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Physics IClass 12

Uniform Circular Motion

Rev. 19-Feb-06 GB

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Newton’s Second Law -Yet Another Review!

Newton’s Second Law:

amFF net

or m

Fa net

The net force and acceleration are always in the same direction becausem is a positive number.

Acceleration is the rate of any change in the velocity vector – eithermagnitude (speed) or direction, or both.

Today’s lecture and activity will stretch our understanding ofacceleration and Newton’s Second Law for a special type of motion.

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Uniform Circular Motion

“U n ifo rm ” c ircu la r m o tio n m ean s th a t th eo b jec t m o v es in a c irc le a t a co n s tan t sp ee d .S o m e d e fin itio n s an d e q u a tio n s :

T = p e rio d = tim e to g o a ro u n d e x ac tly o n ce

r = rad iu s o f c irc le

v = sp eed (sca la r , n o t v ec to r)

Tr2

periodncecircumfere

v

vr2

T

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Acceleration in Circular Motion

When an object travels in a circle, its velocity is constantly changing (in direction at least).

That means the object has a non-zero acceleration even if it moves at constant speed.

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Important Facts AboutVelocity and Acceleration Vectors

v a

v a

Same direction: speeding up.

Opposite directions: slowing down.

Right angles: changing direction, same speed.v

a

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What is the Direction of Acceleration?

Since the speed is not changing, only thedirection of velocity, acceleration must bealways at right angles to velocity. Theacceleration vector points inward, towardthe center of the circle. This is calledcentripetal acceleration from Latin for“to go to or seek the center.”

Like the direction of the velocity vector,the direction of centripetal acceleration isconstantly changing as the object movesaround the circle.

The magnitude of centripetal acceleration is given by rv

a2

.

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“Centripetal Force”“Centrifugal Force”

“Centripetal force” and “centrifugal force” are two of the mostunnecessary and confusing concepts in introductory physics.“Centripetal force” is the net force on any object in circular motion.“Centrifugal force” is a result of measuring the displacement of anobject relative to an accelerated (rotating) observer.

In this course, we will never refer to centripetal or centrifugal force.We will refer only to actual physical forces such as gravity.

It will never be correct to answer any question using centripetalforce or centrifugal force. Instead, use centripetal acceleration andNewton’s Second Law as measured by a non-accelerating observer.

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Example:A Rock on a String

Twirl a 1 kg rock attached to a string in a 1 m radius vertical circle. The speed is 4 m/sec. What forces act on the rock and what are the directions of those forces?

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Using Newton’s Second Law to Solve Problems - Review

1. Identify all forces acting on the object.Pushes or Pulls Friction (if specified)Gravity Normal (Surface) Forces

2. Choose a coordinate system.If you know the direction of acceleration, onecoordinate axis should be in that direction.

3. Draw a “Free-Body Diagram.”We know how to do this now.

4. Express the force vectors in components.This may require trigonometry.

5. Use Newton’s Second Law to write oneequation for each direction considered.We will only consider vertical forces today.

6. Solve the equation(s).

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Case A:Rock at the Top of the Circle

T W = mg

a X

T h e c e n t e r o f t h e c i r c l e i s b e l o w t h e r o c k , s o a c c e l e r a t i o n i s d o w n .

1 . F o r c e s : W e i g h t ( W ) d o w n a n d T e n s i o n ( T ) d o w n .2 . C o o r d i n a t e s : + X d o w n . ( W h y ? )3 . F r e e - b o d y d i a g r a m :4 . X C o m p o n e n t s : ( W ) a n d ( T ) .5 . S e c o n d L a w : ( W ) + ( T ) = m a .

6 . S o l v e :

g

rv

mgmamWamT2

T = 1 ( 1 6 / 1 – 9 . 8 ) = 6 . 2 N .W = 9 . 8 N .

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Case B:Rock at the Bottom of the Circle

T h e c e n t e r o f t h e c i r c l e i s a b o v e t h e r o c k , s o a c c e l e r a t i o n i s u p .

1 . F o r c e s : W e i g h t ( W ) d o w n a n d T e n s i o n ( T ) u p .2 . C o o r d i n a t e s : + X u p . ( W h y ? )3 . F r e e - b o d y d i a g r a m :4 . X C o m p o n e n t s : ( – W ) a n d ( T ) .5 . S e c o n d L a w : ( – W ) + ( T ) = m a .

6 . S o l v e :

g

rv

mgmamWamT2

T = 1 ( 1 6 / 1 + 9 . 8 ) = 2 5 . 8 N .W = 9 . 8 N .

T

W = mg

a X

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Normal Force - A ConceptWe Will Use in Today’s Activity

NW

NW

Elevator Cab

“Normal” force is the force generated by a solidobject to keep other objects from penetrating into it.As the name implies, the direction of this force is atright angles (“normal”) to the surface. The physicalcause of this force is the stretching of chemicalbonds, much like the stretching of a lattice of springs.

Normal force is often equal and opposite to weight,but not always. Consider an elevator cab. How doesthe normal force compare to weight if the cab ismoving at a constant velocity? Accelerating upward?Accelerating downward?

Floor

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Class #12Take-Away Concepts

1. Acceleration (or net force) at a right angle to velocity causes achange of direction but not a change of speed.

2. As an object moves around a circle at a constant speed, itaccelerates toward the center with magnitude given by

rv

a2

3. Do not try to use “centripetal” or “centrifugal” forces to solve aproblem. Use centripetal acceleration and Newton’s 2nd Law.

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Class #12Problems of the Day

___ 1. Imagine you are riding on a train going around a horizontalcurve of radius r at speed v. You are holding an object, ofmass m, stationary (relative to the train) in your hand.Let g be the acceleration constant of gravity and a = v2/r.The net force exerted on the object is:

A) mg upward + ma horizontal toward centerB) mg downward + ma horizontal toward centerC) ma horizontal toward centerD) mg upward + ma horizontal away from centerE) mg downward + ma horizontal away from centerF) ma horizontal away from centerG) zeroH) more information is needed to answer this question

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Class #12Problems of the Day

2. The space shuttle Atlantis is in a circular orbit 100. km (1.00 x 105 m)above the surface of the earth. (This is an example of uniform circularmotion.) At this height, the constant of gravity is g = 9.53 N/kg. The

radius of the earth is 6.37 x 106 m. How long (in minutes) does it take forthe shuttle Atlantis to complete one orbit?

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Activity #12Ferris Wheel Thrill Ride

Objectives of the Activity:

1. More experience with video analysis in LoggerPro.2. Investigate uniform circular motion –

Is the acceleration really directed toward the center?3. What is the subjective experience of traveling in a rapid

vertical circular motion?

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Class #12 Optional Material“Centrifugal” Force?

Turn this way. Feel a forcethis way.

Centrifugal is from Latin for “to flee from the center.”Is centrifugal force a “real” force?

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Accelerated Frames of Reference

Newton’s Second Law applies to an inertial reference frame, meaning areference system for measuring position and time that is not accelerating.

If we wish to use Newton’s Second Law in an accelerating reference frame,we need to add extra terms to the equation that can be considered as forcesoperating on every object that we track using the accelerating referenceframe. These are commonly called inertial forces.

Why would we do a crazy thing like using an accelerated reference frameinstead of an inertial reference frame? In a way, it is built into humannature to view ourselves as sitting still while the rest of the universe zips by.It is often a convenient way to calculate things as long as we are careful.

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The Inertial Forces

RmVm2RmamFAm 0

F

# 1 # 2 # 3 # 4

D o n ’ t w o r r y a b o u t u n d e r s t a n d i n g t h e d e t a i l s o f t h i s e q u a t i o n . T h e i m p o r t a n tt h i n g i s t h a t e a c h t e r m r e p r e s e n t s a d i f f e r e n t t y p e o f i n e r t i a l f o r c e .

1 . T h i s t e r m i s d u e t o l i n e a r ( i n a s t r a i g h t l i n e ) a c c e l e r a t i o n o f t h e r e f e r e n c ef r a m e . F o r e x a m p l e , i n a c a r i f y o u s l a m o n t h e b r a k e s ( a c c e l e r a t i o n t o t h er e a r ) , i t f e e l s l i k e e v e r y t h i n g i n t h e c a r i s t h r o w n f o r w a r d .

2 . C e n t r i f u g a l f o r c e – t h i s t e r m i s d u e t o r o t a t i o n o f t h e r e f e r e n c e f r a m e .3 . C o r i o l i s f o r c e – t h i s t e r m i s d u e t o m o v i n g i n a r o t a t i n g r e f e r e n c e f r a m e .

I f y o u e v e r t r i e d t o w a l k d o w n t h e a i s l e o f a t r a i n a s i t r o u n d e d a c u r v e ,y o u e x p e r i e n c e d t h i s f o r c e .

4 . T h i s t e r m i s d u e t o a c c e l e r a t i o n o f t h e r o t a t i o n o f a r e f e r e n c e f r a m e . I f y o ua r e r i d i n g o n a m e r r y - g o - r o u n d , y o u n e e d t o h o l d o n t i g h t e r a s i t s t a r t s u p .

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DiscussionSwinging a Water Bucket Overhead

Take a sturdy water bucket ½ full of water and swingit in a circle over your head. What will the water doif

1. you swing it quickly?2. you swing it slowly?3. you stop it when it is directly overhead?

Can we relate the concepts and math formulas wehave talked about today to the transition betweensituation #1 and situation #2?