chapter 11 rolling, torque, angular...

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.




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:



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


• 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


• 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


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


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)



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



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


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.



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:



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

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


Angular momentum has

meaning only with respect to

a specified origin

11.7 Angular Momentum


Sample problem: Angular Momentum

11.8 Newton’s 2nd Law in Angular Form


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


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


• 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.



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.



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


• 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.



• 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.



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



Sample problem

11 Table 11.1


11 Table 10.3


Example:


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)



System of Particles


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)


11 Solved Problems


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


11 Solved Problems


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


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


11 Solved Problems


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


(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


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


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


(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.



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)



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).



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

Figure 11-2 © 2014 John Wiley & Sons, Inc. All rights reserved.

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.


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


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


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


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


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


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


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.


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


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


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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