AP Physics 5: Circular and Rotational Motion
A.
Circular Motion
1. constant perimeter (tangential) speed: vt = 2r/T
a. distance = circumference of the circle: 2r
b. time = time for one revolution: T (period)
2. constant inward (centripetal) acceleration: ac = v2/r
3. centripetal force, Fc = mac = mv2/r
a. turning on a road problems
v = 2r/T
ac


when the road is horizontal: Fc = Ff = smg
roads are banked in order to reduce the amount of
friction (component of the Fg is || to Fc)
b. horizontal loop problem (mass on a string)
Ft-x = Fc = mv2/r

Ft
Ft-y = Fg = mg
v = 2r/T





4.
Ft = (Fc2 + Fg2)½
tan = Fg/Fc ( is measured from horizontal)
c. vertical loop problem (mass on a string)
top: Fnet = Fc = Ft + Fg  Ft = Fc – Fg
Fg
Ft
Fg
Ft
bottom: Fnet = Fc = Ft – Fg  Ft = Fc + Fg
if on a roller coaster: Fn = Ft
Newton's law of universal gravity, Fg = GMm/r2
a. G = 6.67 x 10-11 N•m2/kg2
b. M = mplanet and m = msatellite
c. r is the distance, measured from center to center
d. g = GM/r2
e. Fg = Fc: GMm/r2 = mv2/r  v = (GM/r)½
v = 2r/T
m
Fg = Fg
M
r
Earth
Moon
Sun
Mass (kg)
5.98 x 1024
7.35 x 1022
1.99 x 1030
Radius (m)
6.38 x 106
1.74 x 106
6.96 x 108
r from Earth (m)
3.84 x 108
1.50 x 1011
Name __________________________
B.
Newton's Laws—Rotation
1. torque,  = rFr (tau—Greek letter for t)
rotation
90o Fr
r
a.
2.
rcm
r = perpendicular distance from axis of rotation to
rotating force Fr
b. when r is not perpendicular to Fr, then  = rFrsin
c. torque units are m•N (not N•m—work)
First Law: Object remains at rest or uniform rotation as
long as no net torque (net) acts on it
a. measured as the moment of inertia, I = mr2
b.  corrects for mass distribution ( = 1 for a hoop)
c. equilibrium (net = 0)
1. center of mass for a complex system
a. balance point where an upward force
equal to the total weight equals the sum of
all the downward weight torques
Fcm = (m1 + m2 + m3)g
m1
m2
m3
r1
r2
r3
F1
F2
F3
 not rotating  CM = 1 + 2 + 3
rcm(m1 + m2 + m3)g = r1m1g + r2m2g + r3m3g
rcm = (r1m1 + r2m2 + r3m3)/(m1 + m2 + m3)
rcm = (rimi)/mi
2. cantilever problems—how far from the edge
can m2 be placed without tipping?
mass of plank at its center
m1
m2
r1
r2
Fg1
not rotating 
Fg2
1 = 2
r1m1g = r2m2g
r1m1 = r2m2
3. two supports problems—what are the
tensions?
FL
mass of plank at its center
FR
rR
m1
m2
r1
r2
assume left side is point of rotation
not rotating 
R = 1 + 2
rRFR = r1m1g + r2m2g
FR = (r1m1g + r2m2g)/rR
solve for FL: FL + FR = m1g + m2g
3. Second Law: net = I (= rFr,  = mr2,  = a/r)
a. rFr = (mr2)(a/r)  Fr = ma (acceleration at rim)
b. pulley problems—what is the acceleration
m1
rough surface ()
m2
Fp – Ff = m1a + m2a

Fp – m1g = (m1 + m2)a
solve for d, v or t using kinematics
Fp
c. rolling problems—what is the acceleration

Frolling = ma + ma
m
Frolling = (1 +  )ma
Fgsin = (1 + )ma
solve for d, v or t using kinematics
Fgsin

C.
Conservation Laws—Rotation
1. rotational momentum, L = rmv (kg•m2/s)
a. when net = 0, then L = 0
b. change r and/or  will change v
orbiting planet
spinning diver
Kepler's Law
(A1  2 = A3  4)
r1v1 = r2v2
r11v1 = r22v2
2. rotational kinetic energy, Kr = ½mv2 (J)
a. v is the velocity at the rim
b. rolling kinetic energy: Krolling = ½(1 + )mv2
3. mixed linear and rotation motion problems
a. Summary of translational and rotational formulas
Variable
Translational Rotational
Rolling
F = ma
force
Fr = ma
F = (1 + )ma
p = mv
momentum
L = rmv p + L = (1 + r)mv
kinetic energy K = ½mv2 Kr = ½mv2 K = ½(1 + )mv2
b. collision problems (conservation of momentum)—
what is the velocity after the boy jumps on the
merry-go-round?
m
v
convert linear into rotational (r = rM,  = 1)
Lm + LM = L'
M, rM, M
rMmmvm + rMMMvM = (rMmm + rMMM)v'
mvm = (mm + MM)v'
v' = mvm/(mm + MM)
c. pulley problems (conservation of energy)—what is
the velocity after descending h meters?
m1
rough surface ()
Ug3 – Wf1 = K1 + K2 + K3
m3gx – km1gx = ½(m1 + m2 + m3)v'2
v' = [2(m3 – km1)gx/(m1 + m2 + m3)]½
d.
m3

m2
pulley problems (conservation of energy)—what is
the velocity after descending h meters?
Ug1 + Ug2 = Ug1' + Ug2' + K1 + K2 + K3
0 + m2gh = m1gh + 0 + ½(m1 + m2 + m3)v'2
(m2 – m1)gh = ½(m1 + m2 + m3)v'2
v = [2(m2 – m1)gh/(m1 + m2 + m3)]½
m3

m2
m1
e.
h
rolling problems (conservation of energy)—what is
the velocity after descending h meters?
 m
Ug = Krolling

mgh = ½(1 + )mv2
h
gh = ½(1 + )v2
v = [2gh/(1 + )]½
f. cart problems (conservation of energy)—what is
the velocity after descending h meters?
m1
Ug = K + Krolling

 , m2
(m1 + 4m2)gh = ½m1v2 + ½(1 + )4m2v2
(m1 + 4m2)gh = ½(m1 + (1 + )4m2)v2
h

D.
Simple Harmonic Motion (SHM)
1. oscillating mass on a spring
acceleration is NOT constant  kinematic
formulas are invalid
b. displacement, velocity and acceleration oscillate
between +A and –A, where A = amplitude
1. x = +A, when t = 0 (pictured above)
t=0
t = ¼T t = ½T t = ¾T
t=T
time
+A
-A
+A
displacement
0
0
-vmax
+vmax
velocity
0
0
0
-amax
+amax
-amax
acceleration
0
0
2. x = 0, when t = 0 (heading downward)
t=0
t = ¼T t = ½T t = ¾T
t=T
time
-A
+A
displacement
0
0
0
-vmax
+vmax
-vmax
velocity
0
0
+amax
-amax
acceleration
0
0
0
c. maximum acceleration, amax = A(k/m) = vmax2/A
Steps
Algebra
Fs = ma
start with
substitute kA for Fs
kA = ma
solve for a
amax = A(k/m)
d. maximum velocity, vmax = A(k/m)½ = 2A/T
Steps
Algebra
Us = K
start with
substitute ½kA2 for Us and ½mv2 for K ½kA2 = ½mv2
vmax = A(k/m)½
solve for v
e. velocity at x, in terms of vmax: vx = vmax[1 – (x2/A2)]½
Steps
Algebra
Kx + USx = Umax
start with
substitute for Us and K
½mvx2 + ½kx2 = ½kA2
solve for vx2
vx2 = (k/m)(A2 – x2)
2
multiply-divide by A
vx2 = A2(k/m)[(A2/A2) – (x2/A2)]
vx = A(k/m)½[(1 – (x2/A2)]½
square root both sides
substitute vmax for A(k/m)½ vx = vmax[1 – (x2/A2)]½
f. time for one cycle, period, T = 2(m/k)½
Steps
Algebra
vmax = A(k/m)½
start with
substitute 2A/T for vmax
2A/T = A(k/m)½
simplify
2/T = (k/m)½
solve for T
T = 2/(k/m)½
substitute (m/k)½ for 1/(k/m)½
T = 2(m/k)½
g. formulas at midpoint, 0, and extremes, A
midpoint
extreme
x
xmax = A
0
v
0
vmax = A(k/m)½ = 2A/T
a
amax = -A(k/m) = vmax2/A
0
F
F = -ma = -kA
0
U
Umax = ½kA2
0
K
Kmax = ½mv2
0
a.
2.
pendulum
period of the simple pendulum, T = 2(L/g)½
Steps
Algebra
F = kx
start with
substitute mgsinrad for F
mgsinrad = kx
substitute Lrad for x
mgsinrad = kLrad
mg = kL
for small angles sinrad = rad
k = mg/L
solve for k
start with
T = 2(m/k)½
substitute mg/L for k
T = 2(m/mg/L)½
simplify
T = 2(L/g)½
b. notice that m cancels out of the equation, so the
period only depends on the L and g
3. damped harmonic motion
a.
a.
4.
amplitude of oscillating spring or swinging
pendulum will decrease until it stops—damping
b. damping is due to friction and air resistance
1. forces always oppose direction of velocity
2. damping is enhanced if oscillator is placed in
viscous fluid (car shock absorbers)
c. forced damping is accomplished with motors that
are programmed to oppose velocity (earthquake
protected buildings)
resonance
a. object can be set to oscillate by an external
force—forced vibration
b.
c.
when forced vibration matches natural vibration,
then amplitude builds with each vibration—
resonance
examples
1. child swinging
2. building during an earthquake
3. air inside a musical instrument
1.
A. Circular Motion
Centripetal Force Lab
Measure the period of a whirling mass using two
techniques, and then vary the tension and radius to see
their effects on the period.
a. Collect the following data.
Control
string length, L
0.5 m
hanging weight, m1
200 g
stopper mass, m2
Questions 2-10 Briefly explain your answer.
2. When a tetherball is whirling around the pole, the net force
is directed
(A) toward the top of the pole
(B) toward the ground
(C) horizontally away from the pole
(D) horizontally toward the pole
D—net force equals Fc, which is toward the center of the
circle
3.
time (10 orbits), t
Double L
m1
string length, L
time (10 orbits), t
200 g
1.0 m
Half m1
m1
string length, L
time (10 orbits), t
100 g
0.5 m
b.
Consider the free-body diagram.
Ftx

Fty
Ft
m1g
m1g
m2g
m2g
Ftx
Fty
What is Fty equal to?
Which force equals Fc?
What is Ft equal to?
c. Calculate the following from the data.
Double L
Control
You are standing in a bus that makes a sharp left turn.
Which of the following is true?
(A) you lean to the left because of centripetal force
(B) you lean to the right because of inertia
(C) you lean forward because of the net force is forward
(D) you lean to the right because of centrifugal force
B—the turn to the left results in the right side of the bus
being in "front" of you as you continue forward due to
inertia
4. You drive your car too fast around a curve and the car starts
to skid. What is the correct description of this situation?
(A) car's engine is not strong enough to keep the car from
being pushed out
(B) friction between the tires and the road is not strong
enough to keep the car in a circle
(C) car is too heavy to make the turn
(D) none of the above
B—the centripetal force needed to make the turn is
generated by friction between the road and tires, which
is not sufficient at the high speed
5. A steel ball is whirling around in a circle on the end of a
string when the string breaks. Which path will it follow?
Half m1
T1 based on t
A
B C
B—without the tension in the string to supply the
centripetal force, the ball continues in a straight line
Ft based on m1
6.
Fty based on m2
 based on Ft and Fty
Two stones A and B have the same mass. They are tied to
strings and whirled in horizontal circles. The radius of the
circular path for stones A is twice the radius of stone B's
path. If the period of motion is the same for both stones,
what is the tension in cord A (FA) compared to cord B (FB)?
(A) FA = FB
(B) FA = 2FB
(C) FA = ½FB
B—Fc = mv2/r = m(2r/T)2/r = 42mr/T2, since 42m/T2 is
constant  Fc  r
Fc based on Ft and Fty
7.
r based on L and 
You driving along a rural road. Which is true when you are
at the lowest point along a dip in the road?
(A) Fn < Fg
(B) Fn = Fg
(C) Fn > Fg
C—at the bottom of the vertical circle:
Fnet = Fc = Fn – Fg > 0 (upward)  Fn > Fg
v based on Fc
8.
T2 based on v
% T1 and T2 (use
average as standard)
c.
How was the period affected by the following?
Doubling L
Halving m1
You swing a ball on the end of a string in a vertical circle.
Which is true of the centripetal force at the top of the circle?
(A) Fc = Ft + Fg (B) Fc = Ft – Fg (C) Fc = Fg – Ft
A—at the top of the vertical circle, both Ft and Fg are
downward, which is the direction of Fc  Fc = Ft + Fg
9.
Which is stronger the Earth's pull on the Moon or the
Moon's pull on the Earth?
(A) Earth's pull (B) Moon's pull (C) they are equal
C—Newton's third law states that for every force there is
an equal but opposite force  the pulls are equal
10. Is there a net force acting on an astronaut floating in orbit
around the Earth while on a space walk?
(A) yes
(B) no
A—The net force is Fg, which provides the Fc necessary
to keep the astronaut in orbit
11. A car is traveling clockwise on the north side of a circular
track. (r = 50 m) takes 16 s to make one lap. Determine
v v = 2r/T = 2(50 m)/16 s = 20 m/s
16. A 1-kg pendulum bob swings back and forth from a 2-m
string that can support 15 N of tension without breaking.
a. What is the maximum speed that the bob can reach at
the bottom of the swing without breaking the string?
Ft = Fg + Fc = mg + mv2/r
15 N = (1kg)(10 m/s2) +(1 kg)v2/(2 m) v = 3.16 m/s
b.
direction of v
east
ac ac = v2/r = (20 m/s)2/(50 m) = 8 m/s2
direction of ac
south
12. The earth is 1.5 x 1011 m from the sun and makes one
complete circular orbit in 1 year.
a. What is the period of orbit in seconds?
T = 365 days x 24 hr/day x 3600 s/hr = 3.2 x 107 s
b.
What is the earth’s orbital velocity?
v = 2r/T = 2(1.5 x 1011 m)/3.2 x 107 s = 3.0 x 104 m/s
c.
What is the centripetal acceleration of the earth toward
the sun?
Ug + K = Ug' + K' mgh = ½mv2 
(10 m/s2)h = ½(3.16 m/s)2 h = 0.50 m
17. How would the force of gravity be affected if the Earth
a. had the same mass but a smaller radius?
Between Earth and Moon
On the Earth's surface
the same
greater
b. had the same radius but a smaller mass?
Between Earth and Moon
On the Earth's surface
less
Less
18. Determine the acceleration due to gravity on the planet
compared to Earth.
Mass
Radius (x Earth)
Acceleration (x gEarth)
m = mEarth
r = rEarth
g
m = mEarth
r = 2rEarth
¼g
m = mEarth
r = ½rEarth
4g
m = 2mEarth
r = rEarth
2g
aC = v2/r = (3.0 x 104 m/s)2/1.5 x 1011 m = 6 x 10-3 m/s2
13. A driver of a 1000-kg sports car attempts a turn whose
radius of curvature is 50 m on a road where  = 0.8.
a. What is the fastest that the driver can make the turn?
Fc = Ff  mv2/r = mg
v2/50 m = (0.8)(10 m/s2)  v = 20 m/s
b.
Could the driver make the turn at this speed
(1) with a 2,000-kg SUV? Explain
Yes, because mass cancels out of the equation.
(2) when the road is wet? Explain
No, because the coefficient of friction is less.
14. A 2-kg mass is moving at 5 m/s in a horizontal circle of
radius 1 m at the end of a cord.
a. What is the horizontal component of tension?
Ft-x = Fc = mv2/r
Ft-x = (2 kg)(5 m/s)2/(1 m) = 50 N
b.
What is the vertical component of tension?
Ft-y = Fg = mg
Ft-y = (2 kg)(10 m/s2) = 20 N
c.
What is the overall tension in the cord?
Ft = (Ft-x2 + Ft-y2)½
Ft = (502 + 202)½ = 54 N
d.
What angle does the cord make with the horizontal?
tan = Ft-y/Ft-x = Fg/Fc
tan = 20/50   = 22o
15. A 2-kg mass is moving at 5 m/s in a vertical circle of radius
1 m at the end of a cord.
a. What is the tension in the cord at the top of the circle?
Ft = Fc – Fg = mv2/r – mg
Ft = (2 kg)(5 m/s)2/(1 m) – (2 kg)(10 m/s2) = 30 N
b.
What is the tension in the cord at the bottom?
Ft = Fc + Fg = mv2/r + mg
Ft = (2 kg)(5 m/s)2/(1 m) + (2 kg)(10 m/s2) = 70 N
What is the maximum height measured from vertical
that the bob can reach?
m = ½mEarth
r = rEarth
½g
19. What is the acceleration due to gravity (g) on Mars?
( m = 6.4 x 1023 kg, r = 3.4 x 106 m)
g = GM/r2
g = (6.67x10-11N•m2/kg2)(6.4x1023kg)/(3.4x106m)2=3.7m/s2
20. When the Apollo Missions went to the moon they passed a
point where the gravitational attractions from the moon and
the earth are equal. What is the ratio rEarth/rMoon where this
happened if mEarth/mMoon = 100?
GmMm/rM2 = GmEm/rE2
rE2/rM2 = mE/mM = 100  rM/rE = 10
21. Consider the following changes to earth.
I
Increase earth's mass
II Decrease earth's mass
III Increase earth's radius
IV Decrease earth's radius
Which changes would decrease the acceleration
II and III
due to gravity on the earth's surface?
Which changes would increase the acceleration
I and IV
due to gravity on the earth's surface?
Which changes would decrease the acceleration
II only
due to gravity on the moon?
Which changes would increase the acceleration
I only
due to gravity on the moon?
22. The Earth's mass is 81 times the mass of the moon and the
Earth's radius is 4 times the radius of the moon.
a. What is gMoon in terms of gEarth?
g = GM/r2  gMoonrMoon2/mMoon = gEarthrEarth2/mEarth
gMoon(1)2/1 = g(42)/81
gMoon = 16g/81 = 0.2 g
b. What is the mass of a 50 kg person on the Moon?
Mass doesn't change: 50 kg
c.
What is the weight of a 50 kg person on the Moon?
Fg = mg = (50 kg)(0.2) = 10 kg
B. Newton's Laws—Rotation
23. Equilibrium Lab
a. Extend from the table edge a ½-m stick with a 50-g
mass at 0 cm and measure the balance point (CM).
50 g
½-m stick
| rr  r50 
table |
|
|
50 cm
25 cm
0 cm
(1) Collect the following data.
rr
r50
ruler mass, mr
(2) Use the cantilever technique to determine the
mass of the ruler.
(3) Determine the percent difference between the
measured mr (true) and the calculated mr from (2).
b.
Extend from the table edge a ½-m stick with 50-g at
40 cm, 10-g at 15 cm and 20-g at 5 cm and measure
the balance point (rcm).
50 g
10 g
20 g
mr
½-m stick|
|
|
|
|
table
50 40
25 15
5 0 cm
Questions 25-26 Four objects have the same mass and radius.
F
axis of rotation  
(A) hollow cylinder,  = 1 (B) solid cylinder,  = 1/2
(C) hollow ball,  = 2/3
(D) solid ball,  = 2/5
25. Which object would have the greatest moment of inertia?
A—most of the mass is along its rim   in I = mr2 is
the greatest
26. Which object would have the greatest rotational acceleration?
D—Fr = ma: the same force would generate a greater
acceleration for the object with the smallest 
27. 3 identical balls descend 3 identical ramps (except for s).
Ball A slides down ramp A (s = 0), ball B rolls down ramp B
(s = .3) and ball C rolls down ramp C (s = .6). Which is
true of their velocities when the reach the end of their ramp?
(A) vA > vB = vC (B) vA > vB > vC (C) vA = vB = vC
A—sliding is greater (Fgsin = ma) than rolling
(Fgsin = (1 +  )ma), and s doesn't effect acceleration
28. A 1-kg block is hung at the end of a rod 1-m long. The
balance point is 0.25 m from the end holding the block, what
is the mass of the rod?
| 0.25 m | 0.25 m |
center of rod
1 kg
(1) Record the balance point rcm.
(2) Use the center of mass formula to determine rcm.
(A) 0.25 kg (B) 0.5 kg
(C) 1 kg
(D) 2 kg
C—placing all the rod's mass at its center 
rbmbg = rrmrg  (0.25 m)(1 kg) = (0.25 m)(mrod)  mr = 1 kg
29. What is the total mass of the mobile? (rods are massless)
(3) Determine the percent difference between the
measured rcm (true) and the calculated rcm from (2).
1m
B
c.
Explore the relationship between center-of-mass and
balance by performing the following.
(1) Stand with your heels and back against a wall and
try to bend over and touch your toes. Explain
You fall over because your hips can't move back to
maintain a center of mass over your feet.
(2) Stand facing the wall with your toes against the
wall and try to stand on your toes. Explain
You fall back because you can't shift your weight
forward to keep you weight over the ball's of your feet.
(3) Rest a meter stick on two fingers. Slowly bring
your fingers together. Explain
The finger closest to the CM supports more weight 
greater Fn and Ff, which locks that finger in place.
Questions 24-36 Briefly explain your answer.
24. You are using a wrench to loosen a rusty nut. Which will
produce the greatest torque?
A
B
C
D
B—torque is the greatest when rsin is maximum, which
is when the distance is the greatest and the angle is 90o
1m
2m
3m
A
1 kg
(A) 5 kg
(B) 6 kg
(C) 7 kg
(D) 8 kg
B—(1 m)B = (3 m)(1 kg)  B = 3 kg
(1 m)(1 kg + 3 kg) = (2 m)(A)  A = 2 kg
1 kg + 3 kg + 2 kg = 6 kg
30. Consider the two configurations of interlocking blocks on the
edge of a table. Which of the following is true?
A
B
(A) A tips
(B) B tips
(C) both tip (D) neither tip
A—the blocks to right of the table's edge are not
balanced by an equal number of blocks to the left
31. Consider the door as viewed from above.
Determine
a. The torque when F1 = 45 N and r1 = 1 m.
1 = rF1 = (1 m)(45 N) = 45 m•N
b.
The force, F2, where r2 = 0.4 m, that will generate the
same torque as part a.
1 = rF2
45 m•N = (0.4 m)F2  F2 = 113 N
32. A 5-kg disk ( = ½) rolls down a 30o incline. Determine
a. The parallel component of Fg.
Fg-|| = Fgsin= (5 kg)(10 m/s2)sin30 = 25 N
b.
The disk's acceleration at the rim.
Fg-|| = Frolling = (1 +  )ma
25 N = (1 + ½)(5 kg)a  a = 3.3 m/s2
33. A 25-kg box rests on the edge of a merry-go-round (r = 2 m).
a. What is the maximum force of friction between the box
and merry-go-round (s = 0.80)?
a.
What is the student's mass?
mstudent = 35.1 kg + 31.6 kg
mstudent = 66.7 kg
b. What is the distance from her feet to her center-of-mass?
rcm = (r1m1 + r2m2)/(m1 + m2)
rcm = [(172 cm)(35.1 kg) + (0 cm)(31.6 kg)](66.7 kg)
rcm = 90.5 cm
37. A 2200-kg trailer is attached to a stationary truck.
Ff = Fn = (0.80)(250 N) = 200 N
b.
What is the maximum velocity before the box slips off?
Ff = Fc = mv2/r
200 N = (25 kg)v2/(2 m)  v = 4 m/s
c.
What is the acceleration of the 200-kg merry-go-round
( = ½) exerting by a 50-N force along the outer rim?
Determine the
a. normal force on the trailer tires at A.
Fr = ma
50 N = ½(200 kg)a  a = 0.5 m/s2
A = cm
d.
b.
How much time will it take to reach the maximum
velocity before the box slips off of the merry-go-round?
vt = vo + at
4 m/s = 0 + (0.5 m/s2)t  t = 8 s
e. Would this time increase or decrease if  = 1.0?
Increase
34. A 5-m, 75-kg plank is extended 2 m over the edge of a
building. What is the maximum distance that a 25-kg child
walks out from the building's
edge without tipping the plank?
child = plank
(x m)(250 N) = (0.5 m)(750 N)
x = 1.5 m
35. Consider the diagram of the printing press
on a table. Determine
(8 m)FA = (5.5 m)(22000 N)  FA = 15,000 N
normal force on the support B.
FA + FB = Fg
15,000 N + FB = 22,000 N  FB = 7,000 N
38. A 200-N sign hangs from the end of a 5-m pole, which is
held at a 37o angle by a horizontal guy wire.
guy wire
pole
Physics
is Phun
37o
Determine the tension in the guy wire.
cc = c
(5 m)Fguy wiresin37 = (5 m)(200 N)sin53  F = 265 N
a.
F1.
F11500 + 15,000
(20 m)F1 = (10 m)(15,000 N) + (5 m)(150,000 N)
F1 = 45,000 N
b. F2.
C. Conservation Laws—Rotation
39. 1 +  lab
Roll different objects down an incline and calculate the final
velocity and (1 + ) for each and compare the calculated
values with the ideal values.
a. Collect the following data.
Ring
Disk
height, h
height, h
distance, d
distance, d
time, t
time, t
F1 + F2 = 1500g + 15,000g
45,000 N + F2 = 15,000 N + 150,000 N  F2 = 120,000 N
36. A plank is placed on two scales, which are then zeroed. A
172-cm-tall student lies on the plank resulting in the
reading shown.
Ball
Cart
height, h
height, h
distance, d
distance, d
time, t
time, t
b.
Calculate the velocity for each using kinematics.
Ring
Disk
Ball
Cart
c.
Calculate (1 + ) for using conservation of energy.
Ring
Disk
Ball
Cart
d.
Calculate the percent difference with the ideal values.
Ring
Disk
Ball
Cart
2
3/
7/
2
5
Ub = K'b + K'r-p
mbgh = ½mbv2 + ½mpv2 = ½(mb + mp)v2
(1 kg)(10 m/s2)(1 m) = ½(1.0 kg + 1.0 kg)v2 v = 3.2 m/s
44. Two weights (m1 = 0.60 kg, m2 = 0.40 kg) are connected
by a cord that hangs from a pulley ( = ½, M = 0.50 kg).
1
M
m1
40. A hoop, cylinder and sphere roll down a 1-m ramp inclined
30o at the same time that a box slides down a frictionless
ramp that is also 1 m long and inclined 30o.
a. Derive a formula for determining the velocity of each
object when it reaches the bottom of the ramp.
Ug = Krollingand h = 1.0 m(sin) = 0.5 m
mgh = ½(1 +  )mv2 v = [g/(1 +  )]½
b.
What are the velocities of each when they reach the
bottom of the ramp?
Hoop
( = 1)
v = [g/(1 +  )]½ = [10/(1 + 1)]½ = 2.24 m/s
Cylinder
v = [g/(1 +  )]½ = [10/(1 + 1/2)]½ = 2.58 m/s
( = 1/2)
Sphere
v = [g/(1 +  )]½ = [10/(1 + 2/5)]½ = 2.67 m/s
( = 2/5)
Box
( = 0)
v = [g/(1 +  )]½ = [10/(1 + 0)]½ = 3.16 m/s
c. What is the order in which they reach the bottom?
Box-sphere-Cylinder-Hoop
41. Determine the velocity of a Yo-Yo ( = ½) that "rolls"
straight down its string a distance of 0.50 m.
Ug = K'rolling  mgh = ½(1 + ½)mv2
(10 m/s2)(0.50 m) = ¾v2  v = 2.6 m/s
42. A marble ( = 2/5) rolls from rest down a ramp and around
a loop (radius = 10 m). Determine
A
B
H
a.
10 m
the minimum velocity at B.
Fg = Fc
mg = mv2/r  v = (rg)½ = 10 m/s
b. the minimum height H at A.
Ug-A = U'g-B + K'rolling-B  mgH + 0 = mgh + ½(1 +  )mv2
(10 m/s2)H = (10 m/s2)(20 m) + ½( 1 +2/5)(10 m/s)2
H = 27 m
43. A string is attached to a 1.0-kg block and is wrapped round
a pulley ( = ½, m = 2.0 kg). The block is released from
rest and accelerates downward while the pulley rotates.
1m
m2
What is the velocity of m1 after descending 1 m?
Um2 + Um1 = U'm2 + U'm1 + K'r-M + K'm1 + K'm2
m2gh + 0 = 0 + m1gh + ½ Mv2 + ½m1v2 + ½m2v2
gh(m2 – m1) = ½( M + m1 + m2)v2
(10 m/s2)(1.0 m)(0.20 kg) = (0.625 kg)v2 v = 1.8 m/s
45. A string attached to a 20-kg block resting on a table
passes over a pulley ( = ½, m = 4 kg) and attaches to a
14-kg mass hanging over the edge of the table. The 20-kg
box slide along the table ( = 0.25) while the 14-kg mass
descends 1 m.
20 kg
1m
14 kg
What is the hanging mass' velocity after descending 1 m?
Um – Wf = K'b + K'r-p + K'm
mmgh – mbgd = ½mbv2 + ½mpv2 + ½mmv2
(14)(10)(1) – (.25)(20)(10)(1) = (10 + 1 + 7)v2  v = 2.2 m/s
46. What is the angular momentum of a 0.2-kg ball traveling at
9 m/s on the end of a string in a circle of radius 1 m?
L = rmv
L = (1 m)(1)(0.2 kg) (9 m/s) = 1.8 kg•m2/s
47. What is the angular momentum of Earth, m = 6.0 x 1024 kg?
a. about its axis of rotation ( = 2/5, rplanet = 6.4 x 106 m)
L = rmv = r(2/5)m(2r/T) = 4mr2/5T
L = 4(6.0 x 1024 kg)(6.4 x 106 m)2/(5)(60 x 60 x 24 s)
L = 7.1 x 1033 kg•m2/s
b. in its orbit around the Sun ( = 1, rorbit = 1.5 x 1011 m)
L = rmv = r(1)m(2r/T) = 2mr2/T
L = 2(6.0 x 1024 kg)(1.5 x 1011 m)2/(60 x 60 x 24 x 365 s)
L = 2.7 x 1040 kg•m2/s
48. Halley's comet follows an elliptical orbit, where its closest
approach to the sun is 8.9 x 1010 m and its farthest
distance is 5.3 x 1012 m. How many times faster does the
comet travel at its fastest compared to its slowest?
r1v1 = r2v2
v1/v2 = r2/r1 = (5.3 x 1012 m)/(8.9 x 1010 m) = 60
49. A child (m = 42 kg) runs toward a stationary merry-go-round
( = ½, m = 180 kg, r = 1.2 m) along a tangent at 3 m/s. The
child jumps on the merry-go-round and sets it rotating.
3 m/s
42 kg
=½
What is the block's velocity after descending 1 m?
180 kg
1.2 m
What is the speed of the merry-go-round after the child
jumps on?
rC CmCvC + rMMmMvM = (rCCmC + rMMmM)v'
(1)(42 kg)(3 m/s) + 0 = [(1)(42 kg) + (½)(180 kg)]v'
v' = 0.95 m/s
50. The rim of a disk ( = ½, m = M, r = R) rotates at a velocity,
V. A ring ( = 1, m = M, r = R) is dropped on top of the disk.
a. Calculate Ltotal before the ring is dropped on the disk.
Ltotal = rddmdvd + rr rmrvr
Ltotal = (R)(½)(M)(V) + (R)(1)(M)(0) = ½RMV
b.
Calculate the velocity after the ring is dropped.
Ltotal = Ltotal' = (rddmd + rrrmr)v'
½RMV = (½RM + RM)v'  v' = ⅓V
51. Tarzan (100 kg) is on a ledge that is 20 m above Jane (45
kg), who is trapped on a lower ledge. Tarzan grabs a long
vine and swings down from the ledge and grabs Jane, who
is stationary. The two swing over to a rock ledge on the
other side of the river gorge that is 10 m higher than the
rock ledge where Jane is trapped. Assuming the vine is
long enough, can Tarzan and Jane reach the other side?
Ug = Krolling  mgh = ½(1 + ½)mv2
(10 m/s2)(1 m) = ¾v2  v = 3.65 m/s
b.
How much time does it take the disk to travel the 2 m?
d = ½(v + vo)t
2 m = ½(3.65 m/s + 0)t t = 1.1 s
c.
Predict how the following alterations would change the
disk's velocity at and time to reach the base of ramp?
Alteration
Final Velocity
Time
same
same
A 2.0-kg disk is used
A 1.0-kg ring ( = 1) is used
less
more
A 3-m ramp is used, but h = 1 m
same
more
53. A string attached to a 10-kg box resting on a table passes
over a pulley ( = ½, m = 1 kg) and attaches to a 5-kg mass
hanging over the edge of the table. The 10-kg box slide 1 m
along the table ( = 0.3) while the 5-kg mass descends.
1m
a.
How much kinetic energy does the system have at the
point where the 5-kg mass has descended 1 m?
K' = Ug – Wf = mmgh – mbgd
K' = (5 kg)(10 m/s2)(1 m) – (0.3)(10 kg)(10 m/s2)(1 m)
K' = 20 J
b. What is the maximum velocity of the system?
T
J
K' = K'b + K'r-p + K'm = ½mbv2 + ½mpv2 + ½mmv2
20 J = (5 kg + 0.25 kg + 2.5 kg)v2 v = 1.6 m/s
54. Halley's Comet has a velocity of 3.88 x 104 m/s when it is
8.9 x 1010 m from the sun. How fast is it traveling when it is
5.3 x 1012 m from the sun?
a.
Calculate Tarzan's velocity when he grabs Jane.
mgh = ½mv2
v = (2gh)½ = [(2)(10 m/s2)(20 m)]½ = 20 m/s
b.
Calculate the velocity after Tarzan grabs Jane.
rmTvT + rmJvJ = (rmT + rmJ)v'
(100 kg)(20 m/s) + 0 = (145 kg)v'  v' = 13.8 m/s
c.
Calculate how high Tarzan swings to the other side.
mgh = ½mv2
h = v'2/2g = (13.8 m/s)2/(2)(10 m/s2) = 9.5 m
No
d. Did Tarzan and Jane make it?
e. What could Tarzan have done to save Jane?
Start from a higher ledge or with a running start.
f. How high would Tarzan have to start to save Jane?
v' = (2gh)½ = [(2)(10 m/s2)(10 m)]½ = 14 m/s
rmTvT = r(mT + mJ)v'
vT = (145 kg)(14 m/s)/100 kg = 20.3 m/s
h = v2/2g = (20.3 m/s)2/(2)(10 m/s2) = 20.6 m
g. What minimum initial velocity would Tarzan need to
save Jane starting from the original ledge?
½mv2 + mgh = ½mv'2
½v2 + (10 m/s2)(20 m) = ½(20.3)2  v = 3.5 m/s
52. A 1-kg, disk ( = ½) is placed on a 2-m ramp where the top
is 1 m above the base of the ramp. The disk is placed at the
top and rolls down to the base of the ramp.
a. What is the disk's velocity when reaches the base?
(8.9 x 1010 m)(3.88 x 104 m/s) = (5.3 x 1012 m)v2
v2 = 652 m/s
55. What is the angular momentum of the Moon?
(m = 7.35 x 1022 kg, rmoon = 1.74 x 106 m, rorbit = 3.84 x 108 m,
Torbit = Trotation = 2.42 x 106 s)
a. about its axis of rotation ( = 2/5)
L = rmv = r(2/5)m(2r/T) = 4mr2/5T
L = 4(7.35 x 1022 kg)(1.74 x 106 m)2/(5)(2.42 x 106 s)
L = 2.3 x 1029 kg•m2/s
b. in its orbit around the Earth ( = 1)
L = rmv = r(1)m(2r/T) = 2mr2/T
L = 2(7.35 x 1022 kg)(3.84 x 108 m)2/(2.42 x 106 s)
L = 2.8 x 1034 kg•m2/s
56. A student (m = 75 kg) runs at 5 m/s tangentially toward a
merry-go-round ( = ½, m = 150 kg, r = 2 m) rotating at 2
m/s, jumps on the merry-go-round and sets it rotating.
What is the velocity of the student after he jumps on to the
merry-go-round?
rsmsvs + rMmMvM = (rsms + r mmm)v'
(1)(75)(5) + (½)(150)(2)= [(1)(75) + (½)(150)]v' 
v' = 3.5 m/s
57. A 2-kg block and a 1-kg sphere hang from 2-m strings. The
sphere is raised to a horizontal position and swings toward
the block and collides with it.
1 kg
2 kg
a.
What is the sphere's velocity before the collision?
Ug = K  msgh = ½msvs2
vs = (2gh)½ = [2(10 m/s2)(2.0 m)]½ = 6.3 m/s
Assume that the collision is inelastic.
b. What is the sphere-block's velocity after the collision?
rmSvS + rmBvB = r (mS + mB)v'
(1 kg)(6.3 m/s) + 0 = (1 kg +2 kg)v'  v' = 2.1 m/s
c.
What is the maximum height reached after the collision?
Ug = K  (mS + mB)gh = ½(mS + mB)v'2
h = v'2/2g = (2.1 m/s)2/(2)(10 m/s2) = 0.22 m
d.
What is the maximum height reached after the collision
if the block and sphere exchange positions initially?
v' = mBvB/(mB + mS) = 12.6 kg•m/s/3 kg = 4.2 m/s
h = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m
The sphere is raised to a horizontal position initially and
then collides elastically with the block.
e. What are the velocities of the block and sphere after
the collision?
vS + vS' = vB + vB'
6.3 m/s + vS' = 0 + vB'  vS' = vB' – 6.3
rmSvS + rmBvB = rmSvS' + rmBvB'
(1 kg)(6.3 m/s) = (1 kg)(vB' – 6.3) + (2 kg)vB' vB' = 4.2 m/s
vS' = vB' – 6.3 = 4.2 m/s – 6.3 m/s = -2.1 m/s
f. What are the maximum heights reached by the block
and sphere?
hB = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m
hS = v'2/2g = (-2.1 m/s)2/(2)(10 m/s2) = 0.22 m
g.
Was potential energy conserved after the collision?
mSghS + mBghB = (1 kg)(2 m) + 0 = 2 J
mBgh'B + mSgh'S = (2)(0.88) + (1)(0.22) = 1.98 J yes
D. Simple Harmonic Motion (SHM)
58. Simple Harmonic Motion Lab
a. Measure the length L and time t for 10 oscillations of a
spring with different hanging masses m, determine k
using two methods and compare the results.
(1) Collect the following data.
m (kg)
0
0.10
0.20
0.30
0.40
0.50
L (m)
t (s)
(2) Determine the following using length data.
added mass
0.10
0.20
0.30
0.40
0.50
F using the
added weight
x using L
k using F = kx
Average k
(3) Determine the following using time data.
added mass
0.10
0.20
0.30
0.40
T using t
k using
T = 2(m/k)½
Average k
0.50
(4) Calculate the percent difference between the two
values of k using the average as true.
b.
Measure the pendulum period for different releasing
angles  and see which angle gives the most ideal
values for T.
(1) Collect the following data.
L (m)
5o
10o
15o
20o
25o
(o)
64. If the amplitude of a simple harmonic oscillator is doubled,
which quantity will change the most?
(A) T
(B) v
(C) a
(D) K + U
D—E = ½kA2, whereas v = A(k/m)½ and a = A(k/m)
change the same, and T = 2(m/k)½ doesn't change at all
65. A spring with mass m has period T. If m is doubled, what is
the new T?
(A) T/2
(B) T
(C) 2T
(D) 2T
C—T = 2(m/k)½: when m doubles T increases by 2
Questions 66-67 Consider the periods of pendulums A and B,
66. Which period is greater when LA = LB, but mA > mB?
(A) A
(B) B
(C) the same
T1 (s)
C—T = 2(L/g)½: only changing L or g can change T
(2) Calculate the following from the data.
angle
5o
10o
15o
20o
25o
x = L(2/360)
F = mgsin
k = F/x
T2 = 2(m/k)½
% = |T1 – T2|/T1
(3) Calculate T3 = 2(L/g)½.
(4) Which angle produced T closest to the one based
on the pendulum's length?
Questions 59-85 Briefly explain your answer.
Questions 59-62 A spring bob in SHM has amplitude A and
period T.
59. What is the total distance traveled by the bob after time T?
(A) 0
(B) ½A
(C) 2A
(D) 4A
D—starting at 0: down A, up A, up A, down A
67. Which period is greater when mA = mB, but LA > LB?
(A) A
(B) B
(C) the same
A—T = 2(L/g)½: greater L = greater T
68. A grandfather clock has a weight at the bottom of the
pendulum that can be moved up or down. If the clock is
running slow, should the weight be moved up or down?
(A) up
(B) down
(C) neither will work
A—T = 2(L/g)½: to speed up the clock (reduce T), you
need to decrease L by moving the weight up
69-70 Consider the following options.
(A) add mass to the oscillator
(B) move the oscillator to an elevator accelerating down
(C) move the oscillator to an elevator accelerating up
(D) move the oscillator to the Moon
69. Which will decrease the period of a pendulum?
C—T = 2(L/g)½: "g" increases when accelerating up  T
decreases
70. Which will change the period of oscillation on a spring?
A—T = 2(m/k)½: adding mass or changing the spring
will change the period
71. After a pendulum starts swinging, its amplitude gradually
decreases with time because of friction. What happens to
the period of the pendulum during this time
(A) decreases (B) no change (C) increases
B—T = 2(L/g)½: amplitude has no effect on T
60. What is the total displacement after time T?
(A) 0
(B) ½A
(C) 2A
(D) 4A
72. When you sit on a swing, the period of oscillation is T1.
When you stand on the same swing, the period of
oscillation is T2. Which is true?
A—you end where you started
(A) T1 < T2
(B) T1 = T2
(C) T1 > T2
C—T = 2(L/g)½: by standing on the swing, you have
61. How long does it take the bob to travel a distance of 6A?
reduced the distance L (distance from fulcrum to center
(A) ½T
(B) ¾T
(C) 5/4T
(D) 3/2T
of mass), which decreases T
73.
When a 50 kg person sits on a swing, the period of
D—Each A is ¼T  6 x ¼T = 3/2T
oscillation is T1, when a 100 kg person sits on the same
swing, the period of oscillation is T2. Which is true?
62. At what point in the motion is v = 0 and a = 0 simultaneously?
(A) T1 < T2
(B) T1 = T2
(C) T1 > T2
(A) x = 0
(B) 0 < x < A(C) x = A
(D) no point
D—when x = 0, a = 0, but v = max; when x = A, v = 0, but
a = max; at intermediate points a  0 and v  0
63. A mass on the end of a spring oscillates in simple harmonic
motion with amplitude A. If the mass doubles but the
amplitude is not changed, what happen to the total energy?
(A) decrease
(B) no change (C) increase
B—E = ½kA2  mass will not change the total energy
B—T = 2(L/g)½: mass has no effect on T
74. Consider the graph
of one cycle of SHM.
a.
Determine the time in terms of T for each situation.
Maximum up
Zero
Maximum down
½T
¼ T, ¾ T
0 T, 1 T
Acceleration
¾T
0 T, ½ T, 1 T
¼T
Velocity
b. Determine the following when m = 1 kg, k = 100 N/m
and A = 0.1 m.
(1) maximum acceleration
a.
Determine the time (in terms of T) for each of the
following.
Maximum up
Zero
Maximum down
¾T
0 T, ½ T, 1 T
¼T
Acceleration
0 T, 1 T
¼ T, ¾ T
½T
Velocity
b. Determine the following when m = 1 kg, k = 100 N/m
and A = 0.25 m.
(1) maximum acceleration
aA = A(k/m) = (0.1 m)(100 N/m)/(1 kg) = 10 m/s2
aA = A(k/m) = (0.25 m)(100 N/m)/(1 kg) = 25 m/s2
(2) maximum velocity
(2) maximum velocity
vo = A(k/m)½ = (0.1 m)[(100 N/m)/(1 kg)]½ = 1 m/s
vo = A(k/m)½ = (0.25 m)[(100 N/m)/(1 kg)]½ = 2.5 m/s
(3) period
(3) period
T = 2(m/k)½ = 2(1/100)½ = 0.63 s
T = 2(m/k)½ = 2(1/100)½ = 0.63 s
(4) maximum kinetic energy
(4) maximum kinetic energy
Ko = ½mv2 = ½(1 kg)(1 m/s)2 = 0.5 J
Ko = ½mv2 = ½(1 kg)(2.5 m/s)2 = 3.125 J
(5) maximum potential energy
(5) maximum potential energy
UA = ½kA2 = (100 N/m)(0.1 m)2 = 0.5 J
UA = ½kA2 = ½(100 N/m)(0.25 m)2 = 3.125 J
(6) velocity when x = 0.05 m
v = vo[1 – (x2/A2)]½ = (1 m/s)[1 – 0.52/0.12)]½ = 0.866 m/s
c.
Graph the potential energy (----), kinetic energy (•••)
and total energy (––) for one complete oscillation.
(6) velocity when x = 0.20 m
vx = vo[1 – (x2/A2)]½ = 2.5 m/s[1 – (.22/.252)]½ = 1.5 m/s
c.
Graph the potential energy, kinetic energy and total
energy for one complete oscillation.
0.5 J
3J
0J
¼T
¾T
complete the following chart (x = +A at t = 0 s)
t
¼T
½T
¾T
1T
x
0m
-0.1 m
0m
0.1 m
d.
v
-1 m/s
0 m/s
1 m/s
0 m/s
a
0 m/s2
10 m/s2
0 m/s2
-10 m/s2
F
0N
10 N
0N
- 10 N
e. How do the following change if the amplitude is 0.2 m?
Max acceleration
Max velocity
Period
doubles
doubles
remains the same
75. A 1-kg ball on the end of a 1-m string is set in motion by
pulling the ball out so that it is raised 0.015 m. Determine
a. the maximum speed
UG = K  mgh = ½mv2
v = (2gh)½ = [2(10 m/s2)(0.015 m)]½ = 0.55 m/s
b.
the period of oscillation.
T=
2(L/g)½
c.
= 2(1.0 m/10
m/s2)½
0J
¼T
¾T
complete the following chart (x = 0 at t = 0 s)
t
¼T
½T
¾T
x
+0.25 m
0m
-0.25 m
d.
v
a
0 m/s
-25
m/s2
-2.5 m/s
0
m/s2
0 m/s
25
m/s2
1T
0m
2.5 m/s
0 m/s2
F
-25 N
0N
25 N
0N
e. Determine the following values when A = 0.50 m.
(1) maximum acceleration
aA = |-A(k/m)| = (0.50 m)(100 N/m)/(1.0 kg) = 50 m/s2
(2) maximum velocity
vo = A(k/m)½ = (0.50 m)[(100 N/m)/(1.0 kg)]½ = 5 m/s
(3) period
= 2.0 s
What would the period be with the following changes?
m = 4 kg
L=4m
g = 40 m/s2
2.0 s
4.0 s
1.0 s
76. Consider the diagram of one cycle of SHM.
T = 2(m/k)½ = 2(1/100)½ = 0.63 s
77. A 1-kg ball swings from the ceiling on the end of a 2-m
string. The ball starts its swing from a position that is
0.2 m above its lowest point.
a. What is the maximum speed of the ball?
Ug = K  mgh = ½mv2
v = (2gh)½ = [2(10 m/s2)(0.20 m)]½ = 2 m/s
b.
What is the period of oscillation for the pendulum?
T = 2(L/g)½ = 2(2.0 m/10 m/s2)½ = 2.8 s
1.
Practice Multiple Choice (No calculator)
In the diagram, a car travels clockwise at constant speed.
7.
A ball attached to a string is moved at constant speed in a
horizontal circular path. A target is located near the path of
the ball as shown in the diagram.
Which letters represent the directions of the car's velocity,
v, and acceleration, a?
v
a
v
a
(A) A
C
(B) C
B
(C) C
A
(D) D
A
C—In uniform circular motion velocity is tangent to the
circle and acceleration is toward the center
2.
A racing car is moving around the circular track of radius
300 m. At the instant when the car's velocity is directed
due east, its acceleration is 3 m/s2 directed due south.
When viewed from above, the car is moving
(A) clockwise at 30 m/s (B) counterclockwise at 30 m/s
(C) clockwise at 10 m/s d. counterclockwise at 10 m/s
A—Traveling east and turning south  clockwise.
ac = v2/r  v = (acr)½ = [(3 m/s2)(300 m)]½ = 30 m/s
3.
D—Ball simultaneously is directed to the right (due to
the spring) and toward the top (due to the rotation)
A person weighing 800 N on earth travels to another planet
with the same mass as earth, but twice the radius. The
person's weight on this other planet is most nearly
(A) 200 N (B) 400 N (C) 800 N (D) 1600 N
A—Fg = GMm/r2, where r is doubled
 Fg should be (½)2 = ¼ as much  ¼(800 N) = 200 N
5.
B—The released ball will move tangent to the circle at
that spot
8.
The diagram shows a 5.0-kg bucket of water being swung
in a horizontal circle of 0.70-m radius at a constant speed
of 2.0 m/s.
The disk is rotating counterclockwise when the ball is
projected outward at the instant the disk is in the position
shown.
Which of the following best shows the subsequent direction
of the ball relative to the ground?
(A) 
(B) 
(C) 
(D) 
4.
At which point along the ball's path should the string be
released, if the ball is to hit the target?
(A) A
(B) B
(C) C
(D) D
A ball is released from rest at position P swings through
position Q then to position R where the string is again
horizontal.
The centripetal force on the bucket of water is
(A) 5.7 N
(B) 29 N
(C) 14 N
(D) 200 N
B—Fc = mv2/r = (5.0 kg)(2.0 m/s)2/0.70 m = 29 N
Questions 9-10 refer to a ball that is tossed straight up from the
surface of a small asteroid with no atmosphere. The ball
rises to a height equal to the asteroid's radius and then
falls straight down toward the surface of the asteroid.
9. What forces act on the ball while it is on the way up?
(A) a decreasing gravitational force that acts downward
(B) an increasing gravitational force that acts downward
(C) a constant gravitational force that acts downward
(D) a constant gravitational force that acts downward and
a decreasing force that acts upward
A—The force of gravity is the only force acting on the
ball and Fg decreases as r increases (Fg = GMm/r2)
What are the directions of the ball's acceleration at
positions, Q and R?
Q
R
Q
R
(A) 

(B) 

(C) 

(D) 

A—At Q, acceleration is , at R acceleration is 
6.
A 5-kg sphere is connected to a 10-kg sphere by a rod.
The center of mass is closest to
(A) A
(B) B
(C) C
(D) D
B—The center of mass is the balance point, which is
closer to the more massive side
10. The acceleration of the ball at the top of its path is
(A) at its maximum value for the ball's flight
(B) equal to the acceleration at the surface
(C) equal to one-half the acceleration at the surface
(D) equal to one-fourth the acceleration at the surface
D—g = GM/r2, r is doubled  g is (½)2 = ¼ as much
Questions 11-12 A 125-N board is 4 m long and is supported by
vertical chains at each end. A person weighing 500 N is
sitting on the board. The tension in the right chain is 250 N.
11. What is the tension in the left chain?
(A) 250 N (B) 375 N
(C) 500 N (D) 625 N
B—F = F
250 N + FL = 125 N + 500 N  FL = 375 N
12. How far from the left end of the board is the person sitting?
(A) 0.4 m (B) 1.5 m
(C) 2 m
(D) 2.5 m
B—rFperson + rFboard = rFright chain
r(500 N) + (2 m)(125 N) = (4 m)(250 N) 
r = 1.5
13. A square piece of plywood on a horizontal tabletop is
subjected to the two horizontal forces shown above.
C—CM =  (rimi)/ (mi)—assume the rod is 4 m
2 = [(0)(3.5) + (1)(X) + (4)(5)]/(8.5 + X)  X = 3
18. A ball attached to a string is whirled around in a horizontal
circle with radius r, speed v and tension T. If the radius is
increased to 4r and the tension remains the same, then the
speed of the ball is
(A) ¼v
(B) ½v
(C) v
(D) 2v
D—T = Fc = mv2/r = mv'2/r': r increases to 4r without
changing T  v'2 also increase by 4. v;2 = 4  v' = 2v
Where should a third force of magnitude 5 N be applied to
put the piece of plywood into equilibrium?
A B
C
D
19. A 0.4-kg object swings on the end of a string. At the
bottom of the swing, the tension in the string is 6 N. What
is the centripetal force acting on the object at the bottom of
the swing?
(A) 2 N
(B) 4 N
(C) 6 N
(D) 10 N
A—At the bottom of the swing, Fnet = Fc = Ft – Fg
Fc = 6 N – (0.4 kg)(10 m/s2) = 2 N
A—C and D would cancel torque, B would cancel up
and down, but only A will cancel both.
14. The diagram represents two satellites of equal mass, A
and B, in circular orbits around a planet.
20. Two wheels, fixed to each other, are free to rotate about a
frictionless axis perpendicular to the page. Four forces are
exerted tangentially to the rims of the wheels.
The net torque on the system about the axis is
(A) zero
(B) FR
(C) 2FR
(D) 5FR
C—net = cc – c
net = (3R)F + (3R)F + (2R)F – (3R)(2F) = 8RF – 6RF = 2RF
Comparing the gravitational force between satellite and
planet, B's gravitational force compared to A's is
(A) half as great
(B) twice as great
(C) one-fourth as great
(D) four times as great
C—Fg = GMm/r2, where r is doubled
 Fg is (½)2 = ¼ as much.
15. The radius of the earth is approximately 6,000 km. The
acceleration of an astronaut in a perfectly circular orbit
6,000 km above the earth would be most nearly
(A) 0 m/s2 (B) 2.5 m/s2 (C) 5 m/s2 (D) 10 m/s2
B—g = GM/r2, where r is doubled
 g is (½)2 = ¼ as much. ¼(10 m/s2) = 2.5 m/s2
16. A 5-m, 100-kg plank rests on a ledge with 2 m extended out.
How far can a 50-kg person walk out on the plank past the
edge of the building before the plank just begins to tip?
(A) ½ m
(B) 1 m
(C) 3/2 m
(D) 2 m
B—rFperson = rFplank (rplank is 2.5 m – 2 m = 0.5 m)
r(50 kg)(10 m/s2) = (0.5 m)(100 kg)(10 m/s2)  r = 1 m
17. The system is balanced when hanging by the rope.
21. Mars has a mass 1/10 that of Earth and a diameter 1/2 that
of Earth. The acceleration of a falling body near the
surface of Mars is most nearly
(A) g/5
(B) 2g/5
(C) g/2
(D) g
B—g = GM/r2  gE = GME/rE2 and GM = GMM/rM2
gM/gE = (MM/ME)(rE/rM)2 = (1/10)(1/½)2 = 4/10  gM = 2g/5
22. A satellite of mass m and speed v moves in a stable,
circular orbit around a planet of mass M. What is the radius
of the satellite’s orbit?
(A) GM/mv (B) Gv/mM (C) GM/v2 (D) GmM/v
C—Fc = Fg
mv2/r = GMm/r2  r = GM/v2
23. A wheel of radius R is mounted on an axle so that the wheel
is in a vertical plane. Three small objects having masses m,
M, and 2M, respectively, are mounted on the rim.
What is m in terms of M when the wheel is stationary?
(A) 1/2 M
(B) M
(C) 3/2 M
(D) 2 M
C—Rm + (rcos60)M = R(2M)
m + ½M = 2M  m = 3/2M
24. In each case the unknown mass m is balanced by a known
mass M1 or M2.
What is the mass of the fish?
(A) 1.5 kg (B) 2 kg
(C) 3 kg
(D) 6 kg
What is the value of m in terms of the known masses?
(A) M1 + M2
(B) (M1 + M2)/2
(C) M1M2
(D) (M1M2)½
31. A particle of mass, m, moves with a constant speed v
along the dashed line y = a.
D—I1m = l2M1 and l1M2 = l2m  l1/l2 = M1/m = m/M2 
M1M2 = m2  m = (M1M2)½
25. An asteroid moves in an elliptic orbit with the Sun at one
focus.
Which of the following increases as the asteroid moves
from point P in its orbit to point Q?
(A) Speed
(B) Angular momentum
(C) Total energy
(D) Potential energy
D—Energy & angular momentum are conserved,
r  v  (r1v1 = r2v2), Ug increases (Ug = -GMm/r)
26. A satellite S is in an elliptical orbit around a planet P with r1
and r2 being its closest and farthest distances, respectively,
from the center of the planet. If the satellite has a speed v1
at its closest distance, what is its speed at its farthest
distance?
(A) (r1/r2)v1
(B) (r2/r1)v1
(C) (r2 – r1)v1
(D) ½(r1 + r2)v1
When the x-coordinate of the particle is xo, the
magnitude of the angular momentum of the particle with
respect to the origin of the system is
(A) zero
(B) amv
(C) xomv
(D) (vo2 + a2)½mv
B—L = rmv = (a)(1)(m)(v) = amv
Questions 32-36 A block oscillates without friction on the end of
a spring. The minimum and maximum lengths of the
spring as it oscillates are, respectively, xmin and xmax.
The graphs below can represent quantities associated with
the oscillation as functions of the length x of the spring.
(A)
(B)
A—r1v1 = r2v2  v2 = (r1/r2)v1
27. A satellite of mass M moves in a circular orbit of radius R
at a constant speed v. Which must be true?
I. The net force on the satellite is equal to Mv2/R
and is directed toward the center of the orbit.
II. The net work done on the satellite by gravity in
one revolution is zero.
III. The angular momentum of the satellite is a
constant.
(A) I only
(B) III only (C) I and II (D) I, II, and III
D—I is true (Fg = Fc = Mv2/R), II is true (F r  no work
done), and III is true (angular momentum is constant)
Questions 28-29 A sphere of mass M, radius R, and = 2/5, is
released from rest at the top of an inclined plane of height h.
(C)
(D)
32. Which graph can best represent the total mechanical
energy of the block-spring system as a function of x?
C—Total mechanical energy is constant when a
conservative force is involved
33. Which graph can best represent the kinetic energy of the
block as a function of x?
D—The kinetic energy is greatest when the velocity is
greatest (K = ½mv2), which occurs at the midpoint
34. Which graph can best represent the potential energy of the
block as a function of x?
B—Potential energy is greatest when x is greatest
(Ug = ½kx2), which occurs at xmin and xmax,
28. If the plane is frictionless, what is the speed of the center
of mass of the sphere at the bottom of the incline?
(A) (2gh)½ (B) 2Mgh
(C) 2MghR2 (D) 5gh
A—Ug = K (no rolling because of zero friction)
mgh = ½mv2 v = 2gh = (2gh)½
29. If the plane has friction so that the sphere rolls without
slipping, what is the speed at the bottom of the incline?
(A) (2gh)½ (B) 2Mgh
(C) 2MghR2 (D) (10gh/7)½
D—mgh = K + Kr = ½mv2 + ½mv2 = ½(1 +  )mv2
v2 = 2gh/(1 + 2/5)  v = (10gh/7)½
30. For which motions is there a variable force involved?
(A) Constant speed in a straight line
(B) Simple harmonic motion
(C) Constant speed in a circle
(D) Constant acceleration in a straight line
B—A has no force, C and D have a constant force, only
SHM has a variable force ( Fs = kx)
35. Which graph can best represent the acceleration of the
block as a function of x?
B—Acceleration is greatest when force is greatest (Fs
= kx), which occurs at xmin and xmax
36. Which graph can best represent the velocity of the block as
a function of x?
D—Velocity is greatest at the midpoint and is zero at
xmin and xmax (same as K)
37. A block attached to the lower end of a vertical spring
oscillates up and down. The period of oscillation depends
on which of the following?
I. Mass of the block
II. Amplitude of the oscillation
III. Spring constant
(A) I only
(B) II only (C) III only (D) I and III only
D—T = 2(m/k)½  T depends on both m and k, but not
on amplitude, A.
38. When a 1-kg bob is attached to a spring, the period of
oscillation is 2 s. What is the period of oscillation when a
2-kg bob is attached to the same spring?
(A) 0.5 s
(B) 1.0 s
(C) 1.4 s
(D) 2.8 s
44. When a mass m is hung on a spring, the spring stretches a
distance d. If the mass is then set oscillating on the spring,
the period of oscillation is proportional to
(A) (d/g)½ (B) (g/d)½ (C) (d/mg)½ (D) (m2g/d)½
D—T = 2(m/k)½ T m.
The mass is doubled  T = 22 = 2.8 s
39. A pendulum and a mass hanging on a spring both have a
period of 1 s on Earth. They are taken to planet X, which
has twice the gravitational acceleration g as Earth. Which
is true about the periods of the two objects on planet X
compared to their periods on Earth?
(A) Both are shorter.
(B) Both are longer.
(C) The pendulum is longer and the spring is the same.
(D) The pendulum is shorter and the spring is the same.
A—Fs = kx  mg = kd  k = gm/d
T  (m/k)½ = (md/gm)½ = (d/g)½
45. A 3-kg block is hung from a spring, causing it to stretch 12
cm at equilibrium. The 3-kg block is then replaced by a 4kg block, and the new block is released from the spring
when it is unstretched. How far will the 4-kg block fall
before its direction is reversed?
D—The period of a pendulum, T = 2(L/g)½  T  (1/g)½
and it is shorter, but the period of a spring is the same
40. The graph is of the displacement x versus time t for a
particle in simple harmonic motion with a period of 4 s.
Which graph shows the potential energy of the particle as
a function of time t for one cycle of motion?
(A)
(B)
(C)
(D)
(A) 9 cm
(B) 18 cm
(C) 24 cm
(D) 32 cm
D—The 4-kg block stretches the spring 16 cm (Fs  x),
the spring falls another 16 cm before turning around
46. A spring is fixed to the wall at one end. A block of mass M
attached to the other end of the spring oscillates with
amplitude A on a frictionless, horizontal surface. The
maximum speed of the block is v.
The spring constant is
(A) Mg/A
(B) Mgv/2A (C) Mv2/2A (D) M(v/A)2
C—Potential energy is greatest when x is greatest
(Ug = ½kx2), which occurs at 1 s and 3 s.
41. Two identical springs are hung from a horizontal support.
When a 1.2-kg block is suspended from the pair of
springs, each spring is stretched an additional 0.15 m.
D—v = A(k/m)½
v2 = A2k/M  k = Mv2/A2 = M(v/A)2
47. A sphere of mass m1 is attached to a spring. A second
sphere of mass m2 is suspended from a string of length L,
If both spheres have the same period of oscillation, which
of the following is an expression for the spring constant?
(A) L/m1g (B) g/m2L (C) m1L/g (D) m1g/L
D—Tp = 2(L/g)½ = Ts = 2(m/k)½
L/g = m1/k  k = m1g/L
The spring constant of each spring is most nearly
(A) 40 N/m (B) 48 N/m (C) 60 N/m (D) 80 N/m
A—Each spring supports half of the weight (12 N).
Fs = kx  6 N = k(0.15 m)  k = 40 N/m
42. A ball is dropped from a height of 10 m onto a hard surface
so that the collision at the surface may be assumed elastic.
Under such conditions the motion of the ball is
(A) simple harmonic with a period of about 1.4 s
(B) simple harmonic with a period of about 2.8 s
(C) simple harmonic with an amplitude of 5 m
(D) periodic but not simple harmonic
1.
Practice Free Response
A roller coaster ride at an amusement park lifts a car of
mass 700 kg to point A at a height of 90 m above the
lowest point on the track, as shown above. The car starts
from rest at point A, rolls with negligible friction down the
incline and follows the track around a loop of radius 20 m.
Point B, the highest point on the loop, is at a height of 50 m
above the lowest point on the track.
The force on ball is constant, except when it is in
contact with the floor,  not SHM, where force  x..
43. An object swings on the end of a cord as a simple
pendulum with period T. Another object oscillates up and
down on the end of a vertical spring, also with period T. If
the masses of both objects are doubled, what are the new
values for the periods?
Pendulum Spring
Pendulum Spring
(A) T/√2
√2T
(B) T
√2T
(C) T
T
(D) √2T
T
B—Pendulum: T = 2(L/g), T = T
Spring: T = 2(m/k), where T m  T = 2T
a.
P
(1) Indicate on the figure the point P at which the
maximum speed of the car is attained.
(2) Calculate the value vmax of this maximum speed.
Ug = K  mgh = ½mv2
v = (2gh)½ = [2(10 m/s2)(90 m)]½ = 42 m/s
b.
Calculate the speed vB of the car at point B.
v = (2gh)½ = [2(10 m/s2)(40 m)]½ = 28 m/s
c.
(1) Draw and label vectors to represent the forces
acting on the car when it is upside down at point B.
a.
Fg
2.
Fn
Determine the potential energy of the hoop at the top
of the ramp, where Ug = 0 at the floor.
Ug = mg(H + Lsin)
Ug = (0.5 kg)(10 m/s2)[1 m + (2 m)(sin30)] = 10 J
(2) Calculate all the forces identified in (c1).
Fg = mg = (700 kg)(10 m/s2) = 7,000 N
Fc = mv2/r = (700 kg)(28 m/s)2/(20 m) = 27,000 N
Fc = Fg + Fn  Fn = Fc – Fg = 27,000 – 7,000 = 20,000 N
A string attached to a 20-kg block resting on a table
passes over a pulley ( = ½, m = 10 kg) and attaches to a
10-kg mass hanging over the edge of the table. The 20-kg
box slide along the table ( = 0.30) while the 10-kg mass
descends 2 m.
b.
The hoop rolls down the ramp and then onto the floor.
Determine the hoop's
(1) speed at the bottom of the ramp.
Ug = Krolling mgLsin = ½(1 +  )mv2
(10 m/s2)(2 m)sin30 = ½(1 + 1)v2  v = 3.2 m/s
(2) speed just before it hits the floor.
Kfloor = Ktable + Utable  ½mvf2 = ½mvt2 + mgh
vf2 = (3.2 m/s)2 + 2(10 m/s2)(1 m) vf = 5.5 m/s
(3) translational kinetic energy before it hits the floor.
K = ½mv2 = ½(0.5 kg)(5.5 m/s)2 = 7.5 J
2m
Determine the
a. force of friction on the 20-kg block as it slides.
(4) percentage of total energy that is rotational kinetic
energy just before it hits the floor.
F = mbg = (0.30)(20 kg)(10 m/s2) = 60 N
(10 J – 7.5 J)/10 J x 100 = 25 %
b.
c.
force of gravity on the 10 kg mass.
Fg = mmg = (10 kg)(10 m/s2) = 100 N
c.
Ug = Krolling  mgLsin = ½(1 +  )mv2
(10 m/s2)(2 m)sin30 = ½(1 + 2/5)v2 = 7/10 v2  v = 3.8 m/s
net force rotating the pulley.
Fnet = 100 N – 60 N = 40 N
d.
(2) speed just before it hits the floor.
Kfloor = Ktable + Utable  ½mvf2 = ½mvt2 + mgh
vf2 = (3.8 m/s)2 + 2(10 m/s2)(1 m) = 34 m2/s2vf = 5.8 m/s
acceleration of the pulley at the rim.
Fnet = (mb + mp + mm)a
40 N = (20 kg + 5 kg + 10 kg)a  a = 1.1 m/s2
e.
(3) translational kinetic energy before it hits the floor.
velocity when the system has moved 2 m.
K = ½mv2 = ½(0.5 kg)(5.8 m/s)2 = 8.4 J
v2 = vo2 + 2ad
v2 = 02 + 2(1.1 m/s2)(2 m)  v = 2.1 m/s
f.
(4) percentage of total energy that is rotational kinetic
energy just before it hits the floor.
loss of potential energy as the 10-kg mass falls 2 m.
Ug = mmgh = (10 kg)(10 m/s2)(2 m) = 200 J
(10 J – 8.4 J)/10 J x 100 = 16 %
g.
d.
work done by friction as the 20-kg block slides 2 m.
Wf = mbgd = (.30)(20 kg)(10 m/s2)(2 m) = 120 J
h.
velocity of the system
Um – Wf = K'b + K'r-p + K'm = ½mbv2 + ½mpv2 + ½mmv2
200 J – 120 J = ½(20 kg + 5 kg + 10 kg)v2  v = 2.1 m/s
3.
The hoop is replaced by a 0.5 kg solid sphere ( = 2/5),
which rolls down the ramp and then onto the floor.
Determine the sphere's
(1) speed at the bottom of the ramp.
A 0.5-kg hoop ( = 1) rolls from rest at the top of the ramp of
length L = 2 m and angle  = 30o. The table height H = 1 m.
4.
Comparing a hoop ( = 1), disk ( = ½) and a sphere
( = 2/5) just before it lands on the floor, which would
(1) have the greatest % rotational kinetic energy?
The hoop (largest  = largest % rotational energy)
(2) land furthest from the base of the table?
The sphere (smallest  = most translational energy)
(3) have the most kinetic energy just before it landed?
All three would have the same amount of kinetic energy
The graph shows a system in simple harmonic motion.
Complete the chart with either +, 0, or –.
t
0s
1s
2s
x (m)
0
–
0
3s
+
v (m/s)
–
0
+
0
(m/s2)
0
+
0
–
F (N)
0
+
0
–
U (J)
0
+
0
+
K (J)
+
0
+
0
a
5.
the cart collides and sticks with a bumper of mass 3m
attached to a spring, which has a spring constant k.
A 3.0 kg bob swings on the end of a 1.0 m string. The
potential energy U of the object as a function of distance x
from its equilibrium position is shown. This particular
object has a total energy E of 0.4 J.
Given: m = 1 kg, h = 0.50 m, k = 250 N/m, determine
a. the potential energy of the cart at the top of the ramp.
Ug = 2mgh
Ug = 2(1 kg)(10 m/s2)(0.50 m) = 10 J
b.
a.
What is the bob's potential energy when its
displacement is +4 cm from its equilibrium position?
U = 0.05 J
b. What is the greatest distance x for the pendulum bob?
Explain your reasoning.
10 cm, the mass stops at this point because all of the
energy is in the form of potential energy.
c. How much time does it take the pendulum to go from
the greatest +x to the greatest –x?
c.
d.
e.
K=
½mv2
 0.4 J = ½(3.0
kg)v2
the translational kinetic energy of the cart and bumper
just after the collision.
Kt' = ½mv2
Kt'= ½(5 kg)(1.1 m/s)2 = 3.2 J
K = 0.4 J – U = 0.4 J – 0.2 J = 0.2 J
What is the object's speed at x = 0?
the speed of the cart just after the collision.
mcvc + mbvb = (mc + mb)v'
2mvc + 0 = 5mv'  v' = 2/5vc = 1.1 m/s
Determine the bob's kinetic energy when x = -7 cm.
e.
the carts translational kinetic energy before the collision.
Kt = ½mv2
Kt = ½(2 kg)(2.8 m/s)2 = 8 J
T = 2(L/g)½ = 2(1.0 m/10 m/s2)½ = 2.0 s  1.0 s
d.
the speed of the cart at the bottom of the ramp.
Ug = K = ½mv2 + ½(1 +  )mv2 = 5/4 mv2
10 J = 5/4(2 kg)v2 v = 2.8 m/s
f.
the amount that the spring is compressed.
K = Us = ½kx2
3.2 J = ½(250 N/m)x2  x = 0.16 m
 v = 0.5 m/s
6.
The cart of mass m with four wheels of mass m/4 and  =
½ is released from rest and rolls from height h. After
rolling down the ramp and across the horizontal surface,
Unit 5 Answers (Don't look until after you have tried the problem)
2
D—net force equals Fc, which is toward the center of the circle
B—the turn to the left results in the right side of the bus being
in "front" of you as you continue forward due to inertia
B—the centripetal force needed to make the turn is generated
by friction between the road and tires, which is not sufficient
at the high speed
B—without the tension in the string to supply the centripetal
force, the ball continues in a straight line
B—Fc = mv2/r = m(2r/T)2/r = 42mr/T2, since 42m/T2 is
constant  Fc  r
C—at the bottom of the vertical circle:
Fnet = Fc = Fn – Fg > 0 (upward)  Fn > Fg
A—at the top of the vertical circle, both Ft and Fg are
downward, which is the direction of Fc  Fc = Ft + Fg
C—Newton's third law states that for every force there is an
equal but opposite force  the pulls are equal
A—The net force is Fg, which provides the Fc necessary to
keep the astronaut in orbit
v = 2r/T = 2(50 m)/16 s = 20 m/s
east
ac = v2/r = (20 m/s)2/(50 m) = 8 m/s2
south
T = 365 days x 24 hr/day x 3600 s/hr = 3.2 x 107 s
v = 2r/T = 2(1.5 x 1011 m)/3.2 x 107 s = 3.0 x 104 m/s
aC = v2/r = (3.0 x 104 m/s)2/1.5 x 1011 m = 6 x 10-3 m/s2
Fc = Ff  mv2/r = mg  v2/50 m = (0.8)(10 m/s2)  v = 20 m/s
Yes, because mass cancels out of the equation.
No, because the coefficient of friction is less.
Ft-x = Fc = mv2/r = (2 kg)(5 m/s)2/(1 m) = 50 N
Ft-y = Fg = mg = (2 kg)(10 m/s2) = 20 N
Ft = (Ft-x2 + Ft-y2)½ = (502 + 202)½ = 54 N
tan = Ft-y/Ft-x = Fg/Fc = 20/50   = 22o
Ft = Fc – Fg = mv2/r – mg
Ft = (2 kg)(5 m/s)2/(1 m) – (2 kg)(10 m/s2) = 30 N
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Ft = Fc + Fg = mv2/r + mg
Ft = (2 kg)(5 m/s)2/(1 m) + (2 kg)(10 m/s2) = 70 N
Ft = Fg + Fc = mg + mv2/r
15 N = (1kg)(10 m/s2) +(1 kg)v2/(2 m) v = 3.16 m/s
Ug + K = Ug' + K' mgh = ½mv2 
(10 m/s2)h = ½(3.16 m/s)2 h = 0.50 m
the same
greater
less
Less
g
¼g
4g
2g
½g
g = GM/r2= (6.67x10-11N•m2/kg2)(6.4x1023kg)/(3.4x106m)2=3.7m/s2
GmMm/rM2 = GmEm/rE2  rE2/rM2 = mE/mM = 100  rM/rE = 10
II and III
I and IV
II only
I only
g = GM/r2  gMoonrMoon2/mMoon = gEarthrEarth2/mEarth
gMoon(1)2/1 = g(42)/81= 16g/81 = 0.2 g
Mass doesn't change: 50 kg
Fg = mg = (50 kg)(0.2) = 10 kg
B—torque is the greatest when rsin is maximum, which is
when the distance is the greatest and the angle is 90o
A—most of the mass is along its rim   in I =  mr2 is the
greatest
D—Fr =  ma: the same force would generate a greater
acceleration for the object with the smallest 
A—sliding is greater (Fgsin = ma) than rolling
(Fgsin = (1 +  )ma), and s doesn't effect acceleration
C—placing all the rod's mass at its center 
rbmbg = rrmrg  (0.25 m)(1 kg) = (0.25 m)(mrod)  mr = 1 kg
B—(1 m)B = (3 m)(1 kg)  B = 3 kg
(1 m)(1 kg + 3 kg) = (2 m)(A)  A = 2 kg
1 kg + 3 kg + 2 kg = 6 kg
A—the blocks to right of the table's edge are not balanced by
an equal number of blocks to the left
1 = rF1 = (1 m)(45 N) = 45 m•N
1 = rF2  45 m•N = (0.4 m)F2  F2 = 113 N
Fg-|| = Fgsin= (5 kg)(10 m/s2)sin30 = 25 N
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Fg-|| = Frolling = (1 +  )ma  25 N = (1 + ½)(5 kg)a  a = 3.3 m/s2
Ff = Fn = (0.80)(250 N) = 200 N
Ff = Fc = mv2/r 200 N = (25 kg)v2/(2 m)  v = 4 m/s
Fr =  ma  50 N = ½(200 kg)a  a = 0.5 m/s2
vt = vo + at  4 m/s = 0 + (0.5 m/s2)t  t = 8 s
Increase
child = plank  (x m)(250 N) = (0.5 m)(750 N)  x = 1.5 m
F11500 + 15,000
(20 m)F1 = (10 m)(15,000 N) + (5 m)(150,000 N)  F1 = 45,000 N
F1 + F2 = 1500g + 15,000g
45,000 N + F2 = 15,000 N + 150,000 N  F2 = 120,000 N
mstudent = 35.1 kg + 31.6 kg = 66.7 kg
rcm = (r1m1 + r2m2)/(m1 + m2)
rcm = [(172 cm)(35.1 kg) + (0 cm)(31.6 kg)](66.7 kg) = 90.5 cm
A = cm  (8 m)FA = (5.5 m)(22000 N)  FA = 15,000 N
FA + FB = Fg  15,000 N + FB = 22,000 N  FB = 7,000 N
cc = c  (5 m)Fguy wiresin37 = (5 m)(200 N)sin53  F = 265 N
Ug = Krollingand h = 1.0 m(sin) = 0.5 m
mgh = ½(1 +  )mv2 v = [g/(1 +  )]½
v = [g/(1 +  )]½ = [10/(1 + 1)]½ = 2.24 m/s
v = [g/(1 +  )]½ = [10/(1 + 1/2)]½ = 2.58 m/s
v = [g/(1 +  )]½ = [10/(1 + 2/5)]½ = 2.67 m/s
v = [g/(1 +  )]½ = [10/(1 + 0)]½ = 3.16 m/s
Box-sphere-Cylinder-Hoop
Ug = K'rolling  mgh = ½(1 + ½)mv2
(10 m/s2)(0.50 m) = ¾v2  v = 2.6 m/s
Fg = Fc mg = mv2/r  v = (rg)½ = 10 m/s
Ug-A = U'g-B + K'rolling-B  mgH + 0 = mgh + ½(1 +  )mv2
(10 m/s2)H = (10 m/s2)(20 m) + ½( 1 +2/5)(10 m/s)2  H = 27 m
Ub = K'b + K'r-p
mbgh = ½mbv2 + ½ mpv2 = ½(mb +  mp)v2
(1 kg)(10 m/s2)(1 m) = ½(1.0 kg + 1.0 kg)v2 v = 3.2 m/s
Um2 + Um1 = U'm2 + U'm1 + K'r-M + K'm1 + K'm2
m2gh + 0 = 0 + m1gh + ½ Mv2 + ½m1v2 + ½m2v2
gh(m2 – m1) = ½( M + m1 + m2)v2
(10 m/s2)(1.0 m)(0.20 kg) = (0.625 kg)v2 v = 1.8 m/s
Um – Wf = K'b + K'r-p + K'm
mmgh – mbgd = ½mbv2 + ½ mpv2 + ½mmv2
(14)(10)(1) – (.25)(20)(10)(1) = (10 + 1 + 7)v2  v = 2.2 m/s
L = r mv = (1 m)(1)(0.2 kg) (9 m/s) = 1.8 kg•m2/s
L = r mv = r(2/5)m(2r/T) = 4mr2/5T
L = 4(6.0 x 1024 kg)(6.4 x 106 m)2/(5)(60 x 60 x 24 s)
L = 7.1 x 1033 kg•m2/s
L = r mv = r(1)m(2r/T) = 2mr2/T
L = 2(6.0 x 1024 kg)(1.5 x 1011 m)2/(60 x 60 x 24 x 365 s)
L = 2.7 x 1040 kg•m2/s
r1v1 = r2v2  v1/v2 = r2/r1 = (5.3 x 1012 m)/(8.9 x 1010 m) = 60
rC CmCvC + rMMmMvM = (rC CmC + rM MmM)v'
(1)(42 kg)(3 m/s) + 0 = [(1)(42 kg) + (½)(180 kg)]v'
v' = 0.95 m/s
Ltotal = rd dmdvd + rr rmrvr = (R)(½)(M)(V) + (R)(1)(M)(0) = ½RMV
Ltotal = Ltotal' = (rd dmd + rr rmr)v'
½RMV = (½RM + RM)v'  v' = ⅓V
mgh = ½mv2  v = (2gh)½ = [(2)(10 m/s2)(20 m)]½ = 20 m/s
r mTvT + r mJvJ = (r mT + rmJ)v'
(100 kg)(20 m/s) + 0 = (145 kg)v'  v' = 13.8 m/s
mgh = ½mv2  h = v'2/2g = (13.8 m/s)2/(2)(10 m/s2) = 9.5 m
No
Start from a higher ledge or with a running start.
v' = (2gh)½ = [(2)(10 m/s2)(10 m)]½ = 14 m/s
rmTvT = r(mT + mJ)v'
vT = (145 kg)(14 m/s)/100 kg = 20.3 m/s
h = v2/2g = (20.3 m/s)2/(2)(10 m/s2) = 20.6 m
½mv2 + mgh = ½mv'2
½v2 + (10 m/s2)(20 m) = ½(20.3)2  v = 3.5 m/s
Ug = Krolling  mgh = ½(1 + ½)mv2
(10 m/s2)(1 m) = ¾v2  v = 3.65 m/s
d = ½(v + vo)t  2 m = ½(3.65 m/s + 0)t t = 1.1 s
same
same
less
more
same
more
K' = Ug – Wf = mmgh – mbgd
K' = (5 kg)(10 m/s2)(1 m) – (0.3)(10 kg)(10 m/s2)(1 m) = 20 J
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K' = K'b + K'r-p + K'm = ½mbv2 + ½mpv2 + ½mmv2
20 J = (5 kg + 0.25 kg + 2.5 kg)v2 v = 1.6 m/s
(8.9 x 1010 m)(3.88 x 104 m/s) = (5.3 x 1012 m)v2  v2 = 652 m/s
L = r mv = r(2/5)m(2r/T) = 4mr2/5T
L = 4(7.35 x 1022 kg)(1.74 x 106 m)2/(5)(2.42 x 106 s)
L = 2.3 x 1029 kg•m2/s
L = r mv = r(1)m(2r/T) = 2mr2/T
L = 2(7.35 x 1022 kg)(3.84 x 108 m)2/(2.42 x 106 s)
L = 2.8 x 1034 kg•m2/s
r smsvs + r MmMvM = (r sms + r mmm)v'
(1)(75)(5) + (½)(150)(2)= [(1)(75) + (½)(150)]v'  v' = 3.5 m/s
Ug = K  msgh = ½msvs2
vs = (2gh)½ = [2(10 m/s2)(2.0 m)]½ = 6.3 m/s
r mSvS + r mBvB = r (mS + mB)v'
(1 kg)(6.3 m/s) + 0 = (1 kg +2 kg)v'  v' = 2.1 m/s
Ug = K  (mS + mB)gh = ½(mS + mB)v'2
h = v'2/2g = (2.1 m/s)2/(2)(10 m/s2) = 0.22 m
v' = mBvB/(mB + mS) = 12.6 kg•m/s/3 kg = 4.2 m/s
h = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m
vS + vS' = vB + vB'
6.3 m/s + vS' = 0 + vB'  vS' = vB' – 6.3
r mSvS + r mBvB = rmSvS' + r mBvB'
(1 kg)(6.3 m/s) = (1 kg)(vB' – 6.3) + (2 kg)vB' vB' = 4.2 m/s
vS' = vB' – 6.3 = 4.2 m/s – 6.3 m/s = -2.1 m/s
hB = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m
hS = v'2/2g = (-2.1 m/s)2/(2)(10 m/s2) = 0.22 m
mSghS + mBghB = (1 kg)(2 m) + 0 = 2 J
mBgh'B + mSgh'S = (2)(0.88) + (1)(0.22) = 1.98 J yes
D—starting at 0: down A, up A, up A, down A
A—you end where you started
D—Each A is ¼T  6 x ¼T = 3/2T
D—when x = 0, a = 0, but v = max; when x = A, v = 0, but a =
max; at intermediate points a  0 and v  0
B—E = ½kA2  mass will not change the total energy
D—E = ½kA2, whereas v = A(k/m)½ and a = A(k/m) change the
same, and T = 2(m/k)½ doesn't change at all
C—T = 2(m/k)½: when m doubles T increases by 2
C—T = 2(L/g)½: only changing L or g can change T
A—T = 2(L/g)½: greater L = greater T
A—T = 2(L/g)½: to speed up the clock (reduce T), you need to
decrease L by moving the weight up
C—T = 2(L/g)½: "g" increases when accelerating up  T
decreases
A—T = 2(m/k)½: adding mass or changing the spring will
change the period
B—T = 2(L/g)½: amplitude has no effect on T
C—T = 2(L/g)½: by standing on the swing, you have reduced
the distance L (distance from fulcrum to center of mass),
which decreases T
B—T = 2(L/g)½: mass has no effect on T
½T
¼ T, ¾ T
0 T, 1 T
¾T
0 T, ½ T, 1 T
¼T
aA = A(k/m) = (0.1 m)(100 N/m)/(1 kg) = 10 m/s2
vo = A(k/m)½ = (0.1 m)[(100 N/m)/(1 kg)]½ = 1 m/s
T = 2(m/k)½ = 2(1/100)½ = 0.63 s
Ko = ½mv2 = ½(1 kg)(1 m/s)2 = 0.5 J
UA = ½kA2 = (100 N/m)(0.1 m)2 = 0.5 J
v = vo[1 – (x2/A2)]½ = (1 m/s)[1 – 0.52/0.12)]½ = 0.866 m/s
0m
-0.1 m
0m
0.1 m
-1 m/s
0 m/s
1 m/s
0 m/s
0 m/s2
10 m/s2
0 m/s2
-10 m/s2
0N
10 N
0N
- 10 N
doubles
doubles
remains the same
UG = K  mgh = ½mv2
v = (2gh)½ = [2(10 m/s2)(0.015 m)]½ = 0.55 m/s
T = 2(L/g)½ = 2(1.0 m/10 m/s2)½ = 2.0 s
2.0 s
4.0 s
1.0 s
¾T
0 T, ½ T, 1 T
¼T
0 T, 1 T
¼ T, ¾ T
½T
aA = A(k/m) = (0.25 m)(100 N/m)/(1 kg) = 25 m/s2
vo = A(k/m)½ = (0.25 m)[(100 N/m)/(1 kg)]½ = 2.5 m/s
T = 2(m/k)½ = 2(1/100)½ = 0.63 s
Ko = ½mv2 = ½(1 kg)(2.5 m/s)2 = 3.125 J
UA = ½kA2 = ½(100 N/m)(0.25 m)2 = 3.125 J
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vx = vo[1 – (x2/A2)]½ = 2.5 m/s[1 – (.22/.252)]½ = 1.5 m/s
+0.25 m
0m
-0.25 m
0m
0 m/s
-2.5 m/s
0 m/s
2.5 m/s
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-25 N
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aA = |-A(k/m)| = (0.50 m)(100 N/m)/(1.0 kg) = 50 m/s2
vo = A(k/m)½ = (0.50 m)[(100 N/m)/(1.0 kg)]½ = 5 m/s
T = 2(m/k)½ = 2(1/100)½ = 0.63 s
Ug = K  mgh = ½mv2 
v = (2gh)½ = [2(10 m/s2)(0.20 m)]½ = 2 m/s
T = 2(L/g)½ = 2(2.0 m/10 m/s2)½ = 2.8 s
Practice Multiple Choice (No calculator)
C—In uniform circular motion velocity is tangent to the circle
and acceleration is toward the center
A—Traveling east and turning south  clockwise.
ac = v2/r  v = (acr)½ = [(3 m/s2)(300 m)]½ = 30 m/s
D—Ball simultaneously is directed to the right (due to the
spring) and toward the top (due to the rotation)
A—Fg = GMm/r2, where r is doubled
 Fg should be (½)2 = ¼ as much  ¼(800 N) = 200 N
A—At Q, acceleration is , at R acceleration is 
B—The center of mass is the balance point, which is closer to
the more massive side
B—Ball will move tangent to the circle at that spot
B—Fc = mv2/r = (5.0 kg)(2.0 m/s)2/0.70 m = 29 N
A—The force of gravity is the only force acting on the ball
and Fg decreases as r increases (Fg = GMm/r2)
D—g = GM/r2, r is doubled  g is (½)2 = ¼ as much
B—F = F: 250 N + FL = 125 N + 500 N  FL = 375 N
B—rFperson + rFboard = rFright chain
r(500 N) + (2 m)(125 N) = (4 m)(250 N) 
A—C and D would cancel torque, B would cancel up and
down, but only A will cancel both.
C—Fg = GMm/r2, where r is doubled  Fg is (½)2 = ¼ as much.
B—g = GM/r2, where r is doubled
 g is (½)2 = ¼ as much. ¼(10 m/s2) = 2.5 m/s2
B—rFperson = rFplank (rplank is 2.5 m – 2 m = 0.5 m)
r(50 kg)(10 m/s2) = (0.5 m)(100 kg)(10 m/s2)  r = 1 m
C—CM = (rimi)/(mi)—assume the rod is 4 m
2 = [(0)(3.5) + (1)(X) + (4)(5)]/(8.5 + X)  X = 3
D—T = Fc = mv2/r = mv'2/r': r increases to 4r without changing
T  v'2 also increase by 4. v;2 = 4  v' = 2v
A—At the bottom of the swing, Fnet = Fc = Ft – Fg
Fc = 6 N – (0.4 kg)(10 m/s2) = 2 N
C—net = cc – c
net = (3R)F + (3R)F + (2R)F – (3R)(2F) = 8RF – 6RF = 2RF
B—g = GM/r2  gE = GME/rE2 and GM = GMM/rM2
gM/gE = (MM/ME)(rE/rM)2 = (1/10)(1/½)2 = 4/10  gM = 2g/5
C—Fc = Fg  mv2/r = GMm/r2  r = GM/v2
C—Rm + (rcos60)M = R(2M)  m + ½M = 2M  m = 3/2M
D—I1m = l2M1 and l1M2 = l2m  l1/l2 = M1/m = m/M2 
M1M2 = m2  m = (M1M2)½
D—Energy & angular momentum are conserved,
r  v  (r1v1 = r2v2), Ug increases (Ug = -GMm/r)
A—r1v1 = r2v2  v2 = (r1/r2)v1
D—I is true (Fg = Fc = Mv2/R), II is true (F r  no work done),
and III is true (angular momentum is constant)
A—Ug = K (no rolling because of zero friction)
mgh = ½mv2 v = 2gh = (2gh)½
D—mgh = K + Kr = ½mv2 + ½ mv2 = ½(1 +  )mv2
v2 = 2gh/(1 + 2/5)  v = (10gh/7)½
B—A has no force, C and D have a constant force, only SHM
has a variable force ( Fs = kx)
B—L = r mv = (a)(1)(m)(v) = amv
C—Total mechanical energy is constant when a conservative
force is involved
D—The kinetic energy is greatest when the velocity is
greatest (K = ½mv2), which occurs at the midpoint
B—Potential energy is greatest when x is greatest
(Ug = ½kx2), which occurs at xmin and xmax,
B—Acceleration is greatest when force is greatest (Fs = kx),
which occurs at xmin and xmax
D—Velocity is greatest at the midpoint and is zero at xmin and
xmax (same as K)
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D—T = 2(m/k)½  T depends on both m and k, but not on
amplitude, A.
D—T = 2(m/k)½ T m.
The mass is doubled  T = 22 = 2.8 s
D—The period of a pendulum, T = 2(L/g)½  T  (1/g)½ and it
is shorter, but the period of a spring is the same
C—Potential energy is greatest when x is greatest
(Ug = ½kx2), which occurs at 1 s and 3 s.
A—Each spring supports half of the weight (12 N).
Fs = kx  6 N = k(0.15 m)  k = 40 N/m
The force on ball is constant, except when it is in contact with
the floor,  not SHM, where force  x..
B—Pendulum: T = 2(L/g), T = T
Spring: T = 2(m/k), where T m  T = 2T
A—Fs = kx  mg = kd  k = gm/d
T  (m/k)½ = (md/gm)½ = (d/g)½
D—The 4-kg block stretches the spring 16 cm (Fs  x), the
spring falls another 16 cm before turning around
D—v = A(k/m)½
v2 = A2k/M  k = Mv2/A2 = M(v/A)2
D—Tp = 2(L/g)½ = Ts = 2(m/k)½
L/g = m1/k  k = m1g/L
Practice Free Response
Ug = K  mgh = ½mv2
v = (2gh)½ = [2(10 m/s2)(90 m)]½ = 42 m/s
v = (2gh)½ = [2(10 m/s2)(40 m)]½ = 28 m/s
Fg = mg = (700 kg)(10 m/s2) = 7,000 N
Fc = mv2/r = (700 kg)(28 m/s)2/(20 m) = 27,000 N
Fc = Fg + Fn  Fn = Fc – Fg = 27,000 – 7,000 = 20,000 N
F = mbg = (0.30)(20 kg)(10 m/s2) = 60 N
Fg = mmg = (10 kg)(10 m/s2) = 100 N
Fnet = 100 N – 60 N = 40 N
Fnet = (mb +  mp + mm)a
40 N = (20 kg + 5 kg + 10 kg)a  a = 1.1 m/s2
v2 = vo2 + 2ad = 02 + 2(1.1 m/s2)(2 m)  v = 2.1 m/s
Ug = mmgh = (10 kg)(10 m/s2)(2 m) = 200 J
Wf = mbgd = (.30)(20 kg)(10 m/s2)(2 m) = 120 J
Um – Wf = K'b + K'r-p + K'm = ½mbv2 + ½ mpv2 + ½mmv2
200 J – 120 J = ½(20 kg + 5 kg + 10 kg)v2  v = 2.1 m/s
Ug = mg(H + Lsin)
Ug = (0.5 kg)(10 m/s2)[1 m + (2 m)(sin30)] = 10 J
Ug = Krolling mgLsin = ½(1 +  )mv2
(10 m/s2)(2 m)sin30 = ½(1 + 1)v2  v = 3.2 m/s
Kfloor = Ktable + Utable  ½mvf2 = ½mvt2 + mgh
vf2 = (3.2 m/s)2 + 2(10 m/s2)(1 m) vf = 5.5 m/s
K = ½mv2 = ½(0.5 kg)(5.5 m/s)2 = 7.5 J
(10 J – 7.5 J)/10 J x 100 = 25 %
Ug = Krolling  mgLsin = ½(1 +  )mv2
(10 m/s2)(2 m)sin30 = ½(1 + 2/5)v2 = 7/10 v2  v = 3.8 m/s
Kfloor = Ktable + Utable  ½mvf2 = ½mvt2 + mgh
vf2 = (3.8 m/s)2 + 2(10 m/s2)(1 m) = 34 m2/s2vf = 5.8 m/s
K = ½mv2 = ½(0.5 kg)(5.8 m/s)2 = 8.4 J
(10 J – 8.4 J)/10 J x 100 = 16 %
The hoop (largest  = largest % rotational energy)
The sphere (smallest  = most translational energy)
All three would have the same amount of kinetic energy
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U = 0.05 J
10 cm, the mass stops at this point because all of the energy
is in the form of potential energy.
T = 2(L/g)½ = 2(1.0 m/10 m/s2)½ = 2.0 s  1.0 s
K = 0.4 J – U = 0.4 J – 0.2 J = 0.2 J
K = ½mv2  0.4 J = ½(3.0 kg)v2  v = 0.5 m/s
Ug = 2mgh = 2(1 kg)(10 m/s2)(0.50 m) = 10 J
Ug = K = ½mv2 + ½(1 +  )mv2 = 5/4 mv2
10 J = 5/4(2 kg)v2 v = 2.8 m/s
Kt = ½mv2 = ½(2 kg)(2.8 m/s)2 = 8 J
mcvc + mbvb = (mc + mb)v'
2mvc + 0 = 5mv'  v' = 2/5vc = 1.1 m/s
Kt' = ½mv2 = ½(5 kg)(1.1 m/s)2 = 3.2 J
K = Us = ½kx2  3.2 J = ½(250 N/m)x2  x = 0.16 m