Physics 151: Lecture 22 Today’s Agenda

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Physics 151: Lecture 22 Today’s Agenda. Topics Energy and Rotations Ch. 10.8 Intro to Rolling Motion Ch. 11. Lecture 22, ACT 1.

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Physics 151: Lecture 22Today’s Agenda
• Topics
• Energy and Rotations Ch. 10.8
• Intro to Rolling Motion Ch. 11
Lecture 22, ACT 1
• A campus bird spots a member of an opposing football team (maybe qb) in an amusement park. The football player is on a ride where he goes around at angular velocity w at distance R from the center. The bird flies in a horizontal circle above him. Will a dropping the bird releases while flying directly above the person’s head hit him?

a. Yes, because it falls straight down.

b. Yes, because it maintains the acceleration of the bird as it falls.

c. No, because it falls straight down and will land behind the person.

d. Yes, because it mainatins the angular velocity of the bird as it falls.

e. No, because it maintains the tangential velocity the bird had at the instant it started falling.

See text: 10.1

Example:
• You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exertedby gravity is :

a. 0.

b. mgr.

c. 2mgr.

d. a function of the angular velocity.

e. small at first, then increasing as the Frisbee loses the torque given it by your hand.

a)

b)

c)

a = 15 m/s2

See text: 10.8

• A uniform rod of length L=0.5m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is

a) angular speed when it reaches the lowest point ?

b) initial angular acceleration ?

c) initial linear acceleration of its free end ?

L

m

See example 10.14

1.7 s

11m/s

32 N

Example 2
• A rope is wrapped around the circumference of a solid disk (R=0.2m) of mass M=10kg and an object of mass m=10 kg is attached to the end of the rope 10m above the ground, as shown in the figure.
• How long will it take for the object to hit the ground ?
• What will be the velocity of the object when it hits the ground ?
• What is the tension on the cord ?

w

M

T

m

h =10 m

If an object of moment of inertia ICM is rotating in place about its center of mass at angular velocity w then its kinetic energy is

Connection with CM motion
• If an object of mass M is moving at velocity VCMwithout rotating then its kinetic energy is
• But what if the object is both moving and rotating?

animation

Recall text 9.6, systems of particles, CM

Connection with CM motion...
• So for a solid object which rotates about its center of mass and whose CM is moving:

VCM

See text: 11.1

Example Problem
• Last winter break, one of my students was bored and decided to make up a new game to play in January on the frozen lakes. The idea was to throw a yoyo on to the ice while holding one end of the string. At this point the yoyo is lying flat a distance D away from the bank and still has a length of string L wrapped around the circular part. The idea is to pull the string hard enough so that it gets back on the bank before the string completely unwinds and let loose the yoyo. You want to know the relationship between the amount of string still unwound and the distance to the bank. You note that the yoyo has a mass M, forms a disk of radius R, and you can pull a string with a maximum force F.

D

M

A

R

F

See text: 11.1

Example Problem

1. We need to solve for the minimum length of string wrapped around the yoyo when it lies on the ice a distance D from the shore.

• For the center of mass, we will use Newton’s Second Law and the constant acceleration kinematics equations.
• For the rotational motion, we will use the rotational version of Newton’s Second Law and the constant angular acceleration kinematics equations.

Top view

Rolling Motion
• Now consider a cylinder rolling at a constant speed.

VCM

CM

The cylinder is rotating about CM and its CM is moving at constant

speed (vCM). Thus its total kinetic energy is given by :

animation

Rolling Motion
• Consider again a cylinder rolling at a constant speed.

Q

VCM

CM

P

At any instant the cylinder is rotating about point P. Its kinetic

energy is given by its rotational energy about that point.

KTOT = 1/2 IPw2

Rolling Motion
• We can find IP using the parallel axis theorem

Q

VCM

CM

IP = ICM + MR2

P

KTOT = 1/2 (ICM + MR2 ) w2

KTOT = 1/2 ICM w2 + 1/2 M (R2w2 ) = 1/2 ICM w2 + 1/2 M vCM2!

v ?

M

Example :Rolling Motion
• A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ?

M

M

M

M

M

h

M

q

Lecture 22, ACT 4aRolling Motion
• A race !!

Two cylinders are rolled down a ramp. They have the same radius but different masses, M1 > M2. Which wins the race to the bottom ?

A) Cylinder 1

B) Cylinder 2

C) It will be a tie

M1

M2

h

M?

q

Lecture 22, ACT 4bRolling Motion
• A race !!

Two cylinders are rolled down a ramp. They have the same moment of inertia but different radius, R1 > R2. Which wins the race to the bottom ?

A) Cylinder 1

B) Cylinder 2

C) It will be a tie

R1

R2

animation

h

M?

q

Lecture 22, ACT 4cRolling Motion
• A race !!

A cylinder and a hoop are rolled down a ramp. They have the same mass and the same radius. Which wins the race to the bottom ?

A) Cylinder

B) Hoop

C) It will be a tie

M1

M2

animation

h

M?

q

Remember our roller coaster.

Perhaps now we can get the ball to go around the circle without anyone dying.

Note:

1

2

How high do we have to start the ball ?

h

h = 2.7 R = (2R + 1/2R) + 2/10 R

-> The rolling motion added an extra 2/10 R to the height)

See text: 10.1

Example:
• A mass m = 4.0 kg is connected, as shown, by a light cord to a mass M = 6.0 kg, which slides on a smooth horizontal surface. The pulley rotates about a frictionless axle and has a radius R = 0.12 m and a moment of inertia I = 0.090 kg m2. The cord does not slip on the pulley. What is the magnitude of the acceleration of m?

a. 2.4 m/s2

b. 2.8 m/s2

c. 3.2 m/s2

d. 4.2 m/s2

e. 1.7 m/s2

Recap of today’s lecture
• Chapter 10,
• Work/Energy of Rotational Motion
• For next time: Read Chapter 11