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Lecture 16

- Chapter 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

Goals:

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 aboutan axis through its center:
- How do we describe the motion:

(Analogous to the linear case )

x

R

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

Comparison to 1-D kinematics

Angular Linear

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

x = R v = R aT = R

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.

w

1

2

w

2 K= ½ m (2v)2 = ½ m (w 2r)2

System of Particles (Distributed Mass):- Twice the radius, four times the kinetic energy
- The rotation axis matters too!

+

m2

m1

m1

m2

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.

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:

RCM

m2

m1

r2

r1

y

x

(In this case, N = 2)

(12,12)

2m

m

m

(0,0)

(24,0)

Sample calculation:- Consider the following mass distribution:

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.

dm

r

y

where dmis an infinitesimal

mass element.

x

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:

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 = ½m1v12 + ½m2v22 + ½m3v32 + ½m4v42

m4

m1

r1

r4

r2

m3

r3

m2

m1

r1

r4

m3

r2

r3

m2

Rotation & Kinetic Energy- Notice that v1 = w r1 , v2 = w r2 , v3 = w r3 , v4 = w r4
- So we can rewrite the summation:
- We recognize the quantity, moment of inertia orI, and write:

Calculating Moment of Inertia

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:

m

m

L

m

m

Calculating Moment of Inertia...

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

I1= 2mL2

I2= mL2

I= 2mL2

m

m

L

m

m

r

R

L

dm

r

Moments of Inertia- For a continuous solid object we have to add up the mr2contribution for every infinitesimal mass element dm.
- An integral is required to find I:

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:

Use the table…

Ball 2

Exercise Rotational Kinetic Energy- ¼
- ½
- 1
- 2
- 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 ?

Ball 2

Exercise Rotational Kinetic Energy- K2/K1 = ½ m wr22 / ½ m wr12 = 0.22 / 0.12 = 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

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 r = 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 w2

(A)Disk 1

(B)Disk 2

(C)Same

w2

w1

F

F

start

d

finish

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 r = 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 I1w12 = ½ I2w22

w1 = (I2 / I1)½w2 and I2 < I1

(A)Disk 1

(B) Disk 2

(C)Same

w2

w1

F

F

start

d

finish

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