Ch 9. Rotational Dynamics
Forces and Torques
Net force  acceleration.
In pure translational motion, all points on an
What causes an object to have an
object travel on parallel paths.
angular acceleration?  TORQUE
The most general motion is a combination of
translation and rotation.
The amount of torque depends on where
and in what direction the force is applied, as
well as the location of the axis of rotation. 1
9.1 The Action of Forces and Torques on Rigid Objects
DEFINITION OF TORQUE
Magnitude of Torque = (Magnitude of the force) x (Lever arm)
  F
Direction: The torque is positive when the force tends to produce a
counterclockwise rotation about the axis.
SI Unit of Torque: newton x meter (N·m)
2
Action of Forces and Torques
Example 2 The Achilles Tendon
The tendon exerts a force of
magnitude 790 N. Determine the
torque (magnitude and direction) of
this force about the ankle joint.
  F

cos 55 
3.6 10  2 m

  720 N 3.6 10 2 m cos 55
 15 N  m
3
Rigid Objects in Equilibrium
Reasoning Strategy
EQUILIBRIUM OF A RIGID BODY
1. Select the object to which the equations
for equilibrium are to be applied.
A rigid body is in equilibrium if it has zero
translational acceleration and zero
angular acceleration. In equilibrium,
the sum of the externally applied forces is
zero, and the sum of the externally
applied torques is zero.
ax  a y  0
 Fx  0
 Fy  0
 0
  0
2. Draw a free-body diagram that shows
all of the external forces acting on the
object.
3. Choose a convenient set of x, y axes
and resolve all forces into components
that lie along these axes.
4. Apply the equations that specify the
balance of forces at equilibrium. (Set
the net force in the x and y directions
equal to zero.)
5. Select a convenient axis of rotation.
Set the sum of the torques about this
axis equal to zero.
6. Solve the equations for the desired
unknown quantities.
4
Example 3 A Diving Board
A woman whose weight is 530 N is poised
at the right end of a diving board with
length 3.90 m. The board has negligible
weight and is supported by a fulcrum 1.40
m away from the left end. Find the forces
that the bolt and the fulcrum exert on the
board.
  F 
2 2
 W W  0
W W
F2 
2
F2

530 N 3.90 m 

 1480 N
1.40 m
F
y
 F1  F2  W  0
 F1  1480 N  530 N  0
F1  950 N
5
Example 5 Bodybuilding
The arm is horizontal and weighs 31.0 N.
The deltoid muscle can supply 1840 N of
force. What is the weight of the heaviest
dumbbell he can hold?
  W 
a a
 Wd  d  M M  0
 M  0.150 msin 13.0
Wd 
 Wa  a  M M
d
 31.0 N 0.280 m   1840 N 0.150 m sin 13.0

0.620 m
 86.1 N
6
Center of Gravity
The center of gravity of a rigid body is the
point at which its weight can be
considered to act when the torque due to
the weight is being calculated.
W1 x1  W2 x2  
xcg 
W1  W2  
7
Example 6 Center of Gravity of an Arm
The horizontal arm is composed of
three parts: the upper arm (17 N), the
lower arm (11 N), and the hand (4.2
N). Find the center of gravity of the
arm relative to the shoulder joint.
W1 x1  W2 x2  
xcg 
W1  W2  
xcg 
17 N 0.13 m   11 N 0.38 m   4.2 N 0.61 m 
17 N  11 N  4.2 N
 0.28 m
8
9.3 Center of Gravity
Conceptual Example 7 Overloading a Cargo Plane
This accident occurred because the plane was overloaded toward
the rear. How did a shift in the center of gravity of the plane cause
the accident?
9
Newton’s 2nd Law for Rotational Motion
2



mr
  
Net external
torque
FT  maT
  FT r
aT  r
  mr 
2
Moment of Inertia, I
Moment of
inertia
 Moment of   Angular

  

Net external torque  
 inertia
  accelerati on 
  I 
 
I   mr 2
Angular acceleration must be expressed in radians/s2.
10
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
Example 9 The Moment of Inertial
Depends on Where the Axis Is.
(b)
Two particles each have mass and are
fixed at the ends of a thin rigid rod. The
length of the rod is L. Find the moment of
inertia when this object rotates relative to
an axis that is perpendicular to the rod at
(a) one end and (b) the center.
 
I   mr 2  m1r12  m2 r22  mL 2  mL 2
2
2
(a)
m1  m2  m
 
I   mr 2  m1r12  m2 r22  m0  mL 
2
m1  m2  m
2
r1  L 2 r2  L 2
I  12 mL2
r1  0 r2  L
I  mL2
11
Newton’s 2nd Law
Rotational Motion
Example 12
Hoisting a Crate
The combined
moment of inertia of
the dual pulley is
50.0 kg·m2. The
crate weighs 4420
N. A tension of
2150 N is
maintained in the
cable attached to
the motor. Find the
angular acceleration
of the dual pulley.
a y   2
  T11  T2 2  I
T1 1  mg  ma y  2  I
T11  mg  m 2  2  I

F

T
 y 2  mg  may
T2  mg  ma y

T1 1  mg 2
I  m 22

2150 N 0.600 m   451 kg 9.80 m s 2 0.200 m 

 6.3 rad
2
2
46.0 kg  m  451 kg 0.200 m 
s2
12
9.5 Rotational Work and Energy
DEFINITION OF ROTATIONAL WORK
Rotational work done by a constant torque in
turning an object through an angle is
WR  
s  r
W  Fs  Fr
The angle must be in radians.
Units: joule (J)
  Fr
W  
13
9.5 Rotational Work and Energy
DEFINITION OF ROTATIONAL
KINETIC ENERGY
The rotational kinetic energy of
a rigid rotating object is
KE  12 mvT2  12 mr 2 2
KER  12 I 2
vT  r


 mr 
The angular speed must be in rad/s.
KE   12 mr 2 2 
 12
2
2
Units: joule (J)
 12 I 2
14
E  12 mv 2  12 I 2  mgh
9.5 Rotational Work and Energy
Example 13 Rolling Cylinders
ENERGY CONSERVATION
A thin-walled hollow cylinder (mass = mh,
radius = rh) and a solid cylinder (mass =
ms, radius = rs) start from rest at the top
of an incline. Determine which cylinder
has the greatest translational speed upon
reaching the bottom.
1
2
mv2f  12 I 2f  mghf 
 mv  I  mghi
1
2
1
2
2
i
1
2
2
i
mv2f  12 I 2f  mghi
f  vf r
1
2
mv2f  12 I v 2f r 2  mghi
2mgho
vf 
m  I r2
The cylinder with the smaller moment of
inertia will have a greater final
translational speed.
15
9.6 Angular Momentum
DEFINITION OF ANGULAR MOMENTUM
The angular momentum L =
body’s moment of inertia * angular velocity
L  I
The angular speed must be in rad/s.
Units: kg·m2/s
CONSERVATION OF ANGULAR
MOMENTUM
Conceptual Example 14 A Spinning
Skater
An ice skater is spinning with both
arms and a leg outstretched. She
pulls her arms and leg inward and
her spinning motion changes
dramatically.
Use the principle of conservation
of angular momentum to explain
how and why her spinning motion
changes.
Angular momentum remains constant (is
conserved) if net external torque is zero.
16
9.6 Angular Momentum
angular momentum conservation
I AA  I PP
Example 15 Satellite in Elliptical Orbit
A satellite in elliptical orbit around earth.
Its closest approach is 8.37x106m from
earth-center. Its greatest distance is
25.1x106m from earth-center. Speed of
satellite at perigee is 8450 m/s. Find
speed at apogee.
I  mr 2   v r
vA
2 vP
mr
 mrP
rA
rP
2
A
rA v A  rP vP


rP vP
8.37 106 m 8450 m s 
vA 

rA
25.1106 m
 2820 m s
L  I
17