Download Chapter 8:

Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 8: Torque and Angular
Momentum
•Rotational Kinetic Energy
•Rotational Inertia
•Torque
•Work Done by a Torque
•Equilibrium (revisited)
•Rotational Form of Newton’s 2nd Law
•Rolling Objects
•Angular Momentum
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§8.1 Rotational KE and Inertia
For a rotating solid body:
K rot
n
1
1
1
1
2
2
2
 m1v1  m2 v2    mn vn   mi vi2
2
2
2
i 1 2
For a rotating body vi=ri where ri is the distance from the
rotation axis to the mass mi.
1
1 n
1 2
2
2 2
  mi ri     mi ri   I
2  i 1
2
i 1 2

n
K rot
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n
The quantity
I   mi ri 2
i 1
is called rotational inertia or
moment of inertia.
Use the above expression when the number of masses
that make up a body is small. Use the moments of inertia
in table 8.1 for extended bodies.
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Example: (a) Find the moment of inertia of the system below.
The masses are m1 and m2 and they are separated by a
distance r. Assume the rod connecting the masses is
massless.

m1
r1
r2
m2
r1 and r2 are the distances
between mass 1 and the
rotation axis and mass 2
and the rotation axis (the
dashed, vertical line)
respectively.
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Example continued:
Take m1 = 2.00 kg, m2 = 1.00 kg,
r1= 0.33 m , and r2 = 0.67 m.
2
I   mi ri 2  m1r12  m2 r22  0.67 kg m 2
i 1
(b) What is the moment of inertia if the axis is moved so
that is passes through m1?
2
I   mi ri 2  m2 r22  1.00 kg m 2
i 1
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Example (text problem 8.4): What is the rotational inertia of a
solid iron disk of mass 49.0 kg with a thickness of 5.00 cm
and a radius of 20.0 cm, about an axis through its center and
perpendicular to it?
From table 8.1:
1
1
2
2
I  MR  49.0 kg 0.2 m   0.98 kg m 2
2
2
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§8.2 Torque
A rotating (spinning) body will continue to rotate unless it
is acted upon by a torque.
hinge
Q: Where on a door do
you push to open it?
P
u
s
h
A: Far from the hinge.
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Torque method 1:
Top view of door
F
F

Hinge
end
F||
  rF
r = the distance from the rotation axis (hinge) to the point
where the force F is applied.
F is the component of the force F that is perpendicular
to the door (here it is Fsin).
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The units of torque are Nm (not joules!).
By convention:
•When the applied force causes the object to rotate
counterclockwise (CCW) then  is positive.
•When the applied force causes the object to rotate
clockwise (CW) then  is negative.
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Torque method 2:
  r F
r is called the lever arm and F is the magnitude of the
applied force.
Lever arm is the perpendicular distance to the line of action
of the force.
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Top view of door
F
Hinge
end

r

Line of
action of
the force
Lever
arm
r
sin  
r
r  r sin 
The torque is:
  r F
 rF sin 
Same as
before
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Example (text problem 8.12): The pull cord of a lawnmower
engine is wound around a drum of radius 6.00 cm, while the
cord is pulled with a force of 75.0 N to start the engine. What
magnitude torque does the cord apply to the drum?
F=75 N
  r F
R=6.00 cm
 rF
 0.06 m75.0 N   4.5 Nm
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Example: Calculate the torque due to the three forces shown
about the left end of the bar (the red X). The length of the bar
is 4m and F2 acts in the middle of the bar.
F2=30 N
30
F3=20 N
45
X
10
F1=25 N
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
Lever arm
for F2
F2=30 N
30
F3=20 N
45
10
X
F1=25 N
Lever arm
for F3
The lever arms are:
r1  0
r2  2m sin 60  1.73 m
r3  4m sin 10  0.695 m
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Example continued:
1  0
The torques are:  2  1.73 m 30 N   51.9 Nm
 3  0.695 m 20 N   13.9 Nm
The net torque is + 65.8 Nm and is the sum of the above
results.
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§8.3 Work done by a Torque
The work done by a torque  is
W   .
where  is the angle (in radians) the object turns through.
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Example (text problem 8.25): A flywheel of mass 182 kg has a
radius of 0.62 m (assume the flywheel is a hoop).
(a) What is the torque required to bring the flywheel from
rest to a speed of 120 rpm in an interval of 30 sec?
 f  120
rev  2 rad  1 min 


  12.6 rad/sec
min  1 rev  60 sec 
  
  rF  r ma  rmr   mr 

 t 
 f  i 
f 
2
2
  mr 
  29.4 Nm
 mr 
 t 
 t 
2
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Example continued:
(b) How much work is done in this 30 sec period?
W     av t 
 i   f
  
 2

f
t   

 2

t  5600 J

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§8.4 Equilibrium
The conditions for equilibrium are:
F  0
τ  0
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Example (text problem 8.36): A sign is supported by a uniform
horizontal boom of length 3.00 m and weight 80.0 N. A cable,
inclined at a 35 angle with the boom, is attached at a
distance of 2.38 m from the hinge at the wall. The weight of
the sign is 120.0 N. What is the tension in the cable and what
are the horizontal and vertical forces exerted on the boom by
the hinge?
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Example continued:
y
FBD for the bar:
Fy
T
X

Fx
x
wbar
Apply the conditions for
equilibrium to the bar:
Fsb
(1)  Fx  Fx  T cos   0
(2)  Fy  Fy  wbar  Fsb  T sin   0
L
(3)    wbar    Fsb L   T sin  x  0
2
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Example continued:
Equation (3) can be solved for T:
L
wbar    Fsb L 
2
T
x sin 
 352 N
Equation (1) can be solved for Fx:
Fx  T cos   288 N
Equation (2) can be solved for Fy:
Fy  wbar  Fsb  T sin 
 2.00 N
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§8.5 Equilibrium in the Human
Body
Example (text problem 8.43): Find the force exerted by the
biceps muscle in holding a one liter milk carton with the
forearm parallel to the floor. Assume that the hand is 35.0
cm from the elbow and that the upper arm is 30.0 cm long.
The elbow is bent at a right angle and one tendon of the
biceps is attached at a position 5.00 cm from the elbow and
the other is attached 30.0 cm from the elbow. The weight of
the forearm and empty hand is 18.0 N and the center of
gravity is at a distance of 16.5 cm from the elbow.
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Example continued:
Fb
“hinge”
(elbow
joint)
w
  F x  wx
b 1
2
Fca
 Fca x3  0
wx2  Fca x3
Fb 
 130 N
x1
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§8.6 Rotational Form of Newton’s
2nd Law
  I
Compare to
 F  ma
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Example (text problem 8.57): A bicycle wheel (a hoop) of
radius 0.3 m and mass 2 kg is rotating at 4.00 rev/sec. After
50 sec the wheel comes to a stop because of friction. What is
the magnitude of the average torque due to frictional forces?
2


I


MR


rev  2 rad 
i  4.00

  25.1 rad/sec
sec  1 rev 
f  0
  f  i


 0.50 rad/s 2
t
t
 av  MR 2   0.09 Nm
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§8.7 Rolling Objects
An object that is rolling combines translational motion (its
center of mass moves) and rotational motion (points in the
body rotate around the center of mass).
For a rolling object:
K tot  K T  K rot

1 2 1 2
mvcm  I
2
2
If the object rolls without slipping then vcm = R.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: Two objects (a solid disk and a solid sphere) are
rolling down a ramp. Both objects start from rest and from
the same height. Which object reaches the bottom of the
ramp first?
h

The object with the largest linear velocity (v) at the
bottom of the ramp will win the race.
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Example continued:
Apply conservation of mechanical energy:
Ei  E f
U i  Ki  U f  K f
1 2 1 2 1 2 1 v
mgh  0  0  mv  I  mv  I  
2
2
2
2 R
2
1
I 
mgh   m  2 v 2
2
R 
Solving for v:
v
2mgh
I 

m


2 
R 

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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The moments of inertia are:
4
gh
For the disk: vdisk 
3
10
gh
For the sphere: vsphere 
7
1
I disk  mR 2
2
2
I sphere  mR 2
5
Since Vsphere> Vdisk the
sphere wins the race.
Compare these to a box sliding down the ramp. vbox  2 gh
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
How do objects in the previous example roll?
y
N
FBD:
w
x
Both the normal force and the weight act through the center
of mass so =0. This means that the object cannot rotate
when only these two forces are applied.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Add friction:
y
FBD:
s
N
Fs
  F r  I
 F  w sin   F  ma
 F  N  w cos  0
x
s
cm
y

Also need acm = R and
w
x
v 2  v02  2ax
The above system of equations can be solved for v at the
bottom of the ramp. The result is the same as when using
energy methods. (See text example 8.13.)
It is static friction that makes an object roll.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§8.8 Angular Momentum
p
Fnet  lim
t 0 t
p  mv
Units of p are kg m/s
When no net external
forces act, the momentum
of a system remains
constant (pi = pf)
L
 net  lim
t 0 t
L  Iω
Units of L are kg m2/s
When no net external
torques act, the angular
momentum of a system
remains constant (Li = Lf).
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 8.70): A turntable of mass 5.00 kg has
a radius of 0.100 m and spins with a frequency of 0.500
rev/sec. What is the angular momentum? Assume a uniform
disk.
rev  2 rad 
  0.500

  3.14 rad/sec
sec  1 rev 
1

L  I   MR 2   0.079 kg m 2 /s
2

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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 8.79): A skater is initially spinning at a
rate of 10.0 rad/sec with I=2.50 kg m2 when her arms are
extended. What is her angular velocity after she pulls her
arms in and reduces I to 1.60 kg m2?
He/she is on ice, so we can ignore external
torques.
Li  L f
I ii  I f  f
 Ii
 f  
If

 2.50 kg m 2 
i  
10.0 rad/sec   15.6 rad/sec
2


 1.60 kg m 

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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§8.9 The Vector Nature of Angular
Momentum
Angular momentum is a vector. Its direction is defined with
a right-hand rule.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Curl the fingers of your right hand so that they curl in the
direction a point on the object moves, and your thumb will
point in the direction of the angular momentum.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Consider a person
holding a spinning
wheel. When viewed
from the front, the
wheel spins CCW.
Holding the wheel horizontal, they step on to a platform
that is free to rotate about a vertical axis.
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Initially, nothing happens. They then move the wheel so
that it is over their head. As a result, the platform turns
CW (when viewed from above).
This is a result of conserving angular momentum.
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Initially there is no angular momentum about the vertical axis.
When the wheel is moved so that it has angular momentum
about this axis, the platform must spin in the opposite
direction so that the net angular momentum stays zero.
Is angular momentum conserved about the
direction of the wheel’s initial, horizontal axis?
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It is not. The floor exerts a torque on the system (platform +
person), thus angular momentum is not conserved here.
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Summary
•Rotational Kinetic Energy
•Moment of Inertia
•Torque (two methods)
•Conditions for Equilibrium
•Newton’s 2nd Law in Rotational Form
•Angular Momentum
•Conservation of Angular Momentum
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