Rotational Dynamics General Physics I - PowerPoint Presentation

Slide1

Rotational Dynamics

General Physics I

MechanicsPhysics 140

Racquetball Striking a Wall

Mt. Etna

Stewart HallSlide2

Torque

The “rotating effect” of a force is greater, if the force is

greaterThe “rotating effect” of a force is greater, if the force is applied further from the axis of rotationThe “rotating effect” of the force also depends on the

direction of the force: it is maximized by a perpendicular force; no effect of a parallel forceThese three properties lead to the following definition of the torque:

Vector r is directed from the axis to the point where the force is applied:

F

r

axisSlide3

Magnitude of torque:

Magnitude of torqueSlide4

A mass is hanging from the end of a horizontal bar which pivots about an axis through it center, but is being held stationary. The bar is released and begins to rotate. As the bar rotates from horizontal to vertical, the magnitude of the torque on the bar…

Decreases

IncreasesRemains the same

A rotating barSlide5

A mass is hanging from the end of a horizontal bar which pivots about an axis through it center, but is being held stationary. The bar is released and begins to rotate. As the bar rotates from horizontal to vertical, the magnitude of the torque on the bar…

Decreases

IncreasesRemains the same

A rotating bar

and

φ

decreases

F

r

φSlide6

A mass is hanging from the end of a horizontal bar which pivots about an axis through it center, but is being held stationary. The bar is released and begins to rotate. As the bar rotates from horizontal to vertical, the magnitude of the torque on the bar…

Decreases

IncreasesRemains the same

A rotating bar

and

φ

decreases

F

r

φ

Lever arm decreases

FSlide7

Depends on the

axis

TorqueSlide8

Depends on the

axisDepends on where the force is applied

TorqueSlide9

Depends on the

axisDepends on where the force is applied

Specifically, gravitational force (weight) can be viewed as being applied at the center of mass

TorqueSlide10

Depends on the

axisDepends on where the force is applied

Specifically, gravitational force (weight) can be viewed as being applied at the center of massWhat about the direction?

TorqueSlide11

Depends on the

axisDepends on where the force is applied

Specifically, gravitational force (weight) can be viewed as being applied at the center of massWhat about the direction?

As usual for any vector product, direction of the torque is given by the right-hand rule:

TorqueSlide12

Depends on the

axisDepends on where the force is applied

Specifically, gravitational force (weight) can be viewed as being applied at the center of massWhat about the direction?

As usual for any vector product, direction of the torque is given by the right-hand rule:

In practice, we will be speaking about clockwise and counterclockwise torques

TorqueSlide13

The Right Hand Rule: Calculating a Torque

points

out of page

r

F

r

rotates

CCW

into

F

τ

points

out of page

points

into page

r

F

r

rotates

CW

into

F

τ

points

into page

r

r

Axis

AxisSlide14

Problem 10.2

16 N m

19 N m28 N m

52 N m64 N mSlide15

Problem 10.2

16 N m

19 N m28 N m

52 N m64 N m

The two torques are in the opposite directions!

|τ

1

|

= 40 N m |

τ2| = 12 N mSlide16

For rotations, both the moment of inertia and the torques depend on the

axis

(which must be

same

!)

Rotational Dynamics

No total torque = no angular

acceleration

If the total torque exerted by all forces on a rigid body around some axis is not zero, then the angular acceleration of the body is given by

where

I

is the moment of inertia around the

same axis

Analogous to the Newton’s 2nd law for translational motionSlide17

Accelerating two wheels

Two wheels with fixed axles, each have the

same mass M, but wheel 2 has twice the radius of wheel 1. Each is accelerated from rest with a force applied as shown. In order to impart identical angular accelerations to both wheels, how should F2 compare to

F1? Assume that all the mass of the wheels is concentrated in the rims. F2 = ½

F1 F2 = F

1 F2 = 2F1

F2

= 4F1

None of the aboveSlide18

Accelerating two wheels

Two wheels with fixed axles, each have the

same mass M, but wheel 2 has twice the radius of wheel 1. Each is accelerated from rest with a force applied as shown. In order to impart identical angular accelerations to both wheels, how should F2 compare to

F1? Assume that all the mass of the wheels is concentrated in the rims. F2 = ½

F1 F2 = F

1 F2 = 2F1

F2

= 4F1

None of the aboveSlide19

Example

A block of mass

m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia

I, which is free to rotate around an axle passing through its center of mass. The rope does not slip on the disk. Because of gravity the block m falls and the pulley rotates. The magnitude of the torque on the pulley is…

less than mgRequal to mgR

greater than mgRimpossible to predict

m

R

ISlide20

Example

A block of mass

m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia

I, which is free to rotate around an axle passing through its center of mass. The rope does not slip on the disk. Because of gravity the block m falls and the pulley rotates. The magnitude of the torque on the pulley is…

= TRless than mgR

equal to mgRgreater than mgRimpossible to predict

m

R

I

TSlide21

Example

A block of mass

m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia

I, which is free to rotate around an axle passing through its center of mass. The rope does not slip on the disk. Because of gravity the block m falls and the pulley rotates. The magnitude of the torque on the pulley is…

= TRless than mgR - because

T < mg!equal to

mgRgreater than mgR

impossible to predict

m

R

I

T

mg

aSlide22

Example

A block of mass

m is attached to a rope wound around the outer rim of a disk of radius R

and moment of inertia I, which is free to rotate around an axle passing through its center of mass. The rope does not slip. What are the acceleration of the block and the tension in the string?

m

R

I

Hanging block:

T

mg

a

T

α

Pulley:

Constraint:Slide23

m

R

I

T

mg

a

T

αSlide24

Another example: Yo-yo

Work done by gravity goes not only to translational motion, but also to rotation:

GPE = KE

trans + KErot and KErot >>

KEtrans

I

CMSlide25

Another example: Yo-yo

Work done by gravity goes not only to translational motion, but also to rotation:

GPE = KE

trans + KErot and KErot >>

KEtransAlternatively, the torque is proportional to r (small), while the moment of inertia is proportional to R

2 (large), resulting in a very small angular acceleration

I

CMSlide26

Another example: Yo-yo

Work done by gravity goes not only to translational motion, but also to rotation:

GPE = KE

trans + KErot and KErot >>

KEtransAlternatively, the torque is proportional to r (small), while the moment of inertia is proportional to R

2 (large), resulting in a very small angular accelerationAcceleration:

I

CMSlide27

Dynamics of a yo-yo

T

M, I

r

mg

Translational motion:

Rotational motion:

Solving for acceleration:

aSlide28

Dynamics of a yo-yo

T

M, I

r

mg

R

Approximate the yo-yo as a uniform disk of radius R:

Then: