45  θ = 0.47 radians from N
[ or, in degrees,
0.47 radians x (360 degrees)/(2π radians)
= 27° W of N]
Question
Three swimmers can swim equally fast relative to the water. They have a race
to see who can swim across a river in the least time. Relative to the water,
Beth (C) swims perpendicular to the flow, Ann (A) swims upstream, and Carly
(C) swims downstream. Which swimmer wins the race?
A) Ann
B) Beth
C) Carly
Question
Three swimmers can swim equally fast relative to the water. They have a race
to see who can swim across a river in the least time. Relative to the water,
Beth (B) swims perpendicular to the flow, Ann (A) swims upstream, and Carly
(C) swims downstream. Which swimmer wins the race?
A) Ann
B) Beth
C) Carly
correct
Time to get across = width of river/perpendicular component of
velocity.
Beth has the largest perpendicular component of velocity.
Question (seagull)
A seagull flies through the air with a velocity of 10 m/s in the absence of
wind. Assume it can only make the same effort while flying in wind. It
makes a daily round-trip to an island one km from shore. Compare the
time is takes for the seagull to fly on a calm day to the time it takes when
the wind is blowing constantly towards the shore at 5 m/s.
a. The round-trip time is the same with and without the wind
b. The round-trip time is always longer with the wind.
c. It is not possible to calculate this.
Question (seagull)
A seagull flies through the air with a velocity of 10 m/s in the absence of
wind. Assume it can only make the same effort while flying in wind. It
makes a daily round-trip to an island one km from shore. Compare the
time is takes for the seagull to fly on a calm day to the time it takes when
the wind is blowing constantly towards the shore at 5 m/s.
a. The round-trip time is the same with and without the wind
b. The round-trip time is always longer with the wind.
c. It is not possible to calculate this.
Total round trip time in the absence of wind is 2×(1000 m)/(10 m/s) = 200 s.
In the presence of wind, the seagull’s speed going towards shore is 15 m/s and
away from shore is 5 m/s. The time to go out to the island is (1000 m)/(5 m/s) =
200 s, and the time to return is (1000 m)/(15 m/s) = 67 s, so the total time in the
presence of wind is 267 s.
Acceleration vectors
Average acceleration over time interval Δt:


Δv
a =a=
Δt
Instantaneous acceleration:
€
€



lim Δv dv
a(t) =
=
Δt → 0 Δt
dt
Example 3-4 Acceleration for uniform
circular motion.
Initial velocity has magnitude v and points due
east.
Final velocity has same magnitude v and points
due north.
Velocity has changed  particle is accelerating!

Δv
average acceleration =
Δt
2-D Kinematics
Often, 3-D problems can be reduced to 2-D problems:
Choose y axis to be along direction of acceleration.
Choose x axis to be along the “other” direction of motion.
Example: Throwing a baseball (neglecting air resistance)
•  Acceleration is constant (gravity)
•  Choose y axis up: ay = -g
•  Choose x axis along the ground in
the direction of the throw
y
x
3-2 Projectile motion
For projectile motion:
Horizontal acceleration is zero (horizontal velocity is
constant)
Vertical acceleration is -g (magnitude g, directed
downward)
The horizontal and vertical motions are uncoupled,
except that the object stops moving both horizontally
and vertically at the instant it hits the ground (or
some other object).
Without air resistance, an object dropped
from a plane flying at constant speed in a
straight line will
A. Quickly lag behind the plane.
B. Remain vertically under the plane.
C. Move ahead of the plane.
Without air resistance, an object dropped
from a plane flying at constant speed in a
straight line will
A. Quickly lag behind the plane.
B. Remain vertically under the plane.
C. Move ahead of the plane.
There is no acceleration in the horizontal
direction – object continues to travel with
the same horizontal velocity (same as the
plane). Due to gravitational acceleration,
the object accelerates downward, so its
speed downwards increases.
Vertical and horizontal motions are
independent
The vertical positions of
these two balls are the
same at each time.
For projectile motion,
vertical motion and
horizontal motion are
independent.
Fig 3-12
Horizontal range of a projectile
The horizontal range is the product of the horizontal speed (x component of the
velocity) and the total time that the projectile is in the air.
If object starts at height y=0, then T, the time in the air, is determined by finding
when it reaches height y=0 again:
The two solutions of this equation are T=0 (as expected) and T=2v0y/g=2v0sinθ/g.
The horizontal range is then v0xT = (v0cosθ)(2v0sinθ/g) = v02sin(2θ)/g.
Problem: projectile motion in 2D
Two footballs are thrown from the same point on a flat field. Both are thrown at
an angle of 30° above the horizontal. Ball 2 has twice the initial speed of ball 1.
If ball 1 is caught a distance D1 from the thrower, how far away from the thrower
D2 will the receiver of ball 2 be when he catches it?
(a) D2 = 2D1
(b) D2 = 4D1
(c) D2 = 8D1
Problem: projectile motion in 2D
Two footballs are thrown from the same point on a flat field. Both are thrown at
an angle of 30° above the horizontal. Ball 2 has twice the initial speed of ball 1.
If ball 1 is caught a distance D1 from the thrower, how far away from the thrower
D2 will the receiver of ball 2 be when he catches it?
(a) D2 = 2D1
(b) D2 = 4D1
(c) D2 = 8D1
Initial speed doubled  time in the air and horizontal velocity both
double. So horizontal distance traveled goes up by a factor of 4.
range = v02sin(2θ)/g
The range of a projectile depends on initial angle. Starting at
ground level (y=0), the range is maximized for θ=45°.
range = v02sin(2θ)/g
fig 3-17
If a projectile lands at an elevation lower than the initial
elevation, the maximum horizontal displacement is achieved
when the projection angle is somewhat less than 45°.
fig 3-18
A battleship simultaneously fires two shells at enemy ships
from identical cannons. If the shells follow the parabolic
trajectories shown, which ship gets hit first?
1. Ship A
2. Ship B
3. Both at the same time
A
B
A battleship simultaneously fires two shells at enemy ships
from identical cannons. If the shells follow the parabolic
trajectories shown, which ship gets hit first?
1. Ship A
2. Ship B
3. Both at the same time
A
The higher the shell flies, the
longer the flight takes.
B
Example 3-10: ranger and monkey
Ranger aims at monkey, and the
monkey lets go of branch at the
same time that the ranger shoots.
Assume the dart comes out fast
enough to reach the monkey while it
is in the air.
Does the ranger hit the monkey?
fig 3-20
(a) Yes
(b) No
Example 3-10: ranger and monkey (demo)
Ranger aims at monkey, and the
monkey lets go of branch at the
same time that the ranger shoots.
Assume the dart comes out fast
enough to reach the monkey while it
is in the air.
Does the ranger hit the monkey?
fig 3-20
(a) Yes
(b) No
There is one time T at which the horizontal position of the dart is the same as
that of the monkey. If you find the vertical positions of the dart and the monkey
at that time, they are also the same. So the dart hits the monkey.
Special case 2: Circular motion
An object undergoes circular
motion when it is always a
constant distance from a fixed
point.
Ex. 3-11, A swinging pendulum; fig 3-22
Along a circular path, the
velocity is always
changing direction, so
circular motion involves
constant acceleration
(whether or not the
speed is changing).
Acceleration along a circular path
Centripetal acceleration: acceleration that is perpendicular to the velocity
(directed towards center of circle).
Tangential acceleration: acceleration directed parallel to the velocity – results in
a change of the speed of the particle.
fig. 3-23
Uniform circular motion
Uniform circular motion is circular motion at constant
speed (no tangential acceleration). There is still
centripetal acceleration!
What is uniform circular motion?
y
v
R
Motion in a circle with:
•
Constant radius R
•
Constant speed |v|
•
Velocity is NOT constant (direction is changing)
•
There is acceleration!
x
How can we describe uniform circular motion?
In general, one coordinate system is as good
as any other:
y
v
•  Cartesian:
•  (x, y) [position]
•  (vx, vy) [velocity]
θ
R
•  Polar:
•  (R, θ) [position]
•  (vR, ω) [radial velocity, angular velocity]
In uniform circular motion:
R is constant (hence vR=0)
ω (angular velocity) is constant
⇒ Polar coordinates are a natural way to describe uniform circular motion!
x
Polar coordinates
The arc length s (distance along the circumference) is related to the angle
via:
y
s = Rθ, where θ is the angular displacement.
v
The units of θ are radians.
(x,y)
For one complete revolution:
2πR = Rθcomplete
θcomplete= 2π
1 revolution = 2π radians
X = R cos θ
y = R sin θ
θ
R
x
Polar coordinates
In Cartesian coordinates, we say velocity dx/dt=vx.
y
v
⇒  x = vxt
In polar coordinates, angular velocity dθ/dt = ω.
⇒ θ = ωt.
⇒ ω has units of radians/second.
Distance traveled by particle s = vt.
Since s = Rθ = Rωt, we have
v = ωR
R
θ=ωt
s
x
Find acceleration during uniform circular motion

Δv

Δr
€
€

Δv v
 =
Δr r


Δr v Δr
=
Δt r Δt
v2
a=
r
acceleration is directed
towards center of circle
€
Tangential acceleration
If the speed along a circular path is changing, the tangential acceleration is
at =
dv
dt
The tangential acceleration is the time-derivative of the speed.
€
Prob. 3-7
The velocity of a particle is directed towards the east
while the acceleration is directed toward the
northwest, as shown. The particle is:
(a)  speeding up and turning toward the north
(b)  speeding up and turning toward the
south
(c)  slowing down and turning toward the
north
(e)  maintaining constant speed and turning
toward the south
Prob. 3-7
The velocity of a particle is directed towards the east
while the acceleration is directed toward the
northwest, as shown. The particle is:
(a)  speeding up and turning toward the north
(b)  speeding up and turning toward the
south
(c)  slowing down and turning toward the
north
(e)  maintaining constant speed and turning
toward the south
Prob. 3-26
Initial and final velocities of a particle are
as shown. What is the direction of the
average acceleration?
a. mostly up
b. mostly down
Prob. 3-26
Initial and final velocities of a particle are
as shown. What is the direction of the
average acceleration?
a. mostly up
b. mostly down
Problem 3-75 – hitting the monkey
What is the minimum initial
speed of the dart if it is to hit
the monkey before the
monkey hits the ground?
Monkey is d=11.2 m above
the ground; x=50 m, h=10 m.
Note that tanθ=h/x.
Problem 3-75 – hitting the monkey
What is the minimum initial
speed of the dart if it is to hit
the monkey before the
monkey hits the ground?
Monkey is d=11.2 m above
the ground; x=50 m, h=10 m.
Note that tanθ=h/x.
Amount of time it takes for the
monkey to hit the ground is
Δt = 2d/ g
= 2 × 11.2 / 9.8 = 1.5 s
Because dart must move a distance x horizontally in this amount of time, need
x/(v0 cos θ)≤Δt, or


x
x  h2 + x 2 
(10m)2 + (50m)2
v0 ≥
= 
=
= 34m / s

Δt cos θ Δt 
x
(1.5s)

Problem 3-84.
A projectile is fired into the air from
the top of a 200-m cliff above a
valley. Its initial velocity is 60 m/s at
60° above the horizontal. Where
does the projectile land? (Ignore air
resistance.)
Problem 3-84.
A projectile is fired into the air from
the top of a 200-m cliff above a
valley. Its initial velocity is 60 m/s at
60° above the horizontal. Where
does the projectile land? (Ignore air
resistance.)
Find time when projectile hits ground:
Projectile elevation y(t) = h0+(v0sinθ)t-½gt2. Find the time when y(t)=0:
1
− gt 2 + (v 0 sinθ)t + h0 = 0
2
−v 0 sinθ ± v 02 sin2 θ + 2gh0
⇒t=
−g
Positive root is
v 02 sin2 θ + 2gh0 +€v 0 sinθ
t* =
g
v 02 sin2 θ + 2gh0 + v 0 sinθ
Horizontal position at time t* is v0t*cosθ = v 0 cos θ
= 300m
g
€
€
Problem 3-97
A projectile is launched over level ground
at an initial elevation angle of θ. An
observer measures the height of the
projectile at the point of its highest
elevation and measures the angle ϕ
shown in the figure. Show that
tan ϕ = ½ tan θ. (Ignore air resistance.)
Problem 3-97
A projectile is launched over level ground
at an initial elevation angle of θ. An
observer measures the height of the
projectile at the point of its highest
elevation and measures the angle ϕ
shown in the figure. Show that
tan ϕ = ½ tan θ. (Ignore air resistance.)
The initial velocity has a vertical component of vsinθ and a horizontal
component of vcosθ (we don’t know v, but it will drop out of the final answer).
At the point of maximum elevation, the vertical velocity is zero, so the average
vertical velocity over that interval is ½vsinθ; the horizontal velocity is constant
in time, so the average horizontal velocity over the interval is vcosθ. The
angle ϕ thus satisfies
1 v sinθ
2
1
tan φ =
= tanθ.
v cos θ 2
A projectile is fired at an angle of 45º above
the horizontal. If air resistance is neglected,
the line in the graph that best represents the
horizontal displacement of the projectile as a
function of travel time is
A.  1
B.  2
C.  3
D.  4
E.  None of these is correct.
A projectile is fired at an angle of 45º above
the horizontal. If air resistance is neglected,
the line in the graph that best represents the
horizontal displacement of the projectile as a
function of travel time is
A.  1
B.  2
C.  3
D.  4
E.  None of these is correct.
A ball is thrown horizontally from a cliff with
a velocity v0. A graph of the acceleration of
the ball versus the distance fallen could be
represented by curve
A.  1
B.  2
C.  3
D.  4
E.  5
A ball is thrown horizontally from a cliff with
a velocity v0. A graph of the acceleration of
the ball versus the distance fallen could be
represented by curve
A.  1
B.  2
C.  3
D.  4
E.  5
A golfer drives her ball from the tee down
the fairway in a high arcing shot. When the
ball is at the highest point of its flight,
A.  its velocity and acceleration are both zero.
B.  its velocity is zero but its acceleration is
nonzero.
C.  its velocity is nonzero but its acceleration is
zero.
D.  its velocity and acceleration are both nonzero.
E.  Insufficient information is given to answer
correctly.
A golfer drives her ball from the tee down
the fairway in a high arcing shot. When the
ball is at the highest point of its flight,
A.  its velocity and acceleration are both zero.
B.  its velocity is zero but its acceleration is
nonzero.
C.  its velocity is nonzero but its acceleration is
zero.
D.  its velocity and acceleration are both
nonzero.
E.  Insufficient information is given to answer
correctly.
The figure shows the motion diagram for a Human
Cannonball on the descending portion of the flight. Use
the motion diagram to estimate the direction of the
acceleration during the interval between points 1 and 3.
A.
B.
C.
D.
E.
up
down
left
right
diagonal
The figure shows the motion diagram for a Human
Cannonball on the descending portion of the flight. Use
the motion diagram to estimate the direction of the
acceleration during the interval between points 1 and 3.
A.
B.
C.
D.
E.
up
down
left
right
diagonal