Transcript Chapter 11 - Rolling, Torque and Angular Momentum

Chapter 11: Rolling Motion, Torque
and Angular Momentum
• Rolling motion (axis of rotation is moving)
• Kinetic Energy of rolling motion
• Rolling motion on an incline
• Torque
• Angular momentum
• Angular momentum is conserved
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Rolling motion of a particle on a wheel
(Superposition of rolling and linear motion)
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11-2 Rolling motion
Smooth rolling:
There is no slipping
Linear speed of center of mass:
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vCOM
ds R  d


 R 
dt
dt
Lecture notes by Dr. M. S. Kariapper KFUPM
- PHYSICS
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11-2 Rolling motion
The angular velocity of
any point on the wheel is
the same.
The linear speed of any point on the object changes as shown in the
diagram!!
For one instant (bottom), point P has no linear speed.
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Lecture notes by Dr. M. S. Kariapper KFUPM
- PHYSICS
For one instant (top), point P’ has a linear speed of 2·vCOM
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11-3 Kinetic Energy of Rolling
Superposition principle:
Rolling motion
=
Pure translation +
Pure rotation
Kinetic energy
1
1
2
2
K  Mv  I CM  
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Lecture notes by Dr. M. S. Kariapper KFUPM
of rolling motion:
2
2
- PHYSICS
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Sample Problem 11-1
Approximate each wheel on the car Thrust SSC as a disk of uniform thickness
and mass M = 170 kg, and assume smooth rolling. When the car’s speed was
1233 km/h, what was the kinetic energy of each wheel?
vcom 1233 km / h  342.5m / s
  vcom / R
I com  12 MR2
K  12 I com 2  12 M vcom2
 ( 12 )( 12 MR2 )(vcom / R)2 
2
1Mv
com
2
 43 .M vcom2
 43 (170 kg )(342.5 m / s)2
 1.50107 J
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11-4 Forces of Rolling
Friction and Rolling
If the wheel rolls without sliding (smooth rolling)
and is accelerating, then from vcom   R ,
d vcom  a  Rd
com
dt
dt
 acom   R (smooth rolling)
where acom is the linear acceleration of the center of
mass and α is the angular acceleration.
•
•
•
•
The force to provide for macom is the static frictional force (assuming the wheel rolls
without sliding).
Therefore, for a wheel to roll without sliding, the maximum static frictional force, s N
between the wheel and the ground must be greater than macom.
f s and acom point to the right if the wheel if the wheel rotates faster, for example, at the
start of a bicycle race.
Do not assume that f s is equal to the maximum value of s N
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Rolling Down a Ramp
Mg sin  f s  M acom
F  Ma 
(1)
The positive direction here is chosen to be down the plane.
y
Do not assume that f s is at its maximum value of s N. The
value of f s self-adjusts so the body rolls without sliding.
x
R f s  Icom 
  I 
(2)
α is counterclockwise and positive.
acom   R
where acom points down plane +ve
Therefore from (2) 
and substituting this in (1) 
f s  Icom acom
R2
acom 
g sin
1  I com / M R2
Note that a positive acom points down plane.
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Demo
A ring and and disk of equal mass and diameter are rolling
down a frictionless incline.
Both start at the same position; which one will be faster at
the end of the incline?
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Sample Problem 11-2
A uniform ball, of mass M = 6.00 kg and radius R, rolls
smoothly from rest down a ramp at angle  = 30.0°
(a) The ball descends a vertical height h = 1.20 m to
reach the bottom of the ramp. What is its speed at the
bottom?
Wex  K  U  Eth
Icom (ball )  25 MR2
vcom  R
0  K  U  0


0   12 I com 2  12 M vcom2   0  Mg  h 

2
2
 2vcom
v
11 22MR
22  M g h
2
com
1

com
 5 M R
   2  2 M vcom
22 5
  RR 

2  M gh
7 M vcom
10
2
10
vcom  (10
7 ) g h  ( 7 )(9.8 m / s )(1.2 m)
 4.1 m / s
A positive vcom points down plane.
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(b) What are the magnitude and direction of the friction force on the ball as it
rolls down the ramp?
acom 
g sin
g sin

1  I com / MR2 1  25 MR 2 / MR 2
2
 (9.8 m / s ) 2sin30.0
1 5
 3.50 m / s2
A positive acom points down plane.
  I
f s  I com acom
R2
 25 MR2 acom
R2

 f s R  Icom  acom 
 R 
 25 Macom
 25 (6.00 kg )(3.50 m / s2 )
 8.40 N
A positive fs means that the direction we selected for fs (up) is correct!
fsR is a clockwise torque (+ve)
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11-5 Yo-Yo
The yo-yo can be considered as a rolling down a ramp:
•
Instead of rolling down a ramp at angle  with the
horizontal, the yo-yo rolls down a string at angle  = 90°
with the horizontal.
•
Instead of rolling on its outer surface at radius R, the yo-yo
rolls on an axle of radius Ro.
•
Instead of being slowed by frictional force fs, the yo-yo is
slowed by the net force T on it from the string.
So we would again get the same
expression for the acceleration as for
rolling with  = 90°.
acom 
g
1  Icom / M Ro2
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11-6 Torque and the vector product
 
  r F

magnitude :
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





  r F sin
  r F
Fis the component of F perpendicular to r
  r F
r is the moment arm of F (the perpendicular
distance between O and the line of action of F )
Lecture notes by Dr. M. S. Kariapper KFUPM
- PHYSICS
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Sample Problem 11.3
In Fig. 11-11a, three forces, each of
magnitude 2.0 N, act on a particle. The
particle is in the xz plane at point A given by
position vector r , where r = 3.0 m and
 = 30°. Force F1is parallel to the x axis,
force F 2 is parallel to the z axis, and force
F3 is parallel to the y axis
What is the torque, with respect to the origin
O, due to each force?
1  r F1 sin 1  (3.0 m)(2.0 N )(sin150 )  3.0 N  m
 2  r F2 sin 2  (3.0 m)(2.0 N )(sin120 )  5.2 N  m
 3  r F3 sin 3  (3.0 m)(2.0 N )(sin90 )  6.0 N  m
To find the directions of the torques, we use the right hand rule and rotate r into F
through the smaller of the two angles between their directions.
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Lecture notes by Dr. M. S. Kariapper KFUPM
- PHYSICS
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Sample Problem 11.3
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Lecture notes by Dr. M. S. Kariapper KFUPM
- PHYSICS
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11-7 Angular momentum of a particle
l rp
 m(r  v )
• The SI unit of angular momentum l is kg  m2 / s  J  s
.
• Angular momentum is a “vector”, the direction is determined by the right
hand rule.
• The magnitude of angular momentum is l  r m v sin
• where φ is the angle between r and p when these two vectors are
arranged tail to tail.
 l  r p  rmv


magnitude :
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

 l  r p  rmv

Lecture notes by Dr. M. S. Kariapper KFUPM
- PHYSICS
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Sample Problem 11-4
Figure 11-13 shows an overhead view of two
particles moving at constant momentum along
horizontal paths. Particle 1, with momentum
magnitude p1 = 5.0 kg·m/s, has position vector r1 and
will pass 2.0 m from point O. Particle 2, with
momentum magnitude p2 = 2.0 kg·m/s, has position
vector r2 and will pass 4.0 m from point O.
What is the net angular momentum L about point O of the two-particle system?
l1  r1 p1  (2.0 m)(5.0 kg  m / s)  10 kg  m2 / s
The RHR indicates that l1 is positive. 
l1  10 kg  m2 / s
RHR = right hand rule
l2  r 2 p2  (4.0 m)(2.0 kg  m / s)  8.0 kg  m2 / s
The RHR indicates that l2 is negative. 
l2   8.0 kg  m2 / s
L  l1  l2  10 kg  m2 / s (8.0 kg  m2 / s)   2.0 kg  m2 / s
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11-8 Newton’s Second Law in
Angular Form

Fnet  d p (single particle)
dt
 net  d l (single particle)
dt
• Note that the torque  and angular momentum l must be defined with
respect to the same origin.
•
Proof:
l  m (r x v )
d l  m (r x d v  d r x v )  m (r x a  v x v )
dt
dt dt
Because v x v  0 , this leads to
d l  m ( r x a )  r x ma  r x F  (r x F )  
net 
net
dt
Therefore,  net  d l
dt
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11-9 Angular momentum of a system
of Particles
n
L  l1  l 2  l 3  ...  l n   l i (L = total angular momentum)
i 1
d L  n d li
dt 
i 1 d t
•

n
 net,i

i 1
n
 net,i is the net torque on the particle.   net,i is the sum of all the torque
i 1
(internal and external) on the system. However the internal torques sums to
zero. Let  net represent the net external torque on the system.
ith

 net  d L
dt
( system of particles )
• The net external torque  net acting on a system of particles is equal to the
time rate of change of the system's total angular momentum L .
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11-10 Angular momentum of a rigid
object rotating about a fixed axis
We’ll consider an object that is
rotating about the z-axis.
The angular momentum of the
object is given by:
Lz  I  
Note that in this case L and  are along the z axis.
Also note the analog formula for linear momentum p = m·v
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11-11 Conservation of angular momentum
The total angular momentum of a system is constant in both
magnitude and direction if the resultant external torque
acting on the system is zero.

L  constant
If the system undergoes an internal “rearrangement”, then
 
Li  L f  constant
If the object is rotating about a fixed axis (say z-axis), then:
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I ii  I f  f  constant
Lecture notes by Dr. M. S. Kariapper KFUPM
- PHYSICS
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Demo
A students stands still on a rotating platform and
holds two texts on outstretched arms. He brings
the arms closer. What happens? Discuss
A students stands still on a rotatable platform
and holds a spinning wheel. The bicycle wheel is
spinning in the clockwise direction when viewed
from above.
He flips the wheel over.
What happens?
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TABLE 11-1
More Corresponding Variables and Relations for Translational and Rotational Motiona
Translational
Rotational
Force

F
Torque

Linear momentum

p
Angular momentum
Linear momentumb
Linear momentumb
Newton's second lawb
Conservation lawd
a
b
c
d


P (  pi )


P  M vcom
Angular momentumb
Angular momentumc


dP
b
Newton's
second
law
Fnet 
dt

P  a constant Conservation lawd
 
 ( r x F )
  
l ( r x p)


L (  li )
L  I


dL
 net 
dt

L  a constant
See also Table 10-3.
For systems of particles, including rigid bodies.
For a rigid body about a fixed axis, with L being the component along that axis.
For a closed, isolated system.
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P22
Force F = 2i-3k acts on a particle with position vector r = 0.5j2.0k relative to the origin. In unit vector notation, what is the
resulting torque on the pebble about (a) the origin and (b) the
point (2.0, 0, -3.0)?
P72
A uniform solid ball rolls smoothly along a floor and up a
ramp inclined at 15.0°. It is momentarily stops when it has
rolled 1.50 m along the ramp. What was its initial speed?
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P85
In fig. 11.-62, a constant horizontal force Fapp of magnitude
12 N is applied to a uniform solid cylinder by a fishing line
wrapped around the cylinder. The mass of the cylinder is 10
kg, its radius is 0.10 m, and the cylinder rolls smoothly on
the horizontal surface. (a) What is the magnitude of the
acceleration of the com of the cylinder? (b) What is the
magnitude of the angular acceleration of the cylinder about
the com? (c) In unit vector notation, what is the frictional
force acting on the cylinder?
P90
A uniform rod rotates in a horizontal plane about a vertical
axis through one end. The rod is 6.00 m long, weighs 10.0
N, and rotates at 240 rev/min. Calculate (a) its rotational
inertia about the axis of rotation and (b) the magnitude of its
angular momentum about the axis.
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P48
A cockroach of mass 0.17 kg runs counterclockwise around
the rim of a lazy Susan (a circular disk mounted on a veritcal
axle) that has radius 15 cm, rotational inertia 5.0 x 10-3 kgm2,
and frictionless bearings. The cockroach’s speed (relative to
the ground) is 2.0 m/s, and the lazy Susan turns clockwise
with angular velocity wo = 2.8 rad/s. The cockroach finds a
bread crumb on the rim and, of course, stops. (a) What is the
angular speed of the lazy Susan after the cockroach stops?
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