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Chapter 8: Rotational Motion
Radians
1) 1 radian = angle subtended by an arc (l)
whose length is equal to the radius (r)
2)
q=l
r
r
l
Convert the following:
a) 20o to radians (0.35 rad)
b) 20o to revolutions (0.056 rev)
c) 5000o to radians (87.3 rad)
d) 0.75 rev to radians (4.7 rad)
e) 0.40 radians to degrees (23o)
q
3) 3600 = 2p radians = 1 rev
4) Radians are dimensionless
A bird can only see objects that subtend an
angle of 3 X 10-4 rad. How many degrees is
that?
3 X 10-4 rad
360o = 0.017o
2p rad
How small an object can the bird
distinguish flying at a height of
100 m?
q=l
r
q
r
l (approx.)
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At what height would the bird be
able to just distinguish a rabbit that
is 30 cm long (and tasty)?
A tiny laser beam is aimed from the earth to the
moon (3.8 X 108 m). The beam needs to have a
diameter of 2.50 m on the moon. What is the
angle that the beam can have?
q
r
(ANS: 1000 m)
l (approx.)
Convert:
a)
b)
c)
d)
0.0200 rev/s to radians/s (0.126 rad/s)
30.0o/s to radians/s (0.524 rad/s)
1.40 rad/s to rev/s (0.223 rev/s)
3000 rpm to radians/s (314 rad/s)
The Mighty Thor
swings his hammer
at 400 rev/min.
Express this in
radians/s.
400 rev 1 min 2p rad
1 min 60 s 1 rev
= 41.9 rad/s
(6.6 X 10-9 radians)
Determine the Arc length of your vision
Calculate your arc length of vision at 100 meters.
q
l (at 100 m)
Degrees
Revolutions
Radians
Angular Velocity
v = Dx
Dt
w = Dq
Dt
Angular Velocity – radians an object rotates
per second
–
–
All points on an object rotate at the EXACT
same angular velocity
All points on an object DO NOT rotate with
the same linear speed. (the farther out the
faster)
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Merry-Go-Round
Example
wa = wb
(both rotate through 3600
in the same time
period)
va < vb
(Point b travels a longer
distance around the
circle)
vb
If a person at point b
flies off the merrygo-round, will they
travel in a curve or
straight line?
va
a
b
b
Frequency and Period
Angular Acceleration
a = Dv
Dt
a = Dw
Dt
Frequency = Revolutions per second (rev/s)
Period = Time for one complete revolution
1) The angular acceleration of all points on a
circle is the same.
2) All points on an object DO NOT experience the
same linear acceleration. (the farther out the
more acceleration)
Converting between Angular and
Linear Quantities
Linear = Radius X Angular
v = rw
atan = ra
Note the use of atan to differentiate from
centripetal acceleration, ac or ar:
atan
ar
w = 2p f
T=1
f
One child rides a merry-go-round (2 revolutions per
minute) on an inside lion 2.0 m from the center. A
second child rides an outside horse, 3.0 m from the
center.
a) Calculate the frequency and the period. (0.0333 Hz, 30
s)
b) Calculate the angular velocity in rad/s. (0.209 rad/s)
c) Calculate the linear velocity of each rider. (0.419 m/s,
0.628 m/s)
d) Who has a higher angular velocity?
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A clock has a seconds hand that is 7.00 cm long.
a) Calculate the frequency and period. (0.0167 Hz, 60 s)
b) Calculate the angular velocity of the second hand in
radians/s. (0.105 rad/s)
c) Calculate the angular velocity in degrees/s. (6o/s)
d) Calculate the linear velocity at the end of the seconds
hand. (7.35 X 10-3 m/s)
A baseball bat is found to have a linear
acceleration of 13.9 m/s2 at the “sweet spot.”
The sweet spot is at 59 cm from the handle.
a) Calculate the angular acceleration in rad/s2 (a =
23.6 rad/s2)
b) Convert the angular acceleration to rev/s2 (3.75
rev/s2)
A golf club has an angular acceleration of 20.0 rad/s2
and is 114 cm long.
a) Calculate the tangential acceleration of the club. (22.8
m/s2)
b) Convert the angular acceleration to rev/s2. (3.18 rev/s2)
c) Calculate the force the club could give to a 45.93 g golf
ball. (1.05 N)
d) Calculate the velocity of the ball if the club is in
contact with the ball for 50 cm. (4.77 m/s)
Angular Kinematics
A bike wheel starts at 2.0 rad/s. The cyclist
accelerates at 3.5 rad/s2 for the next 2.0 s.
a) Calculate the wheel’s new angular speed (9.0
rad/s)
b) Calculate the number of revolutions. (1.75 revs)
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A DVD rotates from rest to 31.4 rad/s in 0.892 s.
a) Calculate the angular acceleration. (35.2 rad/s2 )
b) How many revolutions did it make? (2.23 rev)
c) If the radius of the disc is 4.45 cm, find the
linear speed of a point on the outside edge of the
disc. (1.40 m/s)
A spinning bike tire of radius 33.0 cm has an
angular speed of 50.0 rad/s. Twenty seconds
later, its angular speed is 150.0 rad/s.
a) Calculate the angular acceleration. (5.00 rad/s2)
b) Calculate the angular displacement over the 20
s. (2000 radians)
c) Calculate the revolutions travelled in the 20 s
(318 rev)
d) Calculate the linear distance the tire travelled in
20 s. (660 m)
A game show wheel with a 90 cm radius is initially
turning at 3.0 rev/s. A point on the outside of the
wheel travels 147 meters before stopping.
a) Calculate how many revolutions it went through.
(26 rev)
b) Calculate the angular deceleration. (-1.09 rad/s2)
c) Calculate how long it took to stop. (17.3 s)
d) Calculate the initial linear speed and linear
acceleration (tangential) on the outside of the
wheel (17.0 m/s, -0.98 m/s2)
e) Calculate the initial centripetal acceleration on a
point at the edge of the wheel. (319 m/s2)
A bicycle slows from vo = 8.4 m/s to rest over a
distance of 115 m. The diameter of each wheel
is 68.0 cm.
a. Calculate the angular velocity of the wheels
before braking starts. (24.7 rad/s)
b. How many revolutions did each wheel undergo?
(HINT: calculate the circumference of the
circle first) (53.8 rev)
c. Calculate the angular acceleration. (-0.903
rad/s2)
d. Calculate the time it took the bike to stop (27.4
s)
A pottery wheel turning with an angular speed
of 30.0 rev/s is brought to rest in 60.0
revolutions.
a) Calculate the radians that the wheel travelled
(377 rad)
b) Convert the initial speed to rad/s (188 rad/s)
c) Calculate the angular acceleration. (-46.9 rad/s2)
d) Calculate the time required to stop. (4.00 s)
e) If the radius of the wheel is 12.0 cm, calculate
the linear distance the outside of the wheel
travelled. (45.2 m)
A curve ball is given an initial angular speed of 40.0
rad/s. In the 0.500 s it takes to reach the catcher, it
has slowed to 36.0 rad/s. The diameter of the ball is
7.50 cm.
a) Calculate the angular acceleration of the ball. (-8
rad/s2)
b) Calculate the number of revolutions the ball made.
(3.02 rev)
c) Calculate the tangential acceleration. (-0.300 m/s2)
d) The ball is thrown at an initial velocity of 36.0 m/s.
Assume that air drag is –1.00 m/s2. Calculate the
distance to home plate. (17.9 m)
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A jet turbine (d=65 cm) is spinning with an
angular velocity of 10 rad/s. It is accelerated
at a rate of 5 rad/ s 2 for 15 s.
a. Calculate the final angular velocity of the
engine. (85 rad/s)
b. Calculate the number of revolutions the turbine
made. (113 rev)
c. Calculate the final linear speed of a point on the
outside of the turbine. (27.6 m/s)
d. Calculate the centripetal acceleration at a point
on the outside of the turbine at the end. (2344
m/s2)
Friction and Rolling Wheels
Rolling uses static friction
– A new part of the wheel/tire is coming in contact with
the road every instant
B
A
Torque
Braking uses kinetic friction
Torque – tendency of a force to rotate a body
about some axis (the force is always
perpendicular to the lever arm)
Point A gets
drug across
the surface
t = Fr
r
pivot
F
A
Torque Sign Conventions
Counter-clockwise
Torque is positive
Torque: Example 1
A wrench is 20.0 cm long and a 200.0 N force is
applied perpendicularly to the end. Calculate the
torque.
20.0 cm
Clockwise
Torque is negative
200.0 N
t = 40.0 m-N
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Torque: Example 2
First we need to resolve the Force vector into x and
y components
200.0 N
Suppose that same 200.0 N force is now applied at
a 60o angle as shown. Calculate the Torque. Is it
greater or less?
200.0 N
60o
Torque: Example 3
The biceps muscle exerts a 700
N vertical force. Calculate
the torque about the elbow.
Only Fy has any effect on the torque
(perpendicular)
Fy = Fsinq = (200.0 N)(sin 60o) = 173.2 N
t = Fr = (173.2 N)(0.20 m) = 34.6 m-N
A force of 200 N acts tangentially on the rim of a
wheel 25 cm in radius.
a) Calculate the torque. (50 mN)
b) Calculate the torque if the force makes an angle
of 40o to a spoke of the wheel. (32.3 mN)
c) If the wheel is mounted vertically, draw a free
body diagram of the wheel if the force is the one
in (a).
t = Fr = (700 N)(0.050 m)
t = 35 m-N
Torque: Example 4
Two wheels, of radii r1 = 30 cm and r2 =50 cm are
connected as shown. Calculate the net torque on
this compound wheel when two 50 N force act as
shown.
50 N
30o
r1
60o
Fx
20.0 cm
r2
Fy
First find the horizontal component of the top
force:
Fx = (50 N)(cos 30o) = 43 N
The top force is pulling clockwise (-) and the
bottom force pulls counterclockwise (+)
St = F1r1 – F2r2
St = (50N)(0.30m) – (43 N)(0.50 m) = -6.5 m-N
50 N
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Two children push on a merry-go-round as
shown. Calculate the net torque on the
merry-go-round if the radius is 2.0 m.
A 60.0 cm diameter wheel is pulled by a 500.0 N
force. The force acts at an angle of 65.0o with
respect to the spoke. Assume that a frictional
force of 300.0 N opposes this force at a radius of
2.00 cm. Calculate the net torque on the wheel.
(+130.0 mN)
(+1800 m N)
Moment of Inertia (I)
• Measure of Rotational Inertia
• An objects resistance to a change in angular
velocity
• Would it be harder to push a child on a
playground merry-go-round or a carousel?
Deriving I
Consider pushing a mass around in a circle (like
the child on a merry-go-round)
F
r
F = ma
a = ra
F = mra
m
t = Fr
t = mrar
t = mr2a
Moment of Inertia: Example 1
• I = moment of inertia
• I = mr2
• More properly
I = Smr2 = m1r12 + m2r22 +….
St= Ia
Calculate the moment of inertia (I) for the barbell
when rotated about point M. We will assume the
barbell is 1.0 m long, and that each weight is a
point mass of 45.4 kg. (Ans: 22.7 kg-m2)
Would it be harder (require more torque) to twirl a
barbell in the middle (pt. M) or the end (Pt. E)
E
M
M
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Moment of Inertia: Example 2
Now calculate I assuming Mr. Fredericks uses his
massive musculature to twirl the barbells from
point E. (Ans: 45.4 kg-m2)
E
Moment of Inertia: Example 3
Calculate I for Violet Beauregarde (40.0 kg, radius =
1.20 m). Use the formulas from the book.
I = 23.0 kg-m2
Calculate the moment of inertia of an 8.00 kg solid
wheel with a radius of 25.0 cm.(Ans: 0.25 kgm2)
A wheel has a moment of inertia of 0.50 kg m2.
a. Calculate the torque (St= Ia) is required to give it an
acceleration of 3 rad/s2. (1.5 mN)
b. Calculate its angular speed (from rest) after 5.00 s. (15
rad/s)
c. Calculate the number of revolutions it goes through in
5.00 s. (37.5 rad, 5.97 rev)
A 15.0 N force is applied to a
cord wrapped around a pulley
of radius 33.0 cm. The pulley
reaches an angular speed (w)
of 30.0 rad/s in 3.00 s.
a) Calculate the angular
acceleration. (10.0 rad/s2)
b) Calculate the torque (4.95 m-N)
c) Calculate the moment of inertia
of the pulley. (0.495 kg-m2)
33.0 cm
15.0 N
A 25.0 kg wheel has a radius of 40.0 cm. A 1.20
kg mass is hung on the end of the wheel by a
string and falls freely.
a) Calculate the moment of inertia of the wheel. (2
kg m2)
b) Setup the equation for all of the forces on the
mass.
c) Setup the equation for the torque on the wheel.
d) Using systems of equations, calculate the
tension in the cord and the tangential
acceleration of the wheel (and bucket) (10.7 N,
0.86 m/s2)
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A 15.0 N force is applied to a cord
wrapped around a pulley of
radius 33.0 cm. The pulley
reaches an angular speed (w) of
30.0 rad/s in 3.00 s. Since this is
a real pulley, there is a frictional
torque (tfr= 1.10 m-N) opposing
rotation.
a) Calculate net torque on the pulley.
(3.85 m-N)
b) Calculate the moment of inertia of
the pulley. (0.385 kg-m2)
Using the same pulley as the
previous problem (I=0.385 kg
m2), hang a 15.0 N bucket (1.53
kg) from the cord. Calculate the
angular and linear acceleration of
the bucket. Also calculate T.
Remember to include the
frictional torque. (tfr= 1.10 m-N)
33.0 cm
33.0 cm
T
15.0 N
We will break this problem into two parts: pulley
and bucket. Let’s first look at the pulley:
St = FTr – tfr
St = Ia
Ia = FTr – tfr
33.0 cm
tfr
FT
(0.385)a = FT(0.33) – 1.10
Now calculate the rotational speed of the pully (w) and the
linear speed of the bucket after 3.00 s.
w=wo + at
w= 0 + at = (6.98 rad/s2)(3.00 s) = 20.9 rad/s
v=rw
v = (0.330 m)(20.9 rad/s) = 6.90 m/s
Now we will deal with the bucket
SF = mg – FT
ma = mg – FT
1.53a = 15 – FT
a= ra1.53ra = 15 – FT
(1.53)(0.33)a = 15 – FT
FT
mg
0.505a = 15 – FT
0.385a = 0.33FT – 1.10
a = 7.07 rad/s2, a = 2.33 m/s2, FT = 11.5 N
A 2.00 kg bucket is attached to a 1.00
kg, 4.00 cm radius cylindrical
pulley. The bucket is suspended
2.00 m above the floor.
a) Calculate the moment of inertia of
the cylinder (1/2 MR2) (8X10-4
kgm2)
b) Calculate the acceleration of the
bucket. (7.84 m/s2)
c) Calculate the angular acceleration
of the pulley. (196 rad/s2)
d) Calculate how long it takes to reach
the floor. (0.714 s)
4.00 cm
T
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a) I = 1/2 MR2 = (1/2)(1.00 kg)(0.04 m)2
I = 8 X 10-4 kgm2
b) Bucket
Pulley
ma = mg –FT
t = FTR
2a = (2)(9.8) – FT
Ia = FTR
2a = 19.6 – FT
a = aR
Ia/R = FTR
8 X 10-4 a= FT(0.04)
0.04
c) a = aR
a = a/R = 7.84/0.04 = 196 rad/s2
d) y = yo + vt + ½at2
y = ½ at2
t = √(2y/a) = 0.714 s
a = -7.84 m/s2
A 1000.0 kg elevator is suspended by a 100.0 kg
cylinder pulley of radius 50.0 cm.
a) Calculate the moment of inertia of the pulley.
(12.5 kgm2)
b) Calculate the linear acceleration if the elevator
drops. (9.33 m/s2)
c) Calculate the angular acceleration of the pulley.
(18.7 rad/s2)
d) Calculate the tension in the cord. (467 N)
e) Are you glad you are not in the elevator?
Translational and Rotational Energy
• Does a rotating helicopter blade have kinetic
energy before the helicopter takes off?
• How about afterwards?
• Does all of the energy of the fuel go into moving
the helicopter?
Deriving the Rotational KE
Translational Speed (v)
– speed of the center of a wheel with respect to the
ground
– Can also be called linear speed
– Use regular K = ½ mv2
KE = S½ mv2
v = rw
KE = S ½ m(rw)2
KE = S ½ mr2w2
I = S mr2
KE = ½ Iw2
Rotational speed (w)
– angular speed of the wheel
– Use K = ½ Iw2
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Calculating Work
A pottery wheel of I = 0.200 kgm2 is accelerated
from rest to 500 rpm.
a) Calculate the final angular velocity in rad/s.
(52.4 rad/s)
b) Calculate the change in kinetic energy (274 J)
c) How much work is done in the process (274 J)
d) If the wheel takes 30 seconds to get up to speed,
calculate the angular acceleration (1.75 rad/s2)
Jupiter has a mass of 1.9 X 1027 kg and a radius of 7.0 X 107
m. Its day is only 9.92 hours long.
a) Calculate Jupiter’s moment of inertia, treating it as a sphere.
(3.72 X 1042 kgm2)
b) Calculate Jupiter’s rotational speed in rad/s. (1.76 X 10-4
rad/s)
c) Calculate Jupiter’s kinetic energy of rotation. (5.76 X 1034 J)
d) Jupiter is 7.78 X 1011 m from the sun. Calculate Jupiter’s
moment of inertia, assuming it is a point mass from the sun.
(1.15 X 1051 kgm2)
e) Calculate Jupiter’s orbital speed around the sun if it takes11.9
year to orbit the sun. (1.67 X 10-8 rad/s)
f) Calculate the kinetic energy about the sun. (1.60 X 1035 J)
Law of Conservation of Mechanical
Energy
(Kt + Kr + Ut)i = (Kt + Kr + Ut)f
Rotational KE: Example 1
What will be the translational speed of a log (100
kg, radius = 0.25 m, I= ½ mr2) as it rolls down a
4 m ramp from rest? (ANS: v = 7.2 m/s)
Rotational work
Wr = tDq
4m
Now we can calculate the angular speed
v = rw so w = v/r
w = 7.23 m/s = 29 rad/s
0.25 m
29 rad
1 rev
=
s
2p rad
Rotational KE: Example 2
What will be the translational speed of Bouncing
Boy (75 kg, radius = 1.2 m) as he rolls down a 3
m ramp from rest? (Treat him as a sphere)
4.6 rev/s
3m
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(KEt + KEr + PEt)i = (KEt + KEr + PEt)f
(0 + 0 + mgy)i = ( ½ mv2 + ½ Iw2 + 0)f
mgy = ½ mv2 + ½ Iw2
2mgy = mv2 + Iw2 (multiplied both sides by 2)
v = rw so w = v/r
2mgy = mv2 + Iv2
r2
2
I = 2/5 mr
2mgy = mv2 + 2mr2v2
5r2
2mgy = mv2 + 2mr2v2
5r2
Now we can calculate the angular speed
What would his speed be if he just slid down the
ramp?
(KEt + KEr + PEt)i = (KEt + KEr + PEt)f
(0 + 0 + mgy)i = ( ½ mv2 + 0 + 0)f
mgy = ½ mv2
gy = ½ v2
v2 = 2gy
v = \/2gy = (2 X 9.8 m/s X 3.00 m)1/2
v = 7.7 m/s
v = rw so w = v/r
w = 6.50 m/s = 5.4 rad/s
1.2 m
5.4 rad
1 rev
=
s
2p rad
0.86 rev/s
2gy = v2 + 2v2
5
2gy = 5v2 + 2v2
5
5
2gy = 7v2
5
v = 10gy = 10(9.8m/s2)(3.00 m) = 6.50 m/s
7
7
Why is this larger than if he rolls?
A wooden dowel (mass = 0.50 kg, r = 1.00 cm)
rolls down a 1.25 m tall ramp.
a) Calculate the moment of inertia (2.5 X 10-5
kgm2)
b) Calculate the linear speed at the bottom. (4.04
m/s)
c) Calculate the angular speed at the bottom. (404
rad/s, 64.3 rev/s)
A bike tire can be modeled as a thin hoop
(I=mr2). The wheel has a diameter of 66.0 cm
and a mass of 2.00 kg. The wheel is allowed to
roll down a ramp and has a linear
(translational) speed of 2.00 m/s at the bottom
of the ramp.
a) Calculate the moment of inertia of the wheel.
(0.218 kgm2)
b) Calculate the height of the ramp (40.8 cm)
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Conservation of Angular Momentum
• The total angular momentum of a rotating body
remains constant if the net torque acting on it is
zero
• Angular Momentum (L) is conserved
• p = mv (linear momentum)
• L = Iw (angular momentum)
• Iiwi = Ifwf
The skater in more detail
Iiwi = Ifwf
I = mr2
For now, let’s just consider the skater’s arms:
Arms out (initial)
Arms in (final)
Her mass will not change
when she moves her arms
mri2wi = mrf2wf
ri2wi = rf2wf
Since rf will be smaller (arms are in) wf must
increase to compensate
“I wanna go fast!”
A child sits on a merry-go-round at point A. If the
child wants to go faster, should he walk towards
the center or the outside?
A
Cons. of Angular Momentum: Ex. 1
A skater holds her arms at a length of 56 cm. She
spins at 9.43 rad/s. What will be her new speed
if she pulls her arms tight to her body, 20 cm?
Iiwi = Ifwf
I = mr2
mri2wi = mrf2wf
ri2wi = rf2wf
wf = ri2wi /rf2
wf = [(0.56 m)2(9.43 rad/s)]/(0.20 m)2 = 74 rad/s
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Cons. of Angular Momentum:
Example 2
A mass m is attached to the end of a string which
passes through a hole in a table. Initially the r1 is
0.80 m and the block moves at 2.4 m/s. What
will be the new speed as the radius of the string
is reduced to 0.48 m?
Old record players rotate at 33.3 rev/min.
Assume a record has a radius of 15.24 cm and
a mass of 100.0 grams.
a) Calculate the angular speed in rad/s (3.49 rad/s)
b) Calculate the moment of inertia. (1/2 mr2) (1.16
X 10-3 kgm2)
c) Calculate angular momentum of the record.
(4.05 X 10-3 kg m2/s)
d) A second record drops on top of the first.
Calculate the new moment of inertia and new
angular speed. (2.32 X 10-3 kg m2, 1.74 rad/s)
Merry-Go-Round
What will happen if you start to walk on a still
merry-go-round? Can you walk around it
without it moving?
Iiwi = Ifwf
I = mr2
mri2wi = mrf2wf
ri2wi = rf2wf
v = rw so w = v/r
ri2vi = rf2vf
ri
rf
rivi = rfvf
vf = rivi
=
(0.80m)(2.4 m/s) = 4.0 m/s
rf
(0.48 m)
Vectors and Angular
Momentum
• Direction of the vector w is
defined by the right-hand
rule.
• L = Iw
• L is defined as going in the
same direction.
• Assume this merry-goround is initially at rest
(Ltotal = 0)
• As the man walks
counterclockwise, the
platform spins clockwise
• Total Angular
Momentum is still zero.
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What will happen if the
student flips the
bicycle wheel while
standing on the merrygo-round?
A merry-go-round (500 kg) has a radius of 1.5 m. A
40.0 kg student is sitting on the merry-go-round as it
spins at 30.0 rev/min.
a) Calculate the angular speed (3.14 rad/s)
b) Calculate the moment of inertia of the merry-go-round
and student, assuming she sits at the edge. (652.5 kg
m2)
c) Calculate the angular momentum. (2049 kgm2/s)
d) A second student (80.0 kg) now jumps on the merrygo-round. Calculate the new moment of inertia and
angular speed. (832.5 kgm2, 2.46 rad/s)
A 20.0 kg box is placed at the top of a 25.0o inclined
plane. It is connected to a rope wrapped around
a 5.00 kg pulley with radius of 20.0 cm.
a) Calculate the acceleration of the box and the
tension in the cord. (3.68 m/s2, 9.21 N)
b) Assume a frictional torque of 2.00 mN. Recalculate
acceleration and tension. (3.24 m/s2, 18.1 N)
2. RSun = 6.5 X 105 km
4. 2900 m
6. 188 rad/s
8. 1.9 rad/s2
10. -260 rad/s2
12. 3300 revolutions
14. a) 1.99 X 10-7 rad/s
b) 7.27 X 10-5 rad/s
4
16. 3.6 X 10 rpm
20. 0.56 rad/s2
22. a) -84 rad/s2
b) 150 rev
24. 38 m
22. a) -84 rad/s2
b) 150 rev
24. 38 m
26. a) 0.58 rad/s2
b) 12 s
30. a) 38 m N
b) 33 m N
32. 650 m N
34. 1.89 kg m2
36. a) 6.1 kg m2
b) 0.61 kg m2
2
38. a) 0.851 kg m b) 0.0720 m N
40.1.99 X 10-3 kg m2 b) 0.0682 m N
42.590 m N, 240 N
46. 2250 kg m2, 8800 mN
42. 590 m N, 240 N
44. a) 7.5 m N
b) 300 N
46. 2250 kg m2, 8800 mN
50. 94 J
52. a) 2.6 X 1029 J b) 2.7 X 1033 J
54.a) 8.37 m/s, 41.8 rad/s b) 2.50 (mass
independent)
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58. a) angular momentum conserved b) 1.6
60. 0.77 kg m2 (pull arms close to body)
62. a) 14 kg m2/s
b) -2.7 m N
66.a) 1.2 rad/s
b) 2000 J, 1200 J (loss of
800J, 40% loss)
32. t = Ffrr = msmgr
t = (0.75)(1/4)(1080 kg)(9.8 m/s2)(0.33 m) = 650 mN
34. I = (2/5)MR2 = (2/5) (12.2)(0.623)2 = 1.89 kg m 2
35. I = MR2 = (1.25)(0.333)2 = 0.139 kg m 2
36. Ix = S MR2 = (1.8)(0.5)2 + (3.1)(0.5)2 + (1.8)(1)2 + (3.1)(1)2
Ix = 6.1 kg m2
Iy = S MR2 = (1.8)(0.25)2 + (3.1)(0.25)2 + (1.8)(0.25)2 +
(3.1)(0.25)2
Iy = 0.61 kg m2
37. 1.20 X 10-10 m
38. a) I = MR2 = (1.05)(0.900)2 = 0.851 kg m2
b) tnet = 0 0 =tapplied – tfriction
tapplied = FfrR = (0.0800 N)(0.900 m) = 0.0720 mN
53.w = (1rev/8s)(2p/1rev) = 0.785 rad/s
I = 1/2mr2 = (1/2)(1640)(8.20)2 = 5.51 X 104 kgm2
W = 1/2Iw2 = (1/2)(5.51 X 104 kgm2)(0.785 rad/s)2
W = 1.70 X 104 J
54. a) 8.37 m/s, 41.8 rad/s b) 2.50 (mass independ.)
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