Physics 207: Lecture 16, Pg 1 Lecture 16Goals: Chapter 12 Chapter 12 Extend the particle model to...

Physics 207: Lecture 16, Pg 1

Lecture 16

Goals:Goals:

• Chapter 12Chapter 12 Extend the particle model to rigid-bodies Understand the equilibrium of an extended object. Analyze rolling motion Understand rotation about a fixed axis. Employ “conservation of angular momentum” concept

Assignment: HW7 due March 25th After Spring Break Tuesday:

Catch up


Rotational Dynamics: A child’s toy, a physics playground or a student’s nightmare

A merry-go-round is spinning and we run and jump on it. What does it do?

What principles would apply?

We are standing on the rim and our “friends” spin it faster. What happens to us?

We are standing on the rim a walk towards the center. Does anything change?


Rotational Variables

Rotation about a fixed axis: Consider a disk rotating about

an axis through its center:

How do we describe the motion:

(Analogous to the linear case )

/R (rad/s) 2

TangentialvTdt

d

dtdxv


Rotational Variables...

Recall: At a point a distance R away from the axis of rotation, the tangential motion: x = R v = R a = R

R

v = R

x

rad)in position (angular 2

1

rad/s)in elocity (angular v

)rad/sin accelation(angular constant

200

0

2

tt

t


Comparison to 1-D kinematics

Angular Linear

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

x = R v = RaT = R

constant

t 0

221

00 tt

constanta

ta 0vv2

21

00 v tatxx

Here aT refers to tangential acceleration


System of Particles (Distributed Mass):

Until now, we have considered the behavior of very simple systems (one or two masses).

But real objects have distributed mass ! For example, consider a simple rotating disk and 2 equal

mass m plugs at distances r and 2r.

Compare the velocities and kinetic energies at these two points.

1 2


System of Particles (Distributed Mass):

Twice the radius, four times the kinetic energy

The rotation axis matters too!

1 K= ½ m v2 = ½ m (r)2

2 K= ½ m (2v)2 = ½ m (2r)2

2212

21 )(K rmmv


A special point for rotationSystem of Particles: Center of Mass (CM)

If an object is not held then it will rotate about the center of mass.

Center of mass: Where the system is balanced ! Building a mobile is an exercise in finding

centers of mass.

m1m2

+m1 m2

+

mobile


System of Particles: Center of Mass

How do we describe the “position” of a system made up of many parts ?

Define the Center of Mass (average position): For a collection of N individual point like particles whose

masses and positions we know:

M

mN

iii

1CM

rR

(In this case, N = 2)

y

x

r2r1

m1m2

RCM


Sample calculation:

Consider the following mass distribution:

(24,0)(0,0)

(12,12)

m

2m

m

RCM = (12,6)

kji CM CM CM1

CM ZYXM

mN

iii

r

R

XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters

YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters

XCM = 12 meters

YCM = 6 meters


System of Particles: Center of Mass

For a continuous solid, convert sums to an integral.

y

x

dm

rr

where dm is an infinitesimal

mass element.

M

dmr

dm

dmr

CM

R


Connection with motion...

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

For a point p rotating:

VCM

nTranslatioRotationTOTALK KK M

mN

iii

1CM

rR

2212

21 )(K ppppR rmvm

p

2CM2

1RotationTOTAL VK MK


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:

K = ½m1v1+ ½m2v2

+ ½m3v3+ ½m4v4

rr1

rr2rr3

rr4

m4

m1

m2

m3

4

1

221 v

iiimK


Rotation & Kinetic Energy

Notice that v1 = r1 , v2 = r2 , v3 = r3 , v4 = r4

So we can rewrite the summation:

We recognize the quantity, moment of inertia or I, and write:

rr1

rr2rr3

rr4

m4

m1

m2

m3

24

1

221

4

1

2221

4

1

221 ][ rrv

iii

iii

iii mmmK

221

Rotational IK

N

iiirm

1

2I


Calculating Moment of Inertia

where r is the distance from the mass to the axis of rotation.

N

iiirm

1

2I

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...

For a single object, I depends on the rotation axis! Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2

L

I = 2mL2I2 = mL2

mm

mm

I1 = 2mL2


Moments of Inertia

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

I = ½ M R2

Some examples of I for solid objects:

RL

rdr

dmr 2I r

dm

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

An integral is required to find I :

Use the table…


Exercise Rotational Kinetic Energy

A. ¼

B. ½

C. 1

D. 2

E. 4

We have two balls of the same mass. Ball 1 is attached to a 0.1 m long rope. It spins around at 2 revolutions per second. Ball 2 is on a 0.2 m long rope. It spins around at 2 revolutions per second.

What is the ratio of the kinetic energy

of Ball 2 to that of Ball 1 ?

221 IK

i

iirm 2I

Ball 1 Ball 2


Exercise Rotational Kinetic Energy

K2/K1 = ½ m r22 / ½ m r1

2 = 2 / 2 = 4

What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ?

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

Ball 1 Ball 2


Exercise Work & Energy

Strings are wrapped around the circumference of two solid disks and pulled with identical forces, F, for the same linear distance, d. Disk 1 has a bigger radius, but both are identical material (i.e. their density = M / V is the same). Both disks rotate freely around axes though their centers, and start at rest. Which disk has the biggest angular velocity after the drop?

W F d = ½ I 2

((A)A) Disk 1

(B)(B) Disk 2

(C)(C) Same

FF

1 2

start

finishd


Exercise Work & Energy

Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same linear distance. Disk 1 has a bigger radius, but both are identical material (i.e. their density = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest. Which disk has the biggest angular velocity after the drop?

W = F d = ½ I1 12 = ½ I2 2

2

1 = (I2 / I1)½ 2 and and I2 < I1

((A)A) Disk 1

(B) (B) Disk 2

(C)(C) Same

FF

1 2

start

finishd


Lecture 16

Assignment: HW7 due March 25th For the next Tuesday:

Catch up

nalTranslatioRotationalTOTALK KK

2CM2

1RotationalTOTAL VK MK

221

Rotational IK

i

iirm 2I


Lecture 16

Assignment: HW7 due March 25th After Spring Break Tuesday: Catch up

Physics 207: Lecture 16, Pg 1 Lecture 16Goals: Chapter 12 Chapter 12 Extend the particle model to...

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