Lecture 19

Chapter 8- Rotational Motion

AssignmentAssignment 88Textbook (Giancoli, 6 th edition), Chapter 7-8:

Due on Thursday, November 19

1. On page 190 of Giancoli, problem 40.



4. The flywheel of a steam engine runs with a constant angular speed of 156 rev/min. When steam is shut off, the friction of the bearings and of the air brings the wheel to rest in 2.20 h.

(a) What is the constant angular acceleration of the wheel, in rev/min2 ?(b) How many rotations will the wheel make before coming to rest?(c) What is the tangential linear acceleration of a point 52.4 cm from the

axis of rotation when the flywheel is turning at 72.5 rev/min(d) What is the magnitude of the total linear acceleration of the point in part

(c)?

Chapter 8

• Angular Quantities

• Constant Angular Acceleration

• Rolling Motion (Without Slipping)

• Centripetal Forces

• Torque

• Rotational Dynamics; Torque and Rotational Inertia

• Rotational Kinetic Energy

• Angular Momentum and Its Conservation

All marks, including assignments, have been posted on the web.

http://ilc2.phys.uregina.ca/~barbi/academic/phys109 /2009/grades-web.pdf

You have until Friday, Nov. 13, to verify that You have until Friday, Nov. 13, to verify that all your marks have been entered and are

correct in the list.

The link will be removed after 1pm on Friday.

Recalling Recalling LastLast LectureLectureRecalling Recalling LastLast LectureLecture

Angular QuantitiesAngular Quantities

� 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 axismove through the same angle in the same time .

� The angle θ in radians is defined:

Where r = radius of the circlel = arc length covered by the angle θ

The angular displacement is what characterizesthe rotational motion.

(8-1)


One radian is defined such that it correspondsto an arc of circle equal to the radius of the circle.Or, if we use eq. 8.1:

Radians are dimensionless.

(8-2)


(8-3)

Angular displacement:

(8-4)


Average angular velocity and instantaneous angular velocity:

Average angular acceleration and instantaneous angular acceleration:

(8-6)(8-5)

Both the velocity and acceleration arethe same for any point in the object .

(8-8)(8-7)


At each angular position, point P will have alinear velocity whose directions are tangent toits circular path .

Tangential linear acceleration :

∆

∆

(8-9)

Centripetal acceleration (radial direction towardsthe center of rotation):

Total acceleration:

(8-10)

(8-11) (8-12)

(8-13)


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: periodis 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.

TodayTodayTodayToday

Linear MomentumLinear Momentum

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

ω =

(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

R 1.57 rad sec 1.2 m 3.0 m 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 2oω ω α θ= +

( )( )

2 22 2 0 850 rev min rev 2 rad 1 min rad241 0.42oω ω π

α−−

= = = − = −

(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= = = − = −

( )12 o tθ ω ω= +

( )2 1 5 0 0 rev2 6 0 s2 1 0 s

8 5 0 rev m in 1 m ino

tθ

ω ω= = =

+

Centripetal Forces (Section 5Centripetal Forces (Section 5--2, textbook)2, textbook)

We can now go back to chapter 5 and discuss the concept of centripetal forces.

For an object to be in uniform circular motion, there must be a net force acting on it.

In fact, we have discussed that there is a radial (centripetal) force associated to a particle (point) moving in circular motion .

According to Newton’s second law, there should exist a net force Frnet such that:

FRnet is known as centripetal force .

The centripetal force points in the direction of the radial acceleration (inwards).

We can use eq. 8-11 ( ) and rewrite 8-16 as:

(8-16)

(8-17)


A misconception is to believe that there exist a force (centrifugal force ) pointing outwards acting on an object when it moves in a circular path.

An example is the case of a person swinging a ball on the end of a string as shown in the figure.

The force on the person’s hand is the reaction of the string to the inward pull the person exerts on it to produce the circular motion.

When the string is released , the ball flies away When the string is released , the ball flies away tangentially as depicted in the figure b below. So, there was no forces in the outward direction, otherwise we would have motion in this direction after the string is released (figure a).


Note: If a particle is developing a circular motion at constant angular velocity ω, its angular acceleration αααα is zerozerozerozero (eq. 8-8), and the tangential acceleration atan is consequently zero (eq. 8-10).

αααα is zerozerozerozero (eq. 8-8),

consequently zero (eq. 8-10).

� The net force will be exclusively due to the radial acceleration and will point inward

On the other hand, if there is a non-zero angular acceleration αααα , then there is a tangential acceleration atan

� The net force will NOT be pointing inward .

(We will come back to it when we discuss torque)

Problem 5-7 (textbook) A ball on the end of a string is revolved at a uniform rate in a vertical circle of radius 72.0 cm, as shown in Fig. 5–33. If its speed is 4.00 m/s and its mass is 0.300 kg, calculate the tension in the string when the ball is

(a) at the top of its path, and

(b) at the bottom of its path.


Problem 5-7

A free-body diagram is shown in the figure. Since the object is moving in a circle with a constant speed, the net force on the object at any point must point to the center of the circle.

(a) Take positive to be downward. Write Newton’s 2nd law in the downward direction.

2

T1 RR F mg F ma m v r= + = = →∑


This is a downward force, as expected.

( ) ( ) ( )R

2

2 2

T1

4.00 m s0.300 kg 9.80 m s 3.73 N

0.720 mF m v r g= − = − =

∑

Problem 5-7

A free-body diagram is shown in the figure. Since the object is moving in a circle with a constant speed, the net force on the object at any point must point to the center of the circle.

(b) Take positive to be upward. Write Newton’s 2nd law in the upward direction.

2

T 2 RF F m g m a m v r= − = = →∑


This is a upward force, as expected.

( ) ( ) ( )2

2 2

T 1

4.00 m s0.300 kg 9.80 m s 9.61 N

0.720 mF m v r g= + = + =

∑

Rolling Motion (Without Slipping)Rolling Motion (Without Slipping)

When a wheel or radius R rolls without slipping along a flat straight path, the points of the wheel in contact with the surface are instantaneously at rest and the wheel rotates about a rotation axis through the contact point .

In fact, the wheel is experiencing both a translational and rotational motion.

The center C of the wheel does not rotate relative to thecontact point. It undergoes only translational motion withvelocity .

R

P

r ω

velocity .

A point P at a distance r from the center of rotation (point ofcontact) will undergo a rotation with angular velocity given by eq. 8-9:

(nonslip condition for the speed of any point at a distance r).

The center of the wheel moves with a linear speed given by:

(nonslip condition for the speed of the center of the wheel)


The linear acceleration of the center of the wheel will be given by:

(nonslip condition for the acceleration of thecenter of the wheel)

R

P

r

center of the wheel)

ω

θ


Let’s now look at this problem from a different perspective.

So far we have placed ourselves on a reference system which is at rest relative to the surface. Let’s now assumea reference system at rest with respect to the center ofthe wheel.

In this system, the surface (and therefore the point of contact)moves with a velocity .

R

As the wheel rotates about its center through and angle θwith angular velocity ω, the point of contact between thewheel and the surface will have moved a distance l (arc of circle)given by eq. 8-1:

The center of the wheel remains directly over the point of contact, therefore it also moves by the sa medistance.

ω

θ

l

l

R

Rotational Kinetic Energy and Moment of InertiaRotational Kinetic Energy and Moment of Inertia

An object can be seen as made of many individual tiny particles located at different positions. In the real word, this is actually the way things are made. The object can then be interpreted as a system of particles.

We have already seen that the total kinetic energy of a system is the sum of the kinetic energy of each of its constituents. So, the kinetic energy of this object can be written as:

If the object is undergoing a rotational motion, we can use eq. 8-9 (noting that the angular velocity is the same for all particles) to write:


Then

⇒

We can now define the following quantity:

(8-18)

(8-19)

Such that:

The quantity I is known as moment of inertia .

(8-19)

(8-20)


Note that in the case of the wheel discussed in the rolling motion without slipping few slides ago, the wheel also undergoes translational motion.

The total kinetic energy will then be the sum of the kineticenergy due to its linear and angular motions:

R

P

r(8-21)

Where

is the total mass of the object.

ω

θ


Moment of inertia depends not only on the mass of each particle, but also on their distribution (position r) in the object.

These two objects have the same mass , but the one on the left has a greater rotational inertia, as so much of its mass is far from the axis of rotation.

We will come back to moment of inertia when we discuss about torque.


The rotational inertia of an object depends not only on its mass distribution but also the location of the axis of rotation

Compare (f) and (g), for example.

We will come back to moment of inertia when we discuss about torque.


Problem 8-45 (textbook) A bowling ball of mass 7.3 kg and radius 9.0 cm rolls without slipping down a lane at 3.3 m/s. Calculate its total kinetic energy.


Problem 8-45

The total kinetic energy is the sum of the translational and rotational kinetic energies. Since the ball is rolling without slipping, the angular velocity is given by

The rotational inertia of a sphere about an axis through its center is

v Rω =

(see table presented two slides ago)225

I m R=

( )( )

22 2 2 2 271 1 1 1 2

total trans rot 2 2 2 2 5 102

2 0.7 7.3 kg 3.3m s 56 J

vKE KE KE mv I mv mR mv

Rω= + = + = + =

= =

Lecture 19

Documents

Transcript of Lecture 19