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Chapter 8
Rotational Motion of Solid Objects
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What Is Rotational
Motion?
What does a yo-yo
have in common with
a merry-go-round?
How can we
describe this
type of motion?
Linear Motion vs.
Rotational Motion
 Linear motion involves an object moving
from one point to another in a straight line.
 Rotational motion involves an object rotating
about an axis.

Examples include a merry-go-round, the rotating
earth, a spinning skater, a top, and a turning
wheel.
 What causes rotational motion?
 Does Newton’s second law apply?
 There is a useful analogy between linear motion and
rotational motion.

Rotational velocity is how fast the object is turning.


Units: revolutions per minute (rpm); degrees per second
Analogous to linear velocity

Rotational displacement is how far the object rotates.



Units: fractions of a complete revolution; degrees; radians
1 complete revolution = 360º = 2 radians
Analogous to linear displacement: the straight-line distance traveled
by an object (including direction of travel)

Rotational acceleration is the rate of change of rotational
velocity.


Units: revolutions per second per second (rev/s2); radians per second
per second (rad/s2)
Analogous to linear acceleration
 Constant acceleration equations for linear and
rotational motion
v  v 0  at
   0  t
1 2
d  v 0 t  at
2

1 2
   0 t  t
2

 Relationship between linear and rotational velocity
On a merry-goround, a rider near
the edge travels a
greater distance in
1 revolution than
one near the
center.
 The outside rider
is therefore
traveling with a
greater linear
speed.

v  r
A merry-go-round is accelerated at a constant rate
of 0.005 rev/s2, starting from rest.
What is its rotational velocity at the end of 1 min?
a)
b)
c)
d)
0.005 rev/s
0.03 rev/s
0.05 rev/s
0.30 rev/s
 = 0.005 rev/s2
0 = 0
t = 60 s
d)  = 0 + t
= 0 + (0.005 rev/s2)(60 s)
= 0.30 rev/s
How many revolutions does the merry-go-round make
in 1 minute?
a)
b)
c)
d)
1.5 rev
3.0 rev
9.0 rev
18.0 rev
 = 0.005 rev/s2
0 = 0
t = 60 s,  = 0.30 rev/s
c)  = 0t + 1/2 t2
= 0 + 1/2 (0.005 rev/s2)(60 s)2
= 9 rev
Torque and Balance
 What causes the merry-go-round to rotate in
the first place?
 What determines whether an object will
rotate?
 If an unbalanced force causes linear motion,
what causes rotational motion?
Torque and Balance
 When is a balance balanced?
 Consider a thin but rigid beam supported by a
fulcrum or pivot point.
 If equal weights are placed at equal distances
from the fulcrum, the beam will not tend to rotate:
it will be balanced.
 To
balance a weight twice as large as a smaller
weight, the smaller weight must be placed twice
as far from the fulcrum as the larger weight.
 Both the weight and the distance from the
fulcrum are important.
 The product of the force and the distance from
the fulcrum is called the torque.
 Torque describes the tendency of a weight to
produce a rotation.
 The
distance from the fulcrum to the point of
application of the force must be measured in a
direction perpendicular to the line of action of the
force.
 This distance is
called the lever arm
or moment arm.
 A longer lever arm
produces a greater
torque.
 For a force F and a
lever arm l, the
resulting torque is:
  Fl
Which of the forces pictured as acting upon
the rod will produce a torque about an axis
perpendicular to the plane of the diagram at
the left end of the rod?
a)
b)
c)
d)
b)
F1
F2
Both.
Neither.
F2 will produce a torque
about an axis at the left
end of the rod. F1 has no
lever arm with respect to
the given axis.
The two forces in the diagram have the same
magnitude. Which orientation will produce the
greater torque on the wheel?
a)
b)
c)
d)
F1
F2
Both.
Neither.
a) F1 provides the larger
torque. F2 has a smaller
component perpendicular
to the radius.
A 50-N force is applied at the end of a wrench
handle that is 24 cm long. The force is applied
in a direction perpendicular to the handle as
shown. What is the torque applied to the nut
by the wrench?
a)
b)
c)
d)
6 N·m
12 N·m
26 N·m
120 N·m
b) 0.24 m  50 N
= 12 N·m
What would the torque be if the force were
applied half way up the handle instead of at
the end?
a)
b)
c)
d)
6 N·m
12 N·m
26 N·m
120 N·m
a) 0.12 m  50 N = 6 N·m
 When
the applied force is not perpendicular to the
crowbar, for example, the lever arm is found by
drawing the perpendicular line from the fulcrum to
the line of action of the force.
 We call torques
that produce
counterclockwise
rotation positive,
and torques that
produce
clockwise rotation
negative.
Two forces are applied to a merry-go-round
with a radius of 1.2 m as shown. What is the
torque about the axle of the merry-go-round
due to the 80-N force?
a)
b)
c)
d)
+9.6 N·m
-36 N·m
+96 N·m
-96 N·m
c) 1.2 m  80 N
= +96 N·m
(counterclockwise)
What is the torque about the axle of the
merry-go-round due to the 50-N force?
a)
b)
c)
d)
+60 N·m
-60 N·m
+120 N·m
-120 N·m
b) -(1.2 m  50 N) = -60 N·m
(clockwise)
What is the net torque acting on the merrygo-round?
a)
b)
c)
d)
e)
a)
+36 N·m
-36 N·m
+96 N·m
-60 N·m
+126 N·m
96 N·m (counterclockwise)
- 60 N·m (clockwise)
= +36 N·m (counterclockwise)
We want to balance a 3-N weight against
a 5-N weight on a beam. The 5-N weight
is placed 20 cm to the right of a fulcrum.
What is the torque produced by the 5-N
weight?
a)
b)
c)
d)
+1 N·m
-1 N·m
+4 N·m
-4 N·m
b) F = 5 N
l = 20 cm = 0.2 m
 = - Fl
= - (5 N)(0.2 m)
= -1 N·m
How far do we have to place the 3-N
weight from the fulcrum to balance the
system?
a)
b)
c)
d)
2 cm
27 cm
33 cm
53 cm
c) F = 3 N
 = +1 N·m
l=/F
= (+1 N·m) / (3 N)
= 0.33 m = 33 cm
 The
center of gravity of an object is the point
about which the weight of the object itself exerts no
torque.
 We can locate the center of gravity by finding the
point where it balances on a fulcrum.

For a more complex object, we locate the center of gravity
by suspending the
object from two
different points,
drawing a line straight
down from the point of
suspension in each
case, and locating the
point of intersection of
the two lines.
 If
the center of
gravity lies below
the pivot point, the
object will
automatically regain
its balance when
disturbed.
 The center of gravity
returns to the
position directly
below the pivot
point, where the
weight of the object
produces no torque.
How far can
the child walk
without tipping
the plank?
 For a uniform plank, its center of gravity is at its geometric
center.
 The pivot point will be the edge of the supporting platform.
 The plank will not tip as long as the counterclockwise torque
from the weight of the plank is larger than the clockwise torque
from the weight of the child.
 The plank will verge on tipping when the magnitude of the
torque of the child equals that of the plank.
An 80-N plank is placed on a dock as shown. The
plank is uniform in density so the center of gravity of
the plank is located at the center of the plank. A
150-N boy standing on the plank walks out slowly
from the edge of the dock. What is the torque
exerted by the weight of the plank about the pivot
point at the edge of the dock?
a)
b)
c)
d)
e)
+80 N·m
-80 N·m
+160 N·m
-160 N·m
+240 N·m
a) 1 m  80 N = +80 N·m
(counterclockwise)
How far from the edge of the dock can the 150-N
boy walk until the plank is just on the verge of
tipping?
0.12 m
b) 0.23 m
c) 0.53m
d) 1.20 m
a)
c) 80 N·m / 150 N = 0.53 m
Rotational Inertia and
Newton’s Second Law
 In linear motion, net force and mass determine the
acceleration of an object.
 For rotational motion, torque determines the rotational
acceleration.
 The rotational counterpart to mass is rotational
inertia or moment of inertia.



Just as mass represents the resistance to a change in linear
motion, rotational inertia is the resistance of an object to
change in its rotational motion.
Rotational inertia is related to the mass of the object.
It also depends on how the mass is distributed about the
axis of rotation.
Simplest example:
a mass at the end of a light rod
 A force is applied to the
mass in a direction
perpendicular to the rod.
 The rod and mass will begin
to rotate about the fixed axis
at the other end of the rod.
 The farther the mass is from
the axis, the faster it moves
for a given rotational
velocity.
Simplest example:
a mass at the end of a light rod
 To produce the same
rotational acceleration, a
mass at the end of the rod
must receive a larger linear
acceleration than one nearer
the axis.
 It is harder to get the system
rotating when the mass is at
the end of the rod than when
it is nearer to the axis.
Rotational Inertia and
Newton’s Second Law
 The resistance to a change in rotational motion
depends on:


the mass of the object;
the square of the distance of the mass from the axis of
rotation.
 For an object with its mass concentrated at a point:
 Rotational inertia = mass x square of distance from axis
I = mr2
 The total rotational inertia of an object like a merry-goround can be found by adding the contributions of all
the different parts of the object.

Rotational Inertia and
Newton’s Second Law
 Newton’s second law for linear motion:
Fnet = ma
 Newton’s second law for rotational motion:

The net torque acting on an object about a given axis is
equal to the rotational inertia of the object about that axis
times the rotational acceleration of the object.
net = I

The rotational acceleration produced is equal to the torque
divided by the rotational inertia.
Example:
a baton with a mass at both ends
 Most of the rotational
inertia comes from the
masses at the ends.
 A torque can be applied at
the center of the rod,
producing a rotational
acceleration and starting
the baton to rotate.
 If the masses were moved
toward the center, the
rotational inertia would
decrease and the baton
would be easier to rotate.
Two 0.2-kg masses are located at either end
of a 1-m long, very light and rigid rod as shown.
What is the rotational inertia of this system
about an axis through the center of the rod?
a)
b)
c)
d)
0.02 kg·m2
0.05 kg·m2
0.10 kg·m2
0.40 kg·m2
c) I = mr2
= (0.2 kg)(0.5m)2 x 2
= 0.10 kg·m2
Rotational
inertias for
more complex
shapes:
Conservation of Angular
Momentum
How do
spinning
skaters or
divers
change
their
rotational
velocities?
Angular Momentum
 Linear momentum is mass (inertia) times linear
velocity:
p = mv
 Angular momentum is rotational inertia times
rotational velocity:
L = I


Angular momentum may also be called rotational
momentum.
A bowling ball spinning slowly might have the same angular
momentum as a baseball spinning much more rapidly,
because of the larger rotational inertia I of the bowling ball.
Conservation of Angular
Momentum
 Linear momentum is conserved if the net external
force acting on the system is zero.
 Angular momentum is conserved if the net external
torque acting on the system is zero.
Inertia m : Fnet  ma
Inertia I :  net  I
p  mv
L  I
If Fnet  0,
p  constant
If  net  0, L  constant
1 2
1 2
KE  mv
KE  I
2
2
Angular momentum is conserved by
changing the angular velocity
 When the masses are
brought in closer to the
student’s body, his
rotational velocity increases
to compensate for the
decrease in rotational
inertia.
 He spins faster when the
masses are held close to
his body, and he spins
more slowly when his arms
are outstretched.
Angular momentum is conserved by
changing the angular velocity
 The diver increases her
rotational velocity by pulling
into a tuck position, thus
reducing her rotational
inertia about her center of
gravity.
Kepler’s Second Law
 Kepler’s second law says
that the radius line from
the sun to the planet
sweeps out equal areas in
equal times.
 The planet moves faster in
its elliptical orbit when it is
nearer to the sun than
when it is farther from the
sun.
Kepler’s Second Law
 This is due to conservation
of angular momentum.
 The gravitational force
acting on the planet
produces no torque about
an axis through the sun
because the lever arm is
zero: the force’s line of
action passes through the
sun.
Kepler’s Second Law
 When the planet moves
nearer to the sun, its
rotational inertia about the
sun decreases.
 To conserve angular
momentum, the rotational
velocity of the planet about
the sun must increase.
Riding a bicycle and
other amazing feats
Why does a
bicycle remain
upright when it
is moving but
promptly falls
over when not
moving?
Angular momentum is a vector
 The direction of the rotational-velocity vector is given
by the right-hand rule.
 The direction of the angular-momentum vector is the
same as the rotational velocity.
Inertia I, rotational velocity 
Angular momentum : L  I
Angular momentum and bicycles
 The wheels have angular
momentum when the
bicycle is moving.
 For straight line motion,
the direction of the
angular-momentum vector
is the same for both
wheels and is horizontal.
 To tip the bike over, the
direction of the vector
must change, requiring a
torque.
Angular momentum and bicycles
 If the bike is not perfectly upright, a gravitational
torque acts about the line of contact of the tires with
the road.
 As the bike begins to fall, it acquires a rotational
velocity and angular
momentum about
this axis.
 If the bike tilts to the
left, the change in
angular momentum
points straight back.
Angular momentum and bicycles
 If the bike is standing still, the gravitational torque
causes the bike to fall.
Angular momentum and bicycles
 When the bike is moving, the change in angular
momentum caused by the gravitational torque adds to
the angular momentum already present from the
rotating tires.
 This causes a
change in the
direction of the
total-angular
momentum vector
which can be
accommodated by
turning the wheel.
A student holds a spinning bicycle wheel
while sitting on a stool that is free to
rotate. What happens if the wheel is
turned upside down?
To conserve
angular momentum,
the original direction
of the angularmomentum vector
must be maintained.
A student holds a spinning bicycle wheel
while sitting on a stool that is free to
rotate. What happens if the wheel is
turned upside down?
The angular
momentum of the
student and stool,
+Ls, adds to that of
the (flipped) wheel,
-Lw, to yield the
direction and
magnitude of the
original angular
momentum +Lw.
A student sits on a stool holding a bicycle wheel with a
rotational velocity of 5 rev/s about a vertical axis. The
rotational inertia of the wheel is 2 kg·m2 about its center and
the rotational inertia of the student and wheel and platform
about the rotational axis of the platform is 6 kg·m2. What is
the initial angular momentum of the system?
a)
b)
c)
d)
e)
10 kg·m2/s upward
25 kg·m2/s downward
25 kg·m2/s upward
60 kg·m2/s downward
60 kg·m2/s upward
e) L = I = (2 kg·m2)(5 rev/s)(2π rad / rev)
≈ 60 kg·m2/s
upward from plane of wheel
If the student flips the axis of the wheel, reversing the
direction of its angular-momentum vector, what is the
rotational velocity of the student and stool about their axis
after the wheel is flipped?
a)
b)
c)
d)
e)
1.67 rad/s
3.33 rad/s
20 rad/s
60 rad/s
120 rad/s
c)  = L / I =2 (60 kg·m2/s) / (6 kg·m2)
≈ 20 rad/s
in direction of original angular velocity
Where does the torque come from that accelerates the
student and the stool?
The student exerts forces on
the handles when he flips the
wheel, producing the torque.