Chapter 8-Rotational Motion - University of Reginauregina.ca/~barbi/academic/phys109/2008/2008/notes/lecture-18.pdf · • Rotational Dynamics; Torque and Rotational Inertia ... Now,

Chapter 8- Rotational Motion

AssignmentAssignment 88

Textbook (Giancoli, 6th edition), Chapter 7-8:

Due on Thursday, November 13, 2008

- Problem 28 - page 189 of the textbook

- Problem 40 - page 190 of the textbook

- Problems 7 and 17 - page 219 of the textbook

Old assignments and midterm exams

(solutions have been posted on

the web)

can be picked up in my office

(LB-212)

All marks, including assignments, have

been posted on the web.

http://ilc2.phys.uregina.ca/~barbi/academic/phys109/marks.pdfhttp://ilc2.phys.uregina.ca/~barbi/academic/phys109/marks.pdf

Please, verify that all your marks have

been entered in the list.

Chapter 8

• Angular Quantities

• Constant Angular Acceleration

• Rolling Motion (Without Slipping)

• Torque

• Rotational Dynamics; Torque and Rotational Inertia

• Collisions in Two or Three Dimensions

• Rotational Kinetic Energy

• Angular Momentum and Its Conservation

• Vector Nature of Angular Quantities

Recalling Recalling LastLast LectureLectureRecalling Recalling LastLast LectureLecture

Momentum is a vector symbolized by the symbol p, and is defined as

The momentum of an object tells how hard (or easy) is to change its state of

motion.

Momentum and Its Relation to ForceMomentum and Its Relation to Force

(7-20)

Eq. 6-22 is another way of expressing Newton’s second law. However, it is a more

general definition because it introduces the situation where the mass may change.

Momentum and Its Relation to ForceMomentum and Its Relation to Force

(7-22)

their surrounding environment, and act for a very short period of time ∆t.

We can use eq. 6-22 and define the impulse on an object as:

Collision and ImpulseCollision and Impulse

(7-23)(7-23)

Conservation of MomentumConservation of Momentum

(7-25)

Equation 6-25 tells that the total momentum of the system (the sum of the

momentum of the two balls) before the collision is equal to the total momentum of

the system after the collision IF the net force acting on the system is zero ��

isolated system. This is known as Conservation of Total Momentum.

The above equation can be extended to include any number of objects such that the

only forces are the interaction between the objects in the system.

Conservation of Energy and Momentum in CollisionsConservation of Energy and Momentum in Collisions

In general, we can identify two different types of collisions:

1. Elastic collision

2. Inelastic collision

In elastic collisions the total kinetic energy of a system is conserved.

� There is no energy dissipate in form of heat or other form of energy.

An example is the collision between the two billiard

balls discussed in the previous slides:balls discussed in the previous slides:

In inelastic collision, there is NO conservation of

kinetic energy.

� Some of the total initial kinetic energy is

transformed into some other form of energy.

(7-26)


Note:

Also…

The total energy (the sum of all energies) in a closed (isolated system)

is ALWAYS conserved.is ALWAYS conserved.


Collision in two or more dimensions

We then have:

(i)

(ii)

It follows then, using the above expressions in (i) and (ii), that:

⇒ This gives a system of two equations

and three variables. if you can measure any

of these variables, the other two can be

calculated from system of equations.

TodayTodayTodayToday

Linear MomentumLinear Momentum

Problem 7-34 (textbook) An internal explosion breaks an object, initially at rest, into

two pieces, one of which has 1.5 times the mass of the other. If 7500 J were released

in the explosion, how much kinetic energy did each piece acquire?

.


Problem 7-34

Use conservation of momentum in one dimension, since the particles will separate

and travel in opposite directions. Call the direction of the heavier particle’s motion the

positive direction. Let A represent the heavier particle, and B represent the lighter

particle. We have

A B1.5m m=

A B0v v= =

B B 2 0

m vp p m v m v v v

′′ ′ ′ ′= → = + → = − = −

The negative sign indicates direction.

Since there was no mechanical energy before the explosion, the kinetic energy of

the particles after the explosion must equal the energy added.

Thus:

2

initial final A A B B A B3

A

0 p p m v m v v vm

′ ′ ′ ′= → = + → = − = −

( ) ( ) ( )( )

22 2 2 25 51 1 1 2 1 1

added 2 2 2 3 2 3 2 3

3 3

added added5 5

1.5

7500 J 4500 J 7500 J 4500 J 3000 J

A B A A B B B B B B B B B

B A B

E KE KE m v m v m v m v m v KE

KE E KE E KE

′ ′ ′ ′ ′ ′ ′ ′= + = + = + = =

′ ′ ′= = = = − = − =

3 3

3.0 10 J 4.5 10 JA B

KE KE′ ′= × = ×

Angular QuantitiesAngular Quantities

We have extensively discussed translational (linear) motion of an object in terms of

its kinematics (displacement), dynamics (forces) and energy.

But we also know that objects can also move following some circular path.

This is called rotational motion.

The basis for the discussions on rotational motion is what we have seen so far

concerning translational motion.

So, I will use the definitions introduced in the previous chapters and apply then to

introduce you to rotational motion.

We will consider only rigid objects � in other words, objects that do not change

shape (or the distances between points in it do not change)


� In purely rotational motion, all points on the object move in circles around the axis

of rotation (“O”) which is perpendicular to this slide.

� The radius of the circle is r.

� All points on a straight line drawn through the axis

move through the same angle in the same time.

� The angle θ in radians is defined:

Where r = radius of the circle

l = arc length covered by the angle θ

The angular displacement is what characterizes

the rotational motion.

For mathematical reasons, it is more convenient

(8-1)


For mathematical reasons, it is more convenient

to define angle not in terms of degrees but

in terms of radians.

One radian is defined such that it corresponds

to an arc of circle equal to the radius of the circle.

Or, if we use eq. 8.1:

Note that radians are dimensionless.

Radian can be related to degrees by observing the

the full length of a circle corresponds to the maximum

arc length, or 2πr. It comprises an angle of 3600.

Using eq. 8.1, we have:

(8-2)


We can also define an object revolution as the length in radians that the object has

travelled. A complete revolution will correspond to the

total length of the circle, or:

As mentioned before, the object rotational displacement

is defined in terms of the angle θ.

(8-3)

Defining a coordinate system as in the figure, the

displacement of a certain point P in the object can be

given by:

(8-4)


In a similar way we did for translational motion, we can define the average angular

velocity and instantaneous angular velocity of this point as:

Similarly, the average acceleration and instantaneous

acceleration can be defined as:

(8-6)(8-5)

Note that given the fact that each point in the object

will be displaced by the same angle in the same

interval of time, both the velocity and acceleration are

the same for any point in the object.

(8-8)(8-7)


Now, the rotational motion of an object or a point in the object can be related to its

translational motion.

For realizing that, you should observe that a point

rotating around a circle will also be subjected to a

translational motion as depicted in the figure.

At each angular position, this point will have a

linear velocity whose directions are tangent to

its circular path.

∆

∆

its circular path.

Note: the direction of the linear velocity changes as the

point undergoes a rotational motion. This is due to the so

called centripetal acceleration. We will come back to this

acceleration in the next slides.


Back to the linear velocity, the figure shows that a change in the rotation angle ∆θ

corresponds to a linear distance traveled ∆l.

With the help of eq. 8-1:∆

∆

(8-9)


Eq. 8-9 says that although ω is the same for every

point in the rotating object, the linear velocity changes

with the distance r of the point to the axis of rotation.

� Therefore, objects farther from the axis of

rotation will move faster.

αααα

(8-9)

If there is an angular acceleration αααα, the angular

velocity ω changes. Therefore, there is also a change

in the linear velocity and thus an acceleration involved

in the process. This acceleration is in the direction of

the velocity and is called tangential linear acceleration:

(8-10)


As already mentioned a couple of slides ago, the velocity changes direction.

But the tangential acceleration is parallel to the velocity, so it is not responsible for the

change in the velocity’s direction.

It turns out that this acceleration is called radial acceleration,

or centripetal acceleration. It is always perpendicular

to the direction of the velocity.

The magnitude of the centripetal acceleration is givenThe magnitude of the centripetal acceleration is given

by magnitude (see textbook, page 107 for details):

Therefore, objects farther from the axis of rotation will

have greater centripetal acceleration.

(8-11)

(8-12)


The total acceleration of the point at a distance r from the axis of the rotation of the

object will be the vector sum the radial (centripetal) and linear (tangential)

accelerations:

We can summarize this discussion with the following table:

(8-13)

We can summarize this discussion with the following table:


Another important quality in rotational motion the frequency of rotation of an object.

The frequency is the number of complete revolutions per second:

Frequencies are measured in hertz.

(8-14)

The time required for a complete revolution is called period, or in other words: period

is the time one revolution takes:

(8-15)

Constant Angular AccelerationConstant Angular Acceleration

The equations of motion for constant angular acceleration are the same as those for

linear motion, with the substitution of the angular quantities for the linear ones.


Problem 8-8 (textbook) A rotating merry-go-round makes one complete revolution

in 4.0 s (Fig. 8–38).

(a) What is the linear speed of a child seated 1.2 m from the center?

(b) What is her acceleration (give components)?


Problem 8-8

The angular speed of the merry-go-round is

(a)

2 rad 4.0 s 1.57 rad sπ =

( )( )1 .5 7 ra d s e c 1 .2 m 1 .9 m sv rω= = =

(b) Ignoring air our other resistance, there is no tangential acceleration (no tangential

forces are applied).Therefore, the acceleration is purely radial.

and

atan

= 0

( )( )1 .5 7 ra d s e c 1 .2 m 1 .9 m sv rω= = =

( ) ( )22 2

R1.57 rad sec 1.2 m 3.0m s towards the centera rω= = =


Problem 8-13 (textbook) A turntable of radius R1 is turned by a circular rubber roller

of radius R2 in contact with it at their outer edges. What is the ratio of their angular

velocities, ω1/ ω2.


Problem 8-13

The tangential speed of the turntable must be equal to the tangential speed of the

roller, if there is no slippage.

1 2 1 1 2 2 1 2 2 1 v v R R R Rω ω ω ω= → = → =


Problem 8-19 (textbook) A cooling fan is turned off when it is running at 850 rev/min.

It turns 1500 revolutions before it comes to a stop.

(a) What was the fan’s angular acceleration, assumed constant?

(b) How long did it take the fan to come to a complete stop?


Problem 8-19

(a)

The angular acceleration can be found from

2 2

2o

ω ω α θ= +

( )( )

2 22 20 850 rev min rev 2 rad 1 min rad

241 0.42o

ω ω πα

−−= = = − = −

(b)

The time to come to a stop can be found from

( ) 2 2241 0.42

2 2 1500 rev min 1 rev 60 s sα

θ= = = − = −

( )1

2 otθ ω ω= +

( )2 1500 rev2 6 0 s2 1 0 s

8 5 0 rev m in 1 m ino

tθ

ω ω= = =

+

Documents

Chapter 8-Rotational Motion - University of Reginauregina.ca/~barbi/academic/phys109/2008/2008/notes/lecture-18.pdf · • Rotational Dynamics; Torque and Rotational Inertia ... Now,