Circular Motion. The Radian Objects moving in circular (or nearly circular) paths are often measured...

Circular Motion

The Radian

Objects moving in circular (or nearly circular) paths are often measured in radians rather than degrees.

In the diagram, the angle θ, in radians, is defined as follows

So, if s = r the angle is 1 rad and if s is equal to the full circumference of the circle, the angle is 2π rad. (In other words, 360° = 2π rad.)

Radians and Degrees

Angular Displacement (θ)

If we consider a small body to be moving round the circle from A to B we say that it has experienced an angular displacement of θ radians. The relation between the (linear) distance moved, d, of the body and the angular displacement θ, is given by

Also, if the angle is small, d is very nearly equal to the magnitude of the linear displacement of the body.

d = rθ

Angular Velocity (ω)

Suppose that the body moved from P1 to P2 in a time t. The linear speed, v, of the body is given by v = d/t.

If we divide d=rθ by t, we have The angular velocity ω is defined as Units of ω are rad/s Therefore,

v = rω  

Merry-Go Round

What happens to your speed as you go to the middle of a merry-go round?

a. The speed remains constant.

b. The speed increases

c. The speed decreases

http://animations.50webs.com/free_cartoons_mobile_animation.htm

Merry-Go Round

What happens to your angular speed as you go to the middle of a merry-go round?

a. The angular speed remains constant.

b. The angular speed increases

c. The angular speed decreases

http://animations.50webs.com/free_cartoons_mobile_animation.htm

http://mocoloco.com/art/upload/2009/12/biondo_merry_go_round.jpg

Angular Acceleration(α)

Previously we assumed that the body moved from P1 to P2 with constant speed. If the linear speed of the body changes then, obviously, the angular speed (velocity) also changes.

The angular acceleration, α, is the rate of change of angular velocity.

So, if the angular velocity changes uniformly from ω1 to ω2 in time t, then we can write:

Now, linear acceleration, a, is given by

Substituting v=rω We find,

t

vva of

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vva of

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Circular Motion Definitions

Time Period, T The time period of a circular motion is the

time taken for one revolution.

Rotational Frequency, f The rotational frequency of a circular motion

is the number of revolutions per unit time. Time period is the inverse of frequency Also, and

Rotational Motion

What is the relationship between Liner and Rotational motion quantities?

Acceleration Equations Revisited

What is the Acceleration?

a. No accelerationb. Acceleration

outwardc. Acceleration toward

the center of the circle

d. Acceleration points tangential to the circle

A plane attached to a string flies in a circle at constant speed

Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

What is the Acceleration?

Even though the speedspeed is constant, velocityvelocity is notnot constant since the direction is changing: acceleration!

If released the plane would travel to A at constant speed.

Therefore, the change in velocity would be a vector from A to P

If θ is very small, this acceleration (Δv/t) points to the center of the circle.

This is called centripetal acceleration

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Centripetal Acceleration

Even though the speedspeed is constant, velocityvelocity is notnot constant since the direction is changing: must be some acceleration!

Consider average acceleration in time Δt

As we shrink Δ t, Δv / Δt dv / dt = aat

vaavg

vv2

t

vv1

vv1vv2

vv

vv1vv2

vv

RRRR vvvv

seems like vv (hence vv/t )

points at the origin!

Centripetal Acceleration

Magnitude:

Direction:- r r (toward center of circle)Since and v = ωr

points toward the center of the circle a = ω2r Since and v = ωr

aa = dvv / dt

We see that a a points

in the - RR direction.

RR

vv2

vv1

vv2

vv

RRRR

vv

RR

Similar triangles: vv

RR

Similar triangles:

But R = vt for small t

So:vt

vR

2 v

vv tR

R

va

2

R

va

2

a = ω2r

Centripetal Acceleration

http://www.lyon.edu/webdata/users/shutton/phy240-fall2003/circle1.gif

Centripetal Acceleration (ac)

‘Centripetal’ means center seeking Vector direction always points toward the

center of the circle Magnitude:

v = speed

r = radius of the circle

r

vac

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http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/ucm.gif

Centripetal Acceleration (ac)

http://physics.csustan.edu/Astro/Help/NEWTON/cpetal2.gif

Example: Centripetal Acceleration

A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F-22 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it?

(a) 500 m

(b) 1000 m

(c) 2000 m

Example: Centripetal Acceleration

Using the definition of ac

Substituting in the values

Simplifying,

d = 2R ~ 2000m

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What is the Net Force?

a. No net forceb. Net Force points

outwardc. Net Force points

toward the center of the circle

d. Net Force points tangential to the circle

A plane attached to a string flies in a circle at constant speed

Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

What is the Net Force?

a. No net forceb. Net Force points

outwardc. Net Force points

toward the center of the circle

d. Net Force points tangential to the circle

A plane attached to a string flies in a circle at constant speed

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What is the Net Force?

Centripetal Force, Fc, is this net force.

Recall, Newton's 2nd Law Fnet=ma

Therefore, Fc = mac

A plane attached to a string flies in a circle at constant speed

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Centripetal Force (Fc)

acts toward the center of the circle depends on mass, speed, and size of

the circle

Net force, provided by another force or interactions of forces

r

vmFc

2

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Centripetal Acceleration Example

When you are driving a car, and you turn the steering wheel sharply to the right in order to turn the car to the right, you "feel" as if a force is pushing your body to the left against the door.

In order for your body to follow the car in the tight circular path, something has to push your body toward the center of the circle-- in this case it is the driver's-side door-- and your tendency otherwise is to travel in a straight line tangent to the circular path

Date Physics

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/rht.html

Centripetal Force Source

What provides the centripetal force (Fc) in the following scenarios?

A car going around a corner at a constant speed.

Friction force between the

tires and the pavement

Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Centripetal Force Source

What provides the centripetal force (Fc) in the following scenarios?

A ball on a string being

twirled in a circle.

Tension force of the string

on the ball

http://www.mansfieldct.org/schools/mms/staff/hand/lawsCentripetalForce_files/image004.jpg

http://www.vast.org/vip/book/LOOPS/HOME7.GIF

Centripetal Force Source

What provides the centripetal force (Fc) in the following scenarios?

The planets orbiting around the sun.

The sun’s force of gravity

on the planets.

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Centripetal Force Source

What provides the centripetal force (Fc) in the following scenarios?

A motorcycle stuntman going around a loop.

Both the normal force of the track and gravity.

Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Centripetal Force Example

A ball attached to a string is twirled in a circle of radius 0.5 m with a speed of 2 m/s. If the ball has a mass of 0.25 kg, what is the tension of the string.

Knowns:

m = 0.25 kg

v = 2 m/s

r = 0.5 m

r

vmFc

2

cnet FFTF 5.0

225.0 2

cFT T = 2 N

http://ww

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The normal reaction, FN, has no component acting towards the center of the circular path.Therefore the required centripetal acceleration is provided by the force of friction, Ff, between the wheel and the road.

Moving in a Straight line on a Horizontal Surface

Moving in a Straight line on a Horizontal Surface

If the force of friction is not strong enough, the vehicle will skid.

Turning on a Banked Surface

The normal reaction, FN, now has a component acting towards the center of the circular path.

If the angle, , is just right, the correct centripetal acceleration can be provided by the horizontal component of the normal reaction.This means that, even if there is very little force of friction the vehicle can still go round the curve with no tendency to skid.Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Angle of Banking

The magnitude of the horizontal component of the normal force is

This force causes the centripetal acceleration, so, the magnitude of NX is also given by

So,

The vertical forces acting on the

vehicle are in equilibrium. Therefore, summing the vertical forces

r

mvF xN

2

sinNxFNF

mgFN cos

r

mvFN

2

sin

Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Angle of Banking

Solving for FN and substituting

into gives:

Simplifying,

This equation allows us to calculate the angle θ needed for a vehicle to go round the curve at a given speed, v, without any tendency to skid.

rg

v 2

tan

mgFN cos

r

mvFN

2

sin r

mvmg 2

cos

sin

Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Indy Physics

Example: What is the minimum speed an Indy 500 car

must go not to slide down the banked curves of the Indianapolis Motor Speedway if the banked angle is 9.2° and the radius is 280.0 m?

NASCAR Physics

What is the minimum speed a NASCAR car must go not to slide down the banked curves of the Bristol Motor Speedway if the banked angle is 36° and the radius is 73.5m?

Loopy

The minimum speed to complete a loop requires: Speed large enough to reach the top of the

loop. At the top of the loop Fnet = Fg + FN

For the minimum speed FN = 0 Therefore, recall So,

at the top of the loop

r

mvFFF cnetg

2min mgFg

Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

grv min

Kepler’s Law of Periods Kepler’s Law of Periods ProofProof

Assumptions:• Must conform to equations for circular

motion• Newton’s Universal Law of Gravity

•Planet rotates in a circular (elliptical) path

Newton’s 3rd Law symmetry

Recall, so

Therefore, rearranging

r

mvmaFF ccnet

2

cnetgravity FFF r

mv

r

GMmFg

2

2

T

rrv

2

T

2

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T

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Law of Periods

Testing the Inverse Square Law of Gravitation

The acceleration due to gravity at the surface of the earth is 9.8m/s2.

If the inverse square relationship for gravity (Fg~1/r2) is correct , then, at a distance ~60 times further away from the center of the earth, the acceleration due to gravity should be

The centripetal acceleration of the moon is given by where the radius of the moon’s orbit is r = 3.84 × 108 m and the time period of the moon’s orbital motion is T = 27.3 days.

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