# Chapter 8 Rotational Equilibrium and Rotational Dynamics ```Chapter 8
Rotational Equilibrium
and
Rotational Dynamics
Wrench Demo
Torque

Torque, t , is the tendency of a force
to rotate an object about some axis
t  Fd
F is the force
 d is the lever arm (or moment arm)
 Units are Newton-m

Direction of Torque

Torque is a vector quantity
Direction determined by axis of twist
 Perpendicular to both r and F
 In 2-d, torque points into or out of paper
 Clockwise torques point into paper.
Defined as negative
 Counter-clockwise torques point out of paper.
Defined as positive

Non-perpendicular forces

Only the ycomponent, Fsinf,
produces a torque
Non-perpendicular forces
t  Fr sin f
F is the force
 r is distance to point where F is
applied
 Φ is the angle between F and r

Torque and Equilibrium

Forces sum to zero (no linear motion)
Fx  0 and Fy  0

Torques sum to zero
(no rotation)
t  0
Meter Stick Demo
Axis of Rotation
Torques require point of reference
 Point of reference can be anywhere



Use same point for all torques
Pick the point to make problem least difficult
Example
Given M = 120 kg.
Neglect the mass of the beam.
a) Find the tension
in the cable
b) What is the force
between the beam and
the wall
Solution
a) Given: M = 120 kg, L = 10 m, x = 7 m
Find: T
Basic formula
t  0
Axis
TL  Mgx  0
Solve for T = 824 N
Solution
b) Given: M = 120 kg, L = 10 m, x = 7 m, T = 824 N
Find f
Basic formula
F
y
0
T  Mg  f  0
Solve for f = 353 N
f
Alternative Solution
b) Given: M = 120 kg, L = 10 m, x = 7 m
Find f
Basic formula
t  0
Mg ( L  x)  fL  0
Solve for f = 353 N
f
Axis
Another Example
Given: W=50 N, L=0.35 m,
x=0.03 m
Find the tension in the muscle
W
x
Basic formula
t  0
L
Fx WL  0
F = 583 N
Center of Gravity
Gravitational force acts on all points
of an extended object
 However, it can be considered as one
net force acting on one point,
the center-of-gravity, X.

 (m g ) x
i
i
 X  mi g
i
i
X
 mi xi
i
m
i
i
Weighted
Average
Example
Consider the 400-kg
beam shown below.
Find TR
Solution
Find TR
Basic formula
t  0
to solve for TR,
choose axis here
 MgX  TR L  0
TR = 1 121 N
X=2m
L=7m
One Last Example
Given: x = 1.5 m, L = 5.0 m,
wbeam = 300 N,
wman = 600 N
Find: T
Fig 8.12, p.228
Slide 17
x
L
Solution
Given: x , L , wbeam ,wman
Find: T
First, find Ty
Basic formula
Fig 8.12, p.228
t  0
Slide 17
 wman x  wbeam L / 2  Ty L  0
Ty = 330 N
Now, find T
Ty  T sin( 53) T = 413 N
x
L
Baton Demo
Moment-of_Inertia Demo
Torque and Angular Acceleration
When t  0, rigid body experiences
angular acceleration
 Relation between t and a is analagous
to relation between F and a

F  ma,
t  Ia
Moment of Inertia
Moment of Inertia

This mass analog is called the moment
of inertia, I, of the object
I   mi ri
i
2
r is defined relative to rotation axis
 SI units are kg m2

I depends on both the mass and its
distribution.
 If an object’s mass is distributed
further from the axis of rotation,
the moment of inertia will be larger.

Demo: Moment of Inertia Olympics
Moment of Inertia of a Uniform Ring
Divide ring into
segments
segment is R

I  mi ri  MR
2
2
Example
What is the moment of inertia of the following
point masses arranged in a square?
Solution
a) Find I about the x-axis?
Given: M2=2, M3=3 kg, L=0.6 m
Basic formula
I   mi ri 2
r = 0.6&middot;sin(45)
First, find distance to
2-kg masses
I  M 2r  M 2r
2
2
=0.72 kg&middot;m2
Solution
b) Find I about the y-axis?
r = 0.6&middot;sin(45)
Same as before,
except you use the
3-kg masses
I  M 3r  M 3r
2
2
=1.08 kg&middot;m2
Solution
c) Find I about the z-axis?
r = 0.6&middot;sin(45)
Use all the masses
I  M 2r  M 2r
2
2
 M 3r  M 3r
2
2
=1.8 kg&middot;m2
Other Moments of Inertia
Example
Treat the spindle as a solid cylinder.
a) What is the moment of Inertia of
the spindle?
b) If the tension in the rope is 10 N,
what is the angular acceleration of
the wheel?
c) What is the acceleration of the
bucket?
d) What is the mass of the bucket?
M
Solution
a) What is the moment of Inertia of
the spindle?
Given: M = 5 kg, R = 0.6 m
Moments of Inertia
2
MR , cylindrica l shell
1
MR 2 , solid cylinder
2
2
MR 2 , solid sphere
5
2
2
MR , spherical shell
3
1
3
1
12
M
1
I  MR 2 = 0.9 kgm2
2
Solution
b) If the tension in the rope is 10 N,
what is a?
Given: I = 0.9 kg m2, T = 10 N, r = 0.6 m
Basic formula
t  rF
t  Ia
rT  Ia
Solve for a
c) What is the acceleration of the
bucket?
Given: r=0.6 m, a = 6.67 rad/s
Basic formula
a  ar
a=4 m/s2
M
Solution
d) What is the mass of the bucket?
Given: T = 10 N, a = 4 m/s2
Basic formula
F  ma
Ma  Mg  T
T
M
g a
M = 1.72 kg
M
Example
A cylindrical space station of
radius R = 12 m and mass
M = 3400 kg is designed
to rotate around the axis
of the cylinder. The space
station’s moment of
inertia is 0.75 MR2. Retrorockets are fired tangentially to the surface of
space station and provide an impulse of 2.9x104 N&middot;s.
a) What is the angular velocity of the space
station after the rockets have finished firing?
b) What is the centripetal acceleration at the
edge of the space station?
Solution
Given: M = 3400, R = 12, I = 0.75 MR2 = 3.67x105
F&middot;t = 2.9x104.
a) Find: w
Basic formula
w
t  Ia  I
t
Basic formula
t  rF
t  t  Iw
RF  t  Iw
Solve for w.
b) Find centripetal acceleration
Basic formula
2
a w r
a w R
2
= 10.8 m/s2
Rotational Kinetic Energy
Each point of a rigid body rotates with angular
velocity w.
1
1
1 2
2
2 2
KE   mi vi   mi ri w  Iw
2
2
2
Including the linear motion
1 2 1 2
KE  mv  Iw
2
2
KE due to rotation
KE of center-of-mass motion
Example
What is the kinetic energy of the Earth due to
the daily rotation?
Given: Mearth=5.98 x1024 kg, Rearth = 6.63 x106 m.
Basic formula
2
w
T
Solid sphere
2
I  MR 2
5
Basic formula
1 2
KE  Iw
2
First, find w
2
w
24  3600
1
2 2
KE  MR w = 2.78 x10-29
5
Example
A solid sphere rolls down a hill of height 40 m.
What is the velocity of the ball when it reaches
the bottom? (Note: We don’t know r or m!)
Basic formula
1 2 1 2
mgh  mv  Iw
2
2
2 2 m cancels
I  mr
1 2 2 2 2
5
gh   v  r w 
2
5

For solidsphere
Basic formula
v  wr
1  2 2 2  2 2 gh
gh   v  v , v 
2
5 
7/5
v = 23.7 m/s
Suitcase Demo
Angular Momentum
L  Iw
L  mvr
Rigid body
Point particle
Analogy between L and p
Angular Momentum
Linear momentum
L = Iw
p = mv
t = L/t
F = p/t
Conserved if no net
outside torques
Conserved if no net
outside forces
Rotating Chair Demo
Angular Momentum and Kepler’s 2nd Law
For central forces, e.g. gravity, t = 0
and L is conserved.
 Change in area in t is:

1
A  r (v t )
2
L  mrv
A
1

L
t 2m
Example
A 65-kg student sprints at
8.0 m/s and leaps onto a
110-kg merry-go-round of
merry-go-round as a
uniform cylinder, find the
resulting angular velocity.
Assume the student lands
on the merry-go-round
while moving tangentially.
Solution
Known: M, R, m, v0
Find: wF
First, find L0
Basic formula
L  mvr
L0  mvR
Next, find Itot
Solid cylinder
I Tot
1
2
 mR  MR
2
2
1
2
I  MR
2
Now, given Itot and L0, find w
Particle
I  MR 2
Basic formula
L  Iw
L0
w
I Tot
Example
Two twin ice skaters separated by 10 meters
skate without friction in a circle by holding onto
opposite ends of a rope. They move around a circle
once every five seconds. By reeling in the rope,
they approach each other until they are separated
by 2 meters.
a) What is the period of the new motion?
b) If each skater had a mass of 75 kg,
what is the work done by the skaters in
pulling closer?
Solution
a) Find: Tf
Given: R0=5 m, Rf=1 m, T0=5 s
Basic formula
L  Iw
2
w
T
Point particle
I  mr 2
First, find expression for initial L
(2 skaters)
2 2
L0  2mR0
T0
Next, apply conservation of L
2
2 2
2mR
 2mR f
T0
Tf
2
Rf
T f  T0 2 = T0/25 = 0.2 s
R0
2
0
Solution
b) Find: W
Given: m=75 kg, R0=5 m, Rf=1 m, T0=5 s, Tf=0.2 s
First, find moments of inertia
I 0  2mR , I f  2mR
2
0
2
f
Next, find values for w
2
2
w0 
, wf 
T0
Tf
Basic formula
1 2
KE  Iw
2
Finally, find change in KE
1
1
2
W  I f w f  I 0w 02 = 7.11x105 J
2
2
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