Download Ch11 - Rolling, Torque, and Angular Momentum

Chapter 11
Rolling, Torque, and Angular
Momentum
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
11-1 Rolling as Translation and Rotation Combined
Learning Objectives
11.01 Identify that smooth
rolling can be considered as
a combination of pure
translation and pure rotation.
11.02 Apply the relationship
between the center-of-mass
speed and the angular speed
of a body in smooth rolling.
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Figure 11-2
11-1 Rolling as Translation and Rotation Combined


We consider only objects that roll smoothly (no slip)
The center of mass (com) of the object moves in a
straight line parallel to the surface

The object rotates around the com as it moves

The rotational motion is defined by:
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Figure 11-3
11-1 Rolling as Translation and Rotation Combined

The figure shows how the velocities of translation and
rotation combine at different points on the wheel
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11-1 Rolling as Translation and Rotation Combined
Figure 11-4
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11-1 Rolling as Translation and Rotation Combined
Figure 11-4
Answer: (a) the same (b) less than
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11-2 Forces and Kinetic Energy of Rolling

Combine translational and rotational kinetic energy:
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11-2 Forces and Kinetic Energy of Rolling

If a wheel accelerates, its angular speed changes

A force must act to prevent slip
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11-2 Forces and Kinetic Energy of Rolling

If slip occurs, then the motion is not smooth rolling!

For smooth rolling down a ramp:
1. The gravitational force is vertically down
2. The normal force is perpendicular to the ramp
3. The force of friction points up the slope
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11-2 Forces and Kinetic Energy of Rolling

We can use this equation to find the acceleration of
such a body

Note that the frictional force produces the rotation

Without friction, the object will simply slide
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11-2 Forces and Kinetic Energy of Rolling
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11-2 Forces and Kinetic Energy of Rolling
Answer: The maximum height reached by B is less than that reached by A.
For A, all the kinetic energy becomes potential energy at h.
Since the ramp is frictionless for B, all of the rotational K stays rotational,
and only the translational kinetic energy becomes potential energy at its maximum
height.
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11-3 The Yo-Yo


As yo-yo moves down string, it
loses potential energy mgh but
gains rotational & translational
kinetic energy
Find linear acceleration of yo-yo
accelerating down its string:
1. Rolls down a “ramp” of angle 90°
2. Rolls on an axle instead of its outer
surface
3. Slowed by tension T rather than friction
11-3 The Yo-Yo

Replacing the values leads us to:
Example Calculate the acceleration of the yo-yo
o
o
M = 150 grams, R0 = 3 mm, Icom = Mr2/2 = 3E-5 kg m2
Therefore acom = -9.8 m/s2 / (1 + 3E-5 / (0.15 * 0.0032))
= - .4 m/s2
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11-4 Torque Revisited


Previously, torque was defined only for a rotating body
and a fixed axis
Now we redefine it for an individual particle that moves
along any path relative to a fixed point

The path need not be a circle; torque is now a vector

Direction determined with right-hand-rule
Figure 11-10
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11-4 Torque Revisited

The general equation for torque is:

We can also write the magnitude as:

Or, using the perpendicular component of force or the
moment arm of F:
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11-4 Torque Revisited
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11-4 Torque Revisited
Answer: (a) along the z direction (b) along the +y direction (c) along the +x direction
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Loosen a bolt
• Which of the three
equal-magnitude
forces in the figure is
most likely to loosen
the bolt?
• Fa?
• Fb?
• Fc?
Loosen a bolt
• Which of the three
equal-magnitude
forces in the figure is
most likely to loosen
the bolt?
• Fb!
Torque
• Let O be the point around
which a solid body will be
rotated.
O
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Torque
Line of
action
• Let O be the point around
which a solid body will be
rotated.
• Apply an external force F
on the body at some point.
O
• The line of action of a force
is the line along which the
force vector lies.
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F applied
Torque
Line of
action
• Let O be the point around
which a body will be
rotated.
• The lever arm (or moment
arm) for a force
L
O
is
The perpendicular distance
from O to the line of action
of the force
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F applied
Torque
Line of
action
• The torque of a force with
respect to O is the product of
force and its lever arm.
t=FL
L
O
• Greek Letter “Tau” = “torque”
• Larger torque if
• Larger Force applied
• Greater Distance from O
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F applied
Torque
t=FL
Larger torque if
• Larger Force applied
• Greater Distance from O
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Torque
Line of
action
• The torque of a force with
respect to O is the product of
force and its lever arm.
t=FL
L
O
• Units: Newton-Meters
• But wait, isn’t Nm = Joule??
F applied
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Torque
• Torque Units: Newton-Meters
• Use N-m or m-N
• OR...Ft-Lbs or lb - feet
• Not…
•
•
•
•
Lb/ft
Ft/lb
Newtons/meter
Meters/Newton
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Torque
• Torque is a
ROTATIONAL
VECTOR!
• Directions:
• “Counterclockwise” (+)
• “Clockwise” (-)
• “z-axis”
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Torque
• Torque is a
ROTATIONAL
VECTOR!
• Directions:
• “Counterclockwise”
• “Clockwise”
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Torque
• Torque is a
ROTATIONAL VECTOR!
• Directions:
• “Counterclockwise” (+)
• “Clockwise” (-)
• “z-axis”
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Torque
• Torque is a
ROTATIONAL
VECTOR!
• But if r = 0, no
torque!
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Torque as a vector
• Torque can be expressed
as a vector using the
vector cross product:
t= r x F
• Right Hand Rule for
direction of torque.
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Torque as a vector
• Torque can be expressed
as a vector using the
vector cross product:
t= r x F
• Where r = vector from axis
of rotation to point of
application of Force
q = angle between r and F
• Magnitude: t = r F sinq
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Line of
action
Lever
arm
O r
q
F applied
Torque as a vector
Line of
action
• Torque can be evaluated
two ways:
t= r x F
t= r * (Fsin q)
Lever
arm
O r
F applied
remember sin q = sin (180-q)
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q
Fsinq
Torque as a vector
• Torque can be evaluated
two ways:
t= r x F
t= (rsin q) * F
The lever arm L = r sin q
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Line of
action
Lever
arm
r sin q
O r
F applied
q
Torque as a vector
Line of
action
• Torque is zero three ways:
t= r x F
t= (rsin q) * F
t= r * (Fsin q)
If q is zero (F acts along r)
If r is zero (f acts at axis)
If net F is zero (other forces!)
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Lever
arm
r sin q
O r
F applied
q
Applying a torque
• Find
t
if F = 900 N and q = 19 degrees
Applying a torque
• Find t if F = 900 N and
q = 19 degrees
11-4 Torque Revisited
Example Calculating net torque:
Figure 11-11
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11-5 Angular Momentum



Angular counterpart to linear
momentum
Note that particle need not
rotate around O to have
angular momentum around it
Unit of angular momentum is
kg m2/s, or J s
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11-5 Angular Momentum



To find direction of angular momentum, use right-hand
rule to relate r & v to the result
To find magnitude, use equation for magnitude of a
cross product:
This is equivalent to:
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11-5 Angular Momentum


Angular momentum has meaning only with respect to
a specified origin and axis.
L is always perpendicular to plane formed by position
& linear momentum vectors
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11-5 Angular Momentum


Angular momentum has meaning only with respect to
a specified origin
It is always perpendicular to the plane formed by the
position and linear momentum vectors
Answer: (a) 1 & 3, 2 & 4, 5
(b) 2 and 3 (assuming counterclockwise is positive)
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11-6 Newton's Second Law in Angular Form


We rewrite Newton's second law as:
Torque and angular momentum must be defined with
respect to the same point (usually the origin)
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11-6 Newton's Second Law in Angular Form

We rewrite Newton's second law as:

Note the similarity to the linear form:
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11-6 Newton's Second Law in Angular Form
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11-6 Newton's Second Law in Angular Form
Answer: (a) F3, F1, F2 & F4 (b) F3 (assuming counterclockwise is positive)
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11-7 Angular Momentum of a Rigid Body

Sum to find angular momentum of a system of particles:

The rate of change of the net angular momentum is:
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11-7 Angular Momentum of a Rigid Body

The rate of change of the net angular momentum is:

In other words, net torque is defined by this change:
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11-7 Angular Momentum of a Rigid Body


We can find the angular momentum of a rigid body
through summation:
The sum is the rotational inertia I of the body
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11-7 Angular Momentum of a Rigid Body

Therefore this simplifies to:
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11-7 Angular Momentum of a Rigid Body

Therefore this simplifies to:
11-7 Angular Momentum of a Rigid Body
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11-7 Angular Momentum of a Rigid Body
Answer: (a) All angular momenta will be the same, because the torque is the same in
each case (b) sphere, disk, hoop
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11-8 Conservation of Angular Momentum


Since we have a new version of Newton's second law,
we also have a new conservation law:
The law of conservation of angular momentum
states that, for an isolated system,
(net initial angular momentum) = (net final angular
momentum)
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11-8 Conservation of Angular Momentum


Since these are vector equations, they are equivalent
to the three corresponding scalar equations
This means we can separate axes and write:
11-8 Conservation of Angular Momentum

If the distribution of mass changes with no external
torque, we have:
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11-8 Conservation of Angular Momentum
Examples

A student spinning on a stool: rotation speeds up when arms
are brought in, slows down when arms are extended

A springboard diver: rotational speed is controlled by tucking
her arms and legs in, which reduces rotational inertia and
increases rotational speed

A long jumper: the angular momentum caused by the torque
during the initial jump can be transferred to the rotation of the
arms, by windmilling them, keeping the jumper upright
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11-8 Conservation of Angular Momentum
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11-8 Conservation of Angular Momentum
Answer: (a) decreases (b) remains the same (c) increases
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The race of the rolling bodies
• Which one wins??
• WHY???
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Acceleration of a yo-yo
• We have translation and rotation, so we use Newton’s
second law for the acceleration of the center of mass
and the rotational analog of Newton’s second law for
the angular acceleration about the center of mass.
•What is a and
T for the yo-yo?
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Acceleration of a rolling sphere
• Use Newton’s second law for the motion of the center
of mass and the rotation about the center of mass.
• What are acceleration & magnitude of friction on ball?
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Work and power in rotational motion
• Tangential Force over angle does work
• Total work done on a body by external torque is equal to the
change in rotational kinetic energy of the body
• Power due to a torque is P = tzz
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Work and power in rotational motion
• Calculating Power from Torque
• Electric Motor provide 10-Nm torque on grindstone, with
I = 2.0 kg-m2 about its shaft.
• Starting from rest, find work W down by motor in 8
seconds and KE at that time.
• What is the average power?
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Work and power in rotational motion
• Calculating Power from Force
• Electric Motor provide 10-N force on block, with m =
2.0 kg.
• Starting from rest, find work W down by motor in
8 seconds and KE at that time.
• What is the average power?
• W = F x d = 10N x distance travelled
• Distance = v0t + ½ at2 = ½ at2
• F = ma => a = F/m = 5 m/sec/sec => distance = 160 m
• Work = 1600 J; Avg. Power = W/t = 200 W
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Angular momentum
• The angular momentum of a rigid body rotating about a
symmetry axis is parallel to the angular velocity and is

given by L = I.
• Units = kg m2/sec (“radians” are implied!)
• Direction = along RHR vector of
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angular velocity.
Angular momentum
• For any system of particles t = dL/dt.
• For a rigid body rotating about the z-axis tz = Iz.
Turbine with I = 2.5 kgm2;  = (40 rad/s3)t2
What is L(t) & L @ t = 3.0 seconds; what is t(t)?
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Conservation of angular momentum
• When the net external torque acting on a system is zero, the total
angular momentum of the system is constant (conserved).
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A rotational “collision”
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Angular momentum in a crime bust
• A bullet (mass = 10g, @ 400 m/s) hits a door (1.00 m wide
mass = 15 kg) causing it to swing. What is  of the door?
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Angular momentum in a crime bust
• A bullet (mass = 10g, @ 400 m/s) hits a door (1.00 m wide
mass = 15 kg) causing it to swing. What is  of the door?
• Lintial = mbulletvbulletlbullet
• Lfinal = I = (I door + I bullet) 
• Idoor = Md2/3
• I bullet = ml2
• Linitial = Lfinal so mvl = I
• Solve for  and KE final
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Gyroscopes and precession
• For a gyroscope, the axis of rotation changes
direction. The motion of this axis is called precession.
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11-9 Precession of a Gyroscope

A non-spinning gyroscope falls
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11-9 Precession of a Gyroscope


A spinning gyroscope instead rotates
around a vertical axis
This rotation is called precession
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Gyroscopes and precession
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Gyroscopes and precession
• For a gyroscope, the axis of
rotation changes direction.
The motion of this axis is
called precession.
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A rotating flywheel
• Magnitude of angular momentum constant, but its direction
changes continuously.
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11-9 Precession of a Gyroscope

Angular momentum of (rapidly spinning) gyroscope is:

Torque can only change direction of L, not magnitude


Only way direction can change along direction of
torque without its magnitude changing is if it rotates
around the central axis
Therefore it precesses instead of toppling over
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11-9 Precession of a Gyroscope

The precession rate is given by:

True for a sufficiently rapid spin rate


Independent of mass, (I is proportional to M) but does
depend on g
Valid for a gyroscope at an angle to the horizontal as
well (a top for instance)
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