physics 111: lecture 17, pg 1 physics 111: lecture 17 today’s agenda l rotational kinematics...

Physics 111: Lecture 17, Pg 1

Physics 111: Lecture 17

Today’s Agenda

Rotational KinematicsAnalogy with one-dimensional kinematics

Kinetic energy of a rotating system Moment of inertiaDiscrete particlesContinuous solid objects

Parallel axis theorem


Rotation

Up until now we have gracefully avoided dealing with the rotation of objects. We have studied objects that slide, not roll.We have assumed pulleys are without mass.

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

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


Lecture 17, Act 1Rotations

Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds.Klyde’s angular velocity is:

(a) the same as Bonnie’s

(b) twice Bonnie’s

(c) half Bonnie’s


Lecture 17, Act 1Rotations

The angular velocity w of any point on a solid object rotating about a fixed axis is the same.Both Bonnie & Klyde go around once (2p radians) every

two seconds.

w

(Their “linear” speed v will be different since v = wr).

BonnieKlyde V21

V


Rotational 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 )

dtd

dtdx

v

Spin round

blackboard


Rotational Variables...

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

angular acceleration:

2

2

dt

d

dt

d

Consider the case when is constant. We can integrate this to

find and as a function of time:

t

0

constant

200 2

1tt


Rotational Variables...

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

And taking the derivative of this we find:a = R

R

v

x 2

00

0

t2

1t

t

constant


Summary (with comparison to 1-D kinematics)

Angular Linear

constant

t0

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 = R a = R


Example: Wheel And Rope

A wheel with radius R = 0.4 m 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 = 4 m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2 radians)

a

R


Wheel And Rope...

Use a = R to find : = a / R = 4 m/s2 / 0.4 m = 10 rad/s2

Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester.

rev80radrot

21

x rad500

20 t

21

t = 0 + 0(10) + (10)(10)2 = 500 rad210

a

R


Rotation & 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:

r1

r2r3

r4

m4

m1

m2

m3


Rotation & Kinetic Energy...

So: but vi = ri

r1

r2r3

r4

m4

m1

m2

m3

v4

v1

v3

v2

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 inertiaabout the rotation axis I has units of kg m2.


Rotation & Kinetic Energy...

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

Point Particle Rotating System

K 12

2I

I m ri ii

2

K mv1

22

v is “linear” velocitym is the mass.

is angular velocityI is the moment of inertiaabout the rotation axis.


Moment 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

Inertia Rods

So where


Calculating 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


Calculating Moment of Inertia...

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

mm

mm

Lr

L/2

2L

2L

2r22

2

2L

m42L

m2L

m2L

m2L

mrmI22222N

1i

2ii

so

I = 2mL2

Using the Pythagorean Theorem



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



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



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

L

I = 2mL2I = mL2

mm

mm

I = 2mL2


Lecture 17, Act 2Moment 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) Ia > Ib > Ic

(b) Ia > Ic > Ib

(c) Ib > Ia > Ic

a

b

c



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


Moments 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

I 1

22MR

Thin hoop of mass M and radius R, about an axis through a diameter.

R

Hoop


Moments of Inertia...

Some examples of I for solid objects:

Solid sphere of mass M and radius R, about an axis through its center.

I 2

52MR

R

I 1

22MR

R

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

Sphere and disk



Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold.Which one has the biggest moment of inertia about an axis

through its center?

same mass & radius

solid hollow

(a) solid aluminum (b) hollow gold (c) same



Moment of inertia depends on mass (same for both) and distance from axis squared, which is bigger for the shell since its mass is located farther from the center.The spherical shell (gold) will have a bigger moment of

inertia.

same mass & radius

ISOLID < ISHELL

solid hollow


Moments of Inertia...

Some examples of I for solid objects (see also Tipler, Table 9-1):

Thin rod of mass M and length L, about a perpendicular axis through its center.

I 1

122ML

L

Thin rod of mass M and length L, about a perpendicular axis through its end.

I 1

32ML

L

Rod


Parallel Axis Theorem

Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass, ICM, is known.

The moment of inertia about an axis parallel to this axis but a distance D away is given by:

IPARALLEL = ICM + MD2

So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis.


Parallel Axis Theorem: Example

Consider a thin uniform rod of mass M and length D. Figure out the moment of inertia about an axis through the end of the rod.

IPARALLEL = ICM + MD2

L

D=L/2M

xCM

ICM ML1

122

We know

IEND ML ML

ML

1

12 2

1

32

22So

which agrees with the result on a previous slide.

ICMIEND


Connection with CM motion

Recall what we found out about the kinetic energy of a system of particles in Lecture 15:

2CM

2iiNET MV

2

1um

2

1K

KREL KCM

KREL CM1

22I

For a solid object rotating about its center of mass, we now see that the first term becomes:

2iiREL um

21

K Substituting ii ru

2ii

2REL rm

21

K but CM2

ii rm I


Connection with CM motion...

So for a solid object which rotates about its center or mass and whose CM is moving:

2CM

2CMNET MV

2

1

2

1K I

VCM

We will use this formula more in coming lectures.


Recap of today’s lecture

Rotational Kinematics (Text: 9-1) Analogy with one-dimensional kinematics

Kinetic energy of a rotating system Moment of inertia (Text: 9-2, 9-3, Table

9-1)Discrete particles (Text: 9-3)Continuous solid objects (Text: 9-3)

Parallel axis theorem (Text: 9-3)

Look at textbook problems Chapter 9: # 7, 11, 27, 31, 33, 37

physics 111: lecture 17, pg 1 physics 111: lecture 17 today’s agenda l rotational kinematics...

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