Download 4.1 The Concepts of Force and Mass

Chapter 9
Rotational Dynamics
9.1 The Action of Forces and Torques on Rigid Objects
In pure translational motion, all points on an
object travel on parallel paths.
The most general motion is a combination of
translation and rotation.
9.1 The Action of Forces and Torques on Rigid Objects
According to Newton’s second law, a net force causes an
object to have an acceleration.
What causes an object to have an angular acceleration?
TORQUE
9.1 The Action of Forces and Torques on Rigid Objects
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.
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)
9.1 The Action of Forces and Torques on Rigid Objects
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.
9.1 The Action of Forces and Torques on Rigid Objects
  F

cos 55 
3.6 10  2 m

790 N
  720 N 3.6 10 2 m cos 55
 15 N  m
9.2 Rigid Objects in Equilibrium
If a rigid body is in equilibrium, neither its linear motion nor its
rotational motion changes.
ax  a y  0
F
x
0
 0
F
y
0
  0
9.2 Rigid Objects in Equilibrium
EQUILIBRIUM OF A RIGID BODY
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.
F
x
0
F
y
0
  0
9.2 Rigid Objects in Equilibrium
Reasoning Strategy
1. Select the object to which the equations for equilibrium are to be applied.
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.
9.2 Rigid Objects in Equilibrium
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.
9.2 Rigid Objects in Equilibrium
  F 
2 2
 W W  0
W W
F2 
2
F2

530 N 3.90 m 

 1480 N
1.40 m
9.2 Rigid Objects in Equilibrium
F
y
 F1  F2  W  0
 F1  1480 N  530 N  0
F1  950 N
9.2 Rigid Objects in Equilibrium
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 dumbell he can hold?
9.2 Rigid Objects in Equilibrium
  W 
a a
 Wd  d  M M  0
 M  0.150 msin 13.0
9.2 Rigid Objects in Equilibrium
 Wa  a  M M
Wd 
d
 31.0 N 0.280 m   1840 N 0.150 m sin 13.0

 86.1 N
0.620 m
9.3 Center of Gravity
DEFINITION OF 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.
9.3 Center of Gravity
When an object has a symmetrical shape and its weight is distributed
uniformly, the center of gravity lies at its geometrical center.
9.3 Center of Gravity
W1 x1  W2 x2  
xcg 
W1  W2  
9.3 Center of Gravity
Example 6 The 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.
9.3 Center of Gravity
W1 x1  W2 x2  
xcg 
W1  W2  
xcg

17 N 0.13 m   11 N 0.38 m   4.2 N 0.61 m 

 0.28 m
17 N  11 N  4.2 N
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?
9.3 Center of Gravity
Finding the center of gravity of an irregular shape.
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
FT  maT
  FT r
aT  r
  mr 
2
Moment of Inertia, I
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
   mr 
2
Net external
torque
Moment of
inertia
 1  m1r12 
 2  m r 
2
2 2

 N  mN rN2 
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
ROTATIONAL ANALOG OF NEWTON’S SECOND LAW FOR
A RIGID BODY ROTATING ABOUT A FIXED AXIS
 Moment of   Angular

  

Net external torque  
 inertia
  accelerati on 
  I 
Requirement: Angular acceleration
must be expressed in radians/s2.
 
I   mr 2
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.
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.
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
(a)
I   mr 2   m1r12  m2 r22  m0  mL 
2
m1  m2  m
I  mL
2
2
r1  0 r2  L
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
 
(b) I   mr 2  m1r12  m2 r22  mL 2 2  mL 2 2
m1  m2  m
I  mL
1
2
2
r1  L 2 r2  L 2
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
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.
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
equal

F

T
 y 2  mg  may
a y   2
  T 
T2  mg  may
1 1
 T2 2  I
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
T1 1  mg  may  2  I
a y   2
T11  mg  m 2  2  I
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
9.5 Rotational Work and Energy
s  r
W  Fs  Fr
  Fr
W  
9.5 Rotational Work and Energy
DEFINITION OF ROTATIONAL WORK
The rotational work done by a constant torque in
turning an object through an angle is
WR  
Requirement: The angle must
be expressed in radians.
SI Unit of Rotational Work: joule (J)
9.5 Rotational Work and Energy
KE  12 mvT2  12 mr 2 2
vT  r

  mr 
KE   12 mr 2 2  12
2
2
 12 I 2
9.5 Rotational Work and Energy
DEFINITION OF ROTATIONAL KINETIC ENERGY
The rotational kinetic energy of a rigid rotating object is
KER  12 I 2
Requirement: The angular speed must
be expressed in rad/s.
SI Unit of Rotational Kinetic Energy: joule (J)
9.5 Rotational Work and Energy
Example 13 Rolling Cylinders
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.
9.5 Rotational Work and Energy
E  12 mv2  12 I 2  mgh
ENERGY CONSERVATION
1
2
1
2
mv2f  12 I 2f  mghf  12 mvi2  12 Ii2  mghi
mv2f  12 I 2f  mghi
f  vf r
9.5 Rotational Work and Energy
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.
9.6 Angular Momentum
DEFINITION OF ANGULAR MOMENTUM
The angular momentum L of a body rotating about a
fixed axis is the product of the body’s moment of
inertia and its angular velocity with respect to that
axis:
L  I
Requirement: The angular speed must
be expressed in rad/s.
SI Unit of Angular Momentum: kg·m2/s
9.6 Angular Momentum
PRINCIPLE OF CONSERVATION OFANGULAR MOMENTUM
The angular momentum of a system remains constant (is
conserved) if the net external torque acting on the system
is zero.
9.6 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.
9.6 Angular Momentum
Example 15 A Satellite in an Elliptical Orbit
An artificial satellite is placed in an
elliptical orbit about the earth. Its point
of closest approach is 8.37x106m
from the center of the earth, and
its point of greatest distance is
25.1x106m from the center of
the earth.
The speed of the satellite at the
perigee is 8450 m/s. Find the speed
at the apogee.
9.6 Angular Momentum
L  I
angular momentum conservation
I AA  I PP
I  mr 2   v r
vA
2 vP
mr
 mrP
rA
rP
2
A
9.6 Angular Momentum
vA
2 vP
mr
 mrP
rA
rP
2
A
rA v A  rP vP
rP vP 8.37 106 m 8450 m s 
vA 

 2820 m s
6
rA
25.110 m