chapter 11 rolling, torque, angular...
TRANSCRIPT
Chapter 11
Rolling, Torque, Angular
Momentum
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
11.2 Rolling as Translational and Rotation Combined
© 2014 John Wiley & Sons, Inc. All rights reserved.
• Motion of Translation : i.e.motion along a straight line
• Motion of Rotation : rotation about a fixed axis
Pure Translation Motion + Pure Rotation Motion = ROLLING MOTION
com Smooth Rolling
of a Disk
Translation of COM Rotation around COM
Although the center of the object moves in a straight line parallel to the
surface, a point on the rim certainly does not.
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11.2 Rolling as Translational and Rotation Combined
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s= R
• Wheel moves through the arc
length of ‘’s’’
• Wheel’s COM moves through a
translational motion
• PATHS ARE EQUAL
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:
11.2 Rolling as Translational and Rotation Combined
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Every point on the wheel
rotates about com with
angular speed of
Everypoint outmost part
of the wheel has linear
speed of ν;
Where ν = R = νcom
All points on the wheel
move to the right with
same linear velocity, νcom
At the bottom of the
wheel (point P), the
portion of the wheel is
stationary
The portion at the top
(point T) moving at a
speed 2νcom
At points B and D, the
speeds are smaller than
the point T.
B D
The figure shows how the velocities of translation and rotation combine at different points on the wheel
com
(Vp = 0)
R
11.3 The Kinetic Energy of Rolling
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• Rolling can be considered as pure rotation around contact point P.
(Vp=0)
Rotational inertia about COM: Icom
Rotational inertia about point P:
IP = Icom + MR2
M : mass of the wheel,
Icom : rotational inertia about an axis through its center of mass
R : the wheel’s radius, at a perpendicular distance h).
(Using the parallel-axis
theorem )
IP = Icom + MR2 K = ½ Icom 2 + ½ M R2 2
inserting vcom (vcom = R), then;
K.E of ROLLING: Rotational kinetic energy + Translational kinetic energy
11.4 The Forces of Rolling: Friction and Rolling
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• A wheel rolls horizontally without sliding while accelerating with
linear acceleration acom.
• A static frictional force fs acts on the wheel at P, opposing its tendency
to slide.
R :the radius of the wheel.
α : angular acceleration
The magnitudes of the linear acceleration acom,
and the angular acceleration a can be related by;
• Rolling is possible when there is friction between the surface and the
rolling object.
• The frictional force provides the torque to rotate the object.
11.4 The Forces of Rolling: Friction and Rolling
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If the wheel slides when the net force acts on it, the frictional force that
acts at P in Figure is a kinetic frictional force, fk.
The motion then is not smooth rolling, and the above relation
(aCOM=aR) does not apply to the motion.
Sliding of Wheel Motion is not Smooth Rolling
fk
11.4 The Forces of Rolling: Rolling Down a Ramp
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A round uniform body of radius R rolls (smoothly fs) down a ramp.
We want to find an expression for the body’s acceleration acom,x down the ramp.
We do this by using Newton’s second law in its linear version (Fnet=Ma) and its
angular version (tnet=Ia).
For smooth rolling
down a ramp:
1. Fg : The gravitational
force is vertically down
2. FN: The normal force
is perpendicular to the
ramp
3. fs :The force of friction
points up the slope
(opposing the sliding)
4. acom=aR where acom is
(–x) and a is (+, ccw)
(see 4)
11.4 The Forces of Rolling: Rolling Down a Ramp
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Ki + Ui = Kf + Uf
Mgh = ½ ICOM ω2 + ½ MvCOM2
vcm = ωR
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
11.4 The Forces of Rolling: Rolling Down a Ramp
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Sample problem: Rolling Down a Ramp
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?
Calculations: Where Icom is the ball’s rotational inertia about an axis through its center of
mass, vcom is the requested speed at the bottom, and is the angular speed there.
Substituting vcom/R for , and 2/5 MR2 for Icom,
b) What are the magnitude and
direction of the frictional force
on the ball as it rolls down the
ramp?
Calculations: First we need to
determine the ball’s acceleration
acom,x :
We can now solve for the frictional force:
11.6 Torque Revisited
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Figure (a) A force F, lying in an x-y plane, acts on a particle at point A. (b) This force produces a
torque t = r x F on the particle with respect to the origin O. By the right-hand rule for vector
(cross) products, the torque vector points in the positive direction of z. Its magnitude is given by
in (b) and by in (c).
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.
11.6 Torque Revisited
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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:
11.6 Torque Revisited
© 2014 John Wiley & Sons, Inc. All rights reserved.
Sample problem: Torque on a particle due to a force
In Figure , three forces, each
magnitude 2.0 N, act on a particle.
The particle is in the xz plane at a
point A given by position vector r,
where r = 3.0 m and = 30o. Force
F1 is parallel to the x axis, force F2
is parallel to the z axis, and F3 is
parallel to the y axis. What is the
torque, about the origion O, due to
each force?
Calculations: Because we want the
torques with respect to the origin O,
the vector required for each cross
product is the given position vector
r. To determine the angle q between
the direction of r and the direction of
each force, we shift the force vectors
of Fig.a, each in turn, so that their
tails are at the origin. Figures b, c,
and d, which are direct views of the
xz plane, show the shifted force
vectors F1, F2, and F3. respectively.
In Fig. d, the angle between the
directions of r and t is 90°.
Now, we find the magnitudes of the
torques to be
11.6 Torque Revisited
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Here we investigate the angular counterpart
to linear momentum
We write:
Note that the particle need not rotate around
O to have angular momentum around it.
The unit of angular momentum is kg m2/s,
or J s
To find the direction of angular momentum,
use the right-hand rule to relate r and v to
the result
To find the magnitude, use the equation for
the magnitude of a cross product:
Which can also be written as:
11.7 Angular Momentum
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Angular momentum has
meaning only with respect to
a specified origin
11.7 Angular Momentum
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Sample problem: Angular Momentum
11.8 Newton’s 2nd Law in Angular Form
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Newton’s 2nd Law for translation
Newton’s 2nd Law for ANGULAR form
PROOF: • We rewrite Newton's second
law as:
The torque and the angular momentum must be defined with
respect to the same point (usually the origin).
11.8 Newton’s 2nd Law in Angular Form
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Sample problem: Torque, Penguin Fall Calculations: The magnitude of l can be found
by using
The perpendicular distance between O and an
extension of vector p is the given distance
D. The speed of an object that has fallen from
rest for a time t is v =gt. Therefore,
To find the direction of we use the right-hand
rule for the vector product, and find that the
direction is into the plane of the figure. The
vector changes with time in magnitude only; its
direction remains unchanged.
(b) About the origin O, what is the torque on the
penguin due to the gravitational force ?
Calculations:
Using the right-hand rule for the vector product
we find that the direction of t is the negative
direction of the z axis, the same as l.
11.9 The Angular Momentum of a System of Particles
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• The total angular momentum L of the system is the (vector) sum of the
angular momenta l of the individual particles (here with label i):
• With time, the angular momenta of individual particles may change
because of interactions between the particles or with the outside. The
rate of change of the net angular momentum is:
• Therefore, the net external torque acting on a system of particles is
equal to the time rate of change of the system’s total angular
momentum L.
11.9 The Angular Momentum of a System of Particles
© 2014 John Wiley & Sons, Inc. All rights reserved.
Note that the torque and angular momentum must
be measured relative to the same origin.
If the center of mass is accelerating, then that
origin must be the center of mass.
a) A rigid body rotates about a z axis with
angular speed . A mass element of mass
Dmi within the body moves about the z axis
in a circle with radius . The mass element
has linear momentum pi and it is located
relative to the origin O by position vector ri.
Here the mass element is shown when is
parallel to the x axis.
b) The angular momentum li, with respect to O,
of the mass element in (a). The z component
liz is also shown.
11.9 The Angular Momentum of a System of Particles
© 2014 John Wiley & Sons, Inc. All rights reserved.
We can find the angular momentum of a rigid body through summation:
The sum is the rotational inertia I of the body.
Therefore this simplifies to:
11.11 Conservation of Angular Momentum
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• Depending on the torques acting on the system, the angular momentum of
the system might be conserved in one or two directions but not in all
directions.
• This means that if external torque along an axis is zero then L is constant.
• 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)
• If the net external torque acting on a system is zero, the angular
momentum L of the system remains constant, no matter what changes
take place within the system.
11.11 Conservation of Angular Momentum
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• If the component of the net external torque on a system along a certain axis is
zero, then the component of the angular momentum of the system along that
axis cannot change, no matter what changes take place within the system.
(a) The student has a relatively large rotational inertia about the rotation
axis and a relatively small angular speed.
(b) By decreasing his rotational inertia, the student automatically increases
his angular speed.
The angular momentum of the rotating system remains unchanged.
11.11 Conservation of Angular Momentum
© 2014 John Wiley & Sons, Inc. All rights reserved.
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.
Examples:
Angular
momentum
conservation
11.11 Conservation of Angular Momentum
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Sample problem
11 Table 11.1
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11 Table 10.3
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Example:
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Rolling Bodies
Torque as a Vector
Direction given by the right-
hand rule
11 Summary
Eq. (11-2)
Eq. (11-18)
Newton's Second Law in
Angular Form
Eq. (11-14)
Angular Momentum of a
Particle
Eq. (11-23)
Eq. (11-5)
Eq. (11-6)
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Angular Momentum of a
System of Particles
Angular Momentum of a
Rigid Body
11 Summary
Conservation of Angular
Momentum
Eq. (11-32)
Eq. (11-33)
Precession of a Gyroscope
Eq. (11-46)
Eq. (11-26)
Eq. (11-29)
Eq. (11-31)
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11 Solved Problems
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1. In Figure, a solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls
without slipping a distance L=6.0 m down a roof that is inclined at the angle ϴ=30o.
(a) What is the angular speed of the cylinder about its center as it leaves the roof?
(b) The roof's edge is at height H=5.0 m. How far horizontally from the roof's edge
does the cylinder hit the level ground?
11 Solved Problems
© 2014 John Wiley & Sons, Inc. All rights reserved.
11 Solved Problems
© 2014 John Wiley & Sons, Inc. All rights reserved.
2. At time t=0, a 3.0 kg particle with velocity v=(5.0 m/s)i –(6.0 m/s)j is at x=3.0 m,
y=8.0 m. It is pulled by a 7.0 N force in the negative x direction. About the origin,
what are (a) the particle's angular momentum, (b) the torque acting on the particle,
and (c) the rate at which the angular momentum is changing?
11 Solved Problems
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3. In Figure, a small 50 g block slides down a frictionless surface through height h=20
cm and then sticks to a uniform rod of mass 100 g and length 40 cm. The rod pivots
about point O through angle ϴ before momentarily stopping. Find ϴ.
11 Solved Problems
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11 Solved Problems
© 2014 John Wiley & Sons, Inc. All rights reserved.
11 Solved Problems
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4. In Figure, a constant horizontal force Fapp of magnitude 12 N is applied to a
uniform solid cylinder by 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 center of mass of the cylinder?
(b) What is the magnitude of the angular acceleration of the cylinder about the center of
mass?
(c) In unit‐vector notation, what is the frictional force acting on the cylinder?
11 Solved Problems
© 2014 John Wiley & Sons, Inc. All rights reserved.
(a) What is the magnitude of the acceleration of the center of mass of the cylinder?
(b) What is the magnitude of the angular acceleration of the cylinder about the center of
mass?
(c) In unit‐vector notation, what is the frictional force acting on the cylinder?
11 Solved Problems
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5. Figure shows a uniform disk, with mass M = 2.5 kg and radius R = 20 cm, mounted
on a fixed horizontal axle. A block with mass m = 1.2 kg hangs from a massless
cord that is wrapped around the rim of the disk. Find the acceleration (a) of the
falling block, the angular acceleration (α) of the disk, and the tension (T ) in the
cord. The cord does not slip, and there is no friction at the axle. (I = 1/2 MR2)
1-8 Scientific Notation
Additional Materials
11.5 The Yo-Yo
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As a yo-yo moves down a string, it loses potential
energy mgh but gains rotational and translational
kinetic energy
11.12 Precession of a Gyroscope
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(b) A rapidly spinning gyroscope, with angular
momentum, L, precesses around the z axis. Its
precessional motion is in the xy plane. This rotation is
called precession
(c) The change in angular momentum, dL/dt, leads to a
rotation of L about O.
(a) A non-spinning gyroscope falls by rotating in an xz
plane because of torque t.
11.12 Precession of a Gyroscope
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The angular momentum of a (rapidly spinning)
gyroscope is:
The torque can only change the direction of L, not
its magnitude, because of (11-43)
The only way its direction can change along the
direction of the torque without its magnitude
changing is if it rotates around the central axis.
Therefore it precesses instead of toppling over.
Eq. (11-43)
11.12 Precession of a Gyroscope
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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).
11.11 Conservation of Angular Momentum
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Sample
problem
Figure 11-20 a shows a student, sitting on a stool
that can rotate freely about a vertical axis. The
student, initially at rest, is holding a bicycle wheel
whose rim is loaded with lead and whose rotational
inertia Iwh about its central axis is 1.2 kg m2. (The
rim contains lead in order to make the value of Iwh
substantial.) The wheel is rotating at an angular
speed wh of 3.9 rev/s; as seen from overhead, the
rotation is counterclockwise. The axis of the wheel
is vertical, and the angular momentum Lwh of the
wheel points vertically upward. The student now
inverts the wheel (Fig. 11-20b) so Lwh that, as seen
from overhead, it is rotating clockwise.
Its angular momentum is now - Lwh.The
inversion results in the student, the stool, and the
wheel’s center rotating together as a composite
rigid body about the stool’s rotation axis, with
rotational inertia Ib= 6.8 kg m2. With what
angular speed b and in what direction does the
composite body rotate after the inversion of the
wheel?
11-2 Rolling as Translation and Rotation Combined
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.
Learning Objectives
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11-3_4 Forces and Kinetic Energy of Rolling
11.03 Calculate the kinetic
energy of a body in smooth
rolling as the sum of the
translational kinetic energy of
the center of mass and the
rotational kinetic energy
around the center of mass.
11.04 Apply the relationship
between the work done on a
smoothly rolling object and its
kinetic energy change.
11.05 For smooth rolling (and
thus no sliding), conserve
mechanical energy to relate
initial energy values to the
values at a later point.
11.06 Draw a free-body
diagram of an accelerating
body that is smoothly rolling
on a horizontal surface or up
or down on a ramp.
Learning Objectives
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11.07 Apply the relationship between the center-of-mass acceleration and the angular acceleration.
11.08 For smooth rolling up or down a ramp, apply the relationship between the object’s acceleration, its rotational inertia, and the angle of the ramp.
11-3_4 Forces and Kinetic Energy of Rolling
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11-5 The Yo-Yo
11.09 Draw a free-body
diagram of a yo-yo moving
up or down its string.
11.10 Identify that a yo-yo is
effectively an object that rolls
smoothly up or down a ramp with an incline angle of 90°.
11.11 For a yo-yo moving up
or down its string, apply the
relationship between the yo-
yo's acceleration and its
rotational inertia.
11.12 Determine the tension in
a yo-yo's string as the yo-yo
moves up or down the
string.
Learning Objectives
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11-6 Torque Revisited
11.13 Identify that torque is a
vector quantity.
11.14 Identify that the point
about which a torque is
calculated must always be
specified.
11.15 Calculate the torque due
to a force on a particle by
taking the cross product of
the particle's position vector
and the force vector, in
either unit-vector notation or
magnitude-angle notation.
11.16 Use the right-hand rule
for cross products to find the
direction of a torque vector.
Learning Objectives
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11-7 Angular Momentum
11.17 Identify that angular
momentum is a vector
quantity.
11.18 Identify that the fixed
point about which an angular
momentum is calculated
must always be specified.
11.19 Calculate the angular
momentum of a particle by
taking the cross product of
the particle's position vector
and its momentum vector, in
either unit-vector notation
or magnitude-angle notation.
11.20 Use the right-hand rule
for cross products to find the
direction of an angular
momentum vector.
Learning Objectives
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11-8 Newton's Second Law in Angular Form
11.21 Apply Newton's second law in angular form to relate the
torque acting on a particle to the resulting rate of change of the
particle's angular momentum, all relative to a specified point.
Learning Objectives
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11-9 Angular Momentum of a Rigid Body
11.22 For a system of
particles, apply Newton's
second law in angular form
to relate the net torque
acting on the system to the
rate of the resulting change
in the system's angular
momentum.
11.23 Apply the relationship
between the angular
momentum of a rigid body
rotating around a fixed axis
and the body's rotational
inertia and angular speed
around that axis.
11.24 If two rigid bodies rotate
about the same axis,
calculate their total angular
momentum.
Learning Objectives
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11-11 Conservation of Angular Momentum
11.25 When no external net torque acts on a system along a
specified axis, apply the conservation of angular momentum
to relate the initial angular momentum value along that axis to
the value at a later instant.
Learning Objectives
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11-9 Precession of a Gyroscope
11.26 Identify that the
gravitational force acting on
a spinning gyroscope
causes the spin angular
momentum vector (and thus
the gyroscope) to rotate
about the vertical axis in a
motion called precession.
11.27 Calculate the precession
rate of a gyroscope.
11.28 Identify that a
gyroscope's precession rate
is independent of the
gyroscope's mass.
Learning Objectives
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