physics 151: lecture 20, pg 1 physics 151: lecture 20 today’s agenda l topics (chapter 10) :...

Physics 151: Lecture 20, Pg 1

Physics 151: Lecture 20Physics 151: Lecture 20Today’s AgendaToday’s Agenda

Topics (Chapter 10) :Rotational Kinematics Ch. 10.1-3Rotational Energy Ch. 10.4Moments of Inertia Ch. 10.5


RotationRotation

Up until now we have gracefully avoided dealing with the rotation of objects. We have studied objects that slide, not roll.We have assumed wheels are massless.

Rotation is extremely important, however, and we need to understand it !

Most of the equations we will develop are simply rotational versions of ones we have already learned when studying linear kinematics and dynamics.


RecallRecallKinematic of Circular Motion:Kinematic of Circular Motion:

RR

vv

st

(x,y)

= t

s = v t

s = R = Rt v = R

x = R cos()= R cos(t)y = R sin()= R sin(t)

= tan-1 (y/x)

avR

2

For uniformcircular motion:

is angular velocity

x

y

Animation


Example:Example:

The angular speed of the minute hand of a clock, in rad/s, is:

a. 1800

b. /60

c. /30

d. e. 120

See text: 10.1


Rotational VariablesRotational Variables

Rotation about a fixed axis:Consider a disk rotating about

an axis through its center:

First, recall what we learned aboutUniform Circular Motion:

(Analogous to )

See text: 10.1

dtd

dtdx

v


Rotational Variables...Rotational Variables...

Now suppose can change as a function of time: We define the

angular acceleration:

2

2

dt

ddtd

See text: 10.1

Consider the case when is constant. We can integrate this to find and as a function of time:

t

0

constant

200 2

1tt


Example:Example:

The graphs below show angular velocity as a function of time. In which one is the magnitude of the angular acceleration constantly decreasing ?

See text: 10.1


Rotational Variables...Rotational Variables...

Recall also that for a point a distanceR away from the axis of rotation:x = Rv = R

And taking the derivative of this we finda = R

R

v

x

See text: 10.2

200

0

t21

t

t

constant

Animation


Summary Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)

Angular Linear

constant

0 t

0 021

2t t

constanta

v v at 0

x x v t at 0 021

2

And for a point at a distance R from the rotation axis:

x = Rv = Ra = R

See text: 10.3


Example: Wheel And RopeExample: Wheel And Rope A wheel with radius R = 0.4m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such

that it has a constant acceleration a = 4m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2 radians)

See text: 10.1

aa

R

rev80radrev

21

x rad 500


Example:Example:

The turntable of a record player has an angular velocity of 8.0 rad/s when it is turned off. The turntable comes to rest 2.5 s after being turned off. Through how many radians does the turntable rotate after being turned off ? Assume constant angular acceleration.

a. 12 radb. 8.0 radc. 10 radd. 16 rade. 6.8 rad

See text: 10.1


Rotation & Kinetic EnergyRotation & Kinetic Energy

Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods).

The kinetic energy of this system will be the sum of the kinetic energy of each piece:

rr1

rr2rr3

rr4

m4

m1

m2

m3

Recall text 9.6, systems of particles, CM


Rotation & Kinetic Energy...Rotation & Kinetic Energy...

So: but vi = ri

rr1

rr2rr3

rr4

m4

m1

m2

m3

vv4

vv1

vv3

vv2

K m vi ii

1

22

K m r m ri ii

i ii

12

12

2 2 2

which we write as:

K 12

2I

I m ri ii

2

Define the moment of inertiamoment of inertia

about the rotation axis I has units of kg m2.

Recall text 9.6, systems of particles, CM


Lecture 20, Lecture 20, Act 1Act 1Rotational Kinetic EnergyRotational Kinetic Energy

I have two basketballs. BB#1 is attached to a 0.1m long rope. I spin around with it at a rate of 2 revolutions per second. BB#2 is on a 0.2m long rope. I then spin around with it at a rate of 2 revolutions per second. What is the ratio of the kinetic energy of BB#2 to that of BB#1?

A) 1/4 B) 1/2 C) 1 D) 2 E) 4

BB#1 BB#2


Rotation & Kinetic Energy...Rotation & Kinetic Energy...

The kinetic energy of a rotating system looks similar to that of a point particle:

Point ParticlePoint Particle Rotating System Rotating System

K 12

2I

I m ri ii

2

K mv1

22

v is “linear” velocity

m is the mass.

is angular velocity

I is the moment of inertia

about the rotation axis.


Moment of InertiaMoment of Inertia

Notice that the moment of inertia I depends on the distribution of mass in the system.The further the mass is from the rotation axis, the bigger the moment of inertia.

For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass).

We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics !

K 12

2I I m ri ii

2

See text: 10.4

So where


Calculating Moment of InertiaCalculating Moment of Inertia

We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is:

I m ri ii

N2

1

where r is the distance from the mass

to the axis of rotation.

Example: Calculate the moment of inertia of four point masses

(m) on the corners of a square whose sides have length L,

about a perpendicular axis through the center of the square:

mm

mm

L

See text: 10.5

See example 10.4 (similar)


Calculating Moment of Inertia...Calculating Moment of Inertia...

The squared distance from each point mass to the axis is:

mm

mm

Lr

L/2

I = 2mL2

See text: 10.5




Now calculate I for the same object about an axis through the center, parallel to the plane (as shown):

mm

mm

L

r

4L

m44L

m4L

m4L

m4L

mrmI22222N

1i

2ii

I = mL2

See text: 10.5




Finally, calculate I for the same object about an axis along one side (as shown):

mm

mm

L

r

2222N

1i

2ii 0m0mmLmLrmI

I = 2mL2

See text: 10.5




For a single object, I clearly depends on the rotation axis !!

L

I = 2mL2I = mL2

mm

mm

I = 2mL2

See text: 10.5



Lecture 20, Lecture 20, Act 2Act 2Moment of InertiaMoment of Inertia

A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Ic respectively.Which of the following is correct:

(a)(a) Ia > Ib > Ic

(b)(b) Ia > Ic > Ib

(c)(c) Ib > Ia > Ic

a

b

c


Lecture 20, Lecture 20, Act 2Act 2Moment of InertiaMoment of Inertia

a

b

c

Label masses and lengths:

m

m m

L

L

Ia m L m L mL 2 2 82 2 2

Calculate moments of inerta:

Ib mL mL mL mL 2 2 2 23

Ic m L mL 2 42 2

So (b) is correct: Ia > Ic > Ib



For a discrete collection of point masses we found:

For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm.

We have to do anintegral to find I :

I m ri ii

N2

1

r

dm

I r dm2

See text: 8-5


Moments of InertiaMoments of Inertia

Some examples of I for solid objects:

Thin hoop (or cylinder) of mass M and radius R, about an axis through its center, perpendicular to the plane of the hoop.

I MR 2

R

See text: 10.5

I r2 dm

I R2dm R2 dm MR2

I 1

22MR

Thin hoop of mass M and radius R,

about an axis through a diameter.

R

see Example 10.5 in the text


Moments of InertiaMoments of Inertia

Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center.


RL

I r2 dm

I1

2MR2

rdr


Moments of Inertia...Moments of Inertia...


Solid sphere of mass M and radius R,

about an axis through its center.

I 2

52MR

R

See text: 10.5

See Table 10.2, Moments of Inertia

Thin spherical shell of mass M and radius R, about an axis through its center.

R

I2

3MR2


Recap of today’s lectureRecap of today’s lecture

Chapter 9, Center of MassElastic CollisionsImpulse

physics 151: lecture 20, pg 1 physics 151: lecture 20 today’s agenda l topics (chapter 10) :...

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