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Ch 9. Rotational DynamicsIn pure translational motion, all points on anobject travel on parallel paths.
The most general motion is a combination oftranslation and rotation.
Forces and Torques
Net force acceleration.
What causes an object to have an angular acceleration? TORQUE
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.
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9.1 The Action of Forces and Torques on Rigid Objects
DEFINITION OF TORQUE
Magnitude of Torque = (Magnitude of the force) x (Lever arm)
FDirection: 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)
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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
m106.355cos 2
mN 15
55cosm106.3N 720 2
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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.
0
0yF0 xF
0 yx aa
0
Reasoning Strategy1. 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.
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Example 3 A Diving BoardA 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. 022 WWF
N 1480m 1.40
m 90.3N 5302 F
22
WWF
021 WFFFy
0N 530N 14801 F
N 9501 F
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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?
0 Mddaa MWW
0.13sinm 150.0M
d
Maad
MWW
m 620.0
0.13sinm 150.0N 1840m 280.0N 0.31
N 1.86
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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.
21
2211
WWxWxWxcg
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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.
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2211
WWxWxWxcg
m 28.0
N 2.4N 11N 17
m 61.0N 2.4m 38.0N 11m 13.0N 17
cgx
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9.3 Center of Gravity
Conceptual Example 7 Overloading a Cargo Plane
This accident occurred because the plane was overloaded towardthe rear. How did a shift in the center of gravity of the plane cause the accident?
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Newton’s 2nd Law for Rotational Motion
TT maF raT
rFT
2mrMoment of Inertia, I
2mrNet externaltorque
Moment of inertia
onacceleratiAngular
inertia ofMoment
torqueexternalNet
I 2mrI
Angular acceleration must be expressed in radians/s2.
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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.
(a)
22222
211
2 0 LmmrmrmmrI
Lrr 21 0mmm 21
2mLI
22222
211
2 22 LmLmrmrmmrI
22 21 LrLr mmm 21
221 mLI
(b)
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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.
ITT 2211 yy mamgTF 2
ymamgT 2
2ya
ImamgT y 211
ImmgT 2211 22
211
mImgT
222
2
srad3.6m 200.0kg 451mkg 46.0
m 200.0sm80.9kg 451m 600.0N 2150
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9.5 Rotational Work and Energy
FrFsW
rs
Fr
W
DEFINITION OF ROTATIONAL WORK
Rotational work done by a constant torque in turning an object through an angle is
RW
The angle must be in radians.
Units: joule (J)
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9.5 Rotational Work and Energy
2221 mrKE
22212
21 mrmvKE T
rvT
22122
21 Imr
221 IKER
DEFINITION OF ROTATIONAL KINETIC ENERGY
The rotational kinetic energy of a rigid rotating object is
The angular speed must be in rad/s.
Units: joule (J)
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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.
mghImvE 2212
21
fff mghImv 2212
21
ENERGY CONSERVATION
iii mghImv 2212
21
iff mghImv 2212
21
rv ff
iff mghrvImv 22212
21
2
2rIm
mghv of
The cylinder with the smaller moment of inertia will have a greater final translational speed.
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9.6 Angular Momentum
DEFINITION OF ANGULAR MOMENTUM
The angular momentum L = body’s moment of inertia * angular velocity
IL The angular speed must be in rad/s.Units: kg·m2/s
CONSERVATION OF ANGULAR MOMENTUM
Angular momentum remains constant (is conserved) if net external torque is zero.
Conceptual Example 14 A Spinning Skater
An ice skater is spinning with botharms and a leg outstretched. Shepulls her arms and leg inward andher spinning motion changesdramatically.
Use the principle of conservationof angular momentum to explainhow and why her spinning motionchanges.
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9.6 Angular Momentum
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.
IL
angular momentum conservation
PPAA II
rvmrI 2
P
PP
A
AA r
vmrrvmr 22
PPAA vrvr
m 1025.1
sm8450m 1037.86
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A
PPA r
vrv
sm2820