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Honors Physics 1Class 15 Fall 2013

Rolling motion

Noninertial reference frames

Fictitious forces

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Rolling motion

We will treat the case where

1)the axis of rotation passes through the center of mass.

2)the axis of rotation does not change direction.

Example = Rolling wheel

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11.2 Rolling

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11.3 The Kinetic Energy of Rolling

If we view the rolling as pure rotation about an axis through P, then

( is the angular speed of the wheel and IP is the rotational inertia of the wheel about the axis through P).

Using the parallel-axis theorem (I P= Icom +Mh2),

(M is the mass of the wheel, Icom is its rotational inertia about an axis through its center of mass, and R is the wheel’s radius, at a perpendiculardistance h).

Using the relation vcom =R, we get:

A rolling object, therefore has two types of kinetic energy: 1.a rotational kinetic energy due to its rotation about its center of mass (=½ Icom2), 2.a translational kinetic energy due to translation of its center of mass (=½ Mv2

com)

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The Forces of Rolling: Friction and Rolling

A wheel rolls horizontally without sliding while accelerating with linear acceleration acom. A static frictional force fs acts on the wheel at P, opposing its tendency to slide.

The magnitudes of the linear acceleration acom, and the angular acceleration can be related by:

where R is the radius of the wheel.

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The Forces of Rolling: Rolling Down a Ramp

A round uniform body of radius R rolls down a ramp. The forces that act on it are the gravitational force Fg, a normal force FN, and a frictional force fs pointing up the ramp.

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Sample problem: Rolling Down a Ramp

A uniform ball, of mass M=6.00 kg and radius R, rolls smoothly from rest down a ramp at angle =30.0°.a) The ball descends a vertical height h=1.20 m to reach the bottom of the ramp. What is its speed at the bottom?Calculations: Where Icom is the ball’s rotational inertia about an axis through its center of mass, vcom is the requested speed at the bottom, and is the angular speed there.Substituting vcom/R for , and 2/5 MR2 for Icom,

b) What are the magnitude and direction of the frictional force on the ball as it rolls down the ramp?

Calculations: First we need to determine the ball’s acceleration acom,x :

We can now solve for the frictional force:

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A wheel and a cube, both of mass M are made to race down a slope.The wheel rolls without slipping. The cube slides without rolling.

Which one gets to the bottom first?

Which one has greater KE at the bottom?

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Energy analysis of rolling down a ramp

a) A cubical block with mass M and side 2R is allowed to slide without rolling down a ramp/slide from height h to height 0. How fast does the center of mass travel at the bottom of the slope?

b) A wheel with mass M and radius R is allowed to roll without slipping on a ramp of height h to the level ground at height 0. How fast does its center of mass travel at the bottom of the ramp?

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The Yo-Yo

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Galilean transformationsConsider measurements of the motion of a mass m made

from each of two coordinate systems.

For simplicity let’s assume that corresponding axes are parallel and the scale units (time and distance) are the same.

The origins of the two systems are displaced by S so

r r S

If the two systems move at a fixed velocity with respect to one another,

then

Observer sees the object moving with velocity .

Observer sees the object moving with velocity

V

dSV

dtdr

vdtd

v

.

r drV

dt dt

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Galilean transformations 2

2

2

2 2

2 2

Observer measures acceleration .

Observer measures acceleration .

If V is constant then and

and Newton's second law is the same.

d ra

dt

d r d r dVa

dtdt dt

a a

F ma ma F

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XKCD LINK

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Linearly Accelerating FrameFictitious forces

How is Newton’s second law affected by measurement in accelerating frames?

assuming constant

Assuming the frame is inertial, then the frame observer

measures an additional force.

fict

a a A A

F ma

F ma mA F mA

F F F

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Example: Hanging mass in a carA small weight of mass m hangs from a string in a car which accelerates

at rate A=1/4g.

a) What is the static angle of the string from the vertical?

b) What is the tension in the string?

T

Tfict

mg

2 2 1/2

cos 0

sin 0

tan

( )

T mg

T mA

mA A

mg g

T m g A

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Principle of Equivalence

In the example problem, we treated acceleration A in the same way as we treated gravitational acceleration.

The Principle of Equivalence states that there is no way to distinguish locally* between a gravitational acceleration and an acceleration of the coordinate system.

*Locally means that we don’t look outside the system for the cause of an acceleration or gravity.

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Rotating motion and fictitious forces

Consider two reference frames, one rotating about the z axis, the other inertial. Origins coincide. Z axes coincide.What fictitious forces are observed in the rotating frame?We already have some experience with this from analyzing forces in the spinning terror ride and centripetal acceleration.

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Rotating frames

Let's consider the time derivatives of a changing particle position

in inertial and rotating (primed parameters) frames.

Let the x,y,z and x',y',z' coordinates of the particle coincide at t=0.

At a later time ( ) ( ) so ( ) ( ).

Since observer' and observer(inertial) started in the same position,

' ( ) '( )

From the drawing we see that ' '( ) - ( )

r t t r t r r r t t r t

r r t t r t

r r r t r t

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Rotating frames

We can easily calculate the difference between ( ) and '( ) :

'( ) ( )

'so or

This approach was general and applies to the time derivative of

any vect

inertial rot

r t r t

r t r t r t

dr drr v v r

dt dt

or, so we can apply it to and .

a

and using we have

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From

in ininertial in

in rot

in rot

in rot rotrot

rot in rot

v a

dv dvv

dt dt

v v r

da v r v r

dt

a a v r

which, 2fict rotF m v m r

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Fictitious forces in rotating frames

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with

centrifugal force

m2 Coriolis force

fict rot

rot in

rot

F m v m r

v v r

m r

v