CIRCULAR MOTION

Linear Motion

d – distance (in meters)v – velocity (in meters/second)a – acceleration (in meters/second2)

Distance = 2r

Linear/Tangential Velocity

Objects moving in a circle still have a linear velocity = distance/time.This is often called tangential velocity, since the direction of the linear velocity is tangent to the circle.

v

Angular Motion

– angular displacement (in radians) – angular velocity (in radians/second) – angular acceleration (in radians/second2)r – radius of circle (in meters)

Circular Motion Terms

The point or line that is the center of the circle is the axis of rotation.If the axis of rotation is inside the object, the object is rotating (spinning).If the axis of rotation is outside the object, the object is revolving.

Angular Velocity

Objects moving in a circle also have a rotational or angular velocity, which is the rate angular position changes.Rotational velocity is measured in degrees/second, rotations/minute (rpm), etc.Common symbol, (Greek letter omega)

D𝜔=

∆𝜃∆ 𝑡

Rotational & Linear VelocityIf an object is rotating:

All points on the object have the same rotational (angular) velocity.All points on the object do not have the same linear (tangential) velocity.

Linear velocity of a point depends on:The rotational velocity of the point.

More rotational velocity means more linear velocity.The distance from the point to the axis of rotation.

More distance from the axis means more linear velocityIn symbols:

v = r

vr

Linear to Angular

d = rvT = raT = r

r

Angular & Linear Velocity

If an object is rotating: All points on the object have the same Angular velocity.All points on the object do not have the same linear (tangential) velocity.

Angular & Linear Velocity

Linear velocity of a point depends on:The Angular velocity of the point.

More rotational velocity means more linear velocity.

The distance from the point to the axis of rotation.

More distance from the axis means more linear velocity.

Acceleration

As an object moves around a circle, its direction of motion is constantly changing.Therefore its velocity is changing.Therefore an object moving in a circle is constantly accelerating.

Centripetal Acceleration

The acceleration of an object moving in a circle points toward the center of the circle.This is called a centripetal (center pointing) acceleration. ac

Centripetal Acceleration

2

2

rarva

c

Tc

locityangular vepathcircular of radiusr

onaccelerati lcentripetaaSpeed Tangential

c

Tv

The centripetal acceleration depends on:The velocity of the object.The radius of the circle.

A spinning ride at a carnival has an angular acceleration of 0.50 rad/s2. How far from the center is a rider who has a tangential acceleration of 3.3 m/s2?

mrrrαaT

6.6)50(.3.3

Circular Motion #1

What is the tire’s angular acceleration if the tangential acceleration at a radius of 0.15m is 9.4 x 10-2 m/s2?

2/63.

15.094.

srad

raT

Circular Motion #2

A test car moves at a constant speed of 10 m/s around a circular road of radius 50 m. Find the car’s A) centripetal acceleration and B) angular speed.

sradrvB /2.0

5010)

222

/250

)10() smrvaA c

Circular Motion #3

A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s2, what is its tangential speed?

?/05.8

2.48:

2

T

c

vsma

mrGivens

smv

rav

T

cT

/7.192.4805.8

2.4805.82

Circular Motion #4

A race car moves along a circular track at an angular speed of 0.512 rad/s. If the car’s centripetal acceleration is 15.4 m/s2, what is the distance between the car and the center of the track?

?/4.15

/512.:

2

rsma

sradGivens

c

mar

rrr

rva

c

c

7.58)512(.

4.15

)(

22

222

Circular Motion #5

A piece of clay sits 0.20 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 20.5 rad/s, what is the magnitude of the centripetal acceleration of the piece of clay on the wheel?

?20.

/5.20:

camr

sradGivens

2

2

2

/05.84

)5.20)(20(.

sm

rac

Circular Motion #6

22

2

221

if

i

if

tt

t

time(s)t)on(rad/sacceleratiangular

nt(rad)displacemeangular

/s)locity(radangular ve final /s)locity(radangular ve initial

2

f

i

Equations for Circular Motion

Angular Acceleration

– angular velocity (radians/second) – angular acceleration (radians/second2)t – time (seconds)

=

A ventilator fan is turning at 600 rev/min when the power is cut off, and it turns 1000 rev while coasting to a stop. Calculate the angular acceleration and the time required to stop.

??1000

min/0min/600

trev

revrev

f

i

Circular Motion #7

2

2

22

min/180

3600002000)1000(2)600(0

2

rev

if

min33.3)180(6000

tt

tif

A bicycle wheel rotates with a constant angular acceleration of 3.5 rad/s2. If the initial speed of the wheel is 2 rad/s at t = 0 s. a) Through what angle does the wheel rotate in 2 s? b) what is the angular speed at t = 2 s?

Circular Motion #8

rad

tti

1174

)2)(5.3(21)2(2

21

2

2

srad

tif

/9722)5.3(2

A potter’s wheel moves from rest to an angular speed of 0.20 rev/s in 30s. Find the angular acceleration in rad/s2.

2/?

30/0

/256.11

220.0

srad

stsrad

sradrevrad

srev

i

f

Circular Motion #9

2/041.

)30(0256.1

srad

tif

A dentist’s drill starts from rest. After 3.20 seconds of constant angular acceleration it turns at a rate of 2.51 x 104 rev/min. a) find the drill’s angular acceleration. b) Determine the angle (radians) through which the drill rotates during this period.

??2.3

/0

/13.262760min1

12

min25100

stsrad

sradsrev

radrev

i

f

Circular Motion #10

2/821

)2.3(013.2627

srad

tif

rad

tti

52.4203

)2.3)(821(210

21

2

2

A floppy disk in a computer rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s.A) What is the angular acceleration of the disk, assuming angular acceleration is uniform? B) How many revolutions does the disk make while coming up to speed? C) Find v if r = 4.45cm D) Find at if r = 4.45cm.

??892.

/4.31/0

st

sradsrad

f

i

Circular Motion #11

2/2.35

)892(.04.31

)

srad

tA if

revradrevrad

ttB i

23.22114

)892)(.2.35(210

21) 22

smsradmrvC /4.1)/4.31)(0445(.)

22 /57.1)/2.35)(0445(.) smsradraD t

Centripetal Force

Newton’s Second Law says that if an object is accelerating, there must be a net force on it.For an object moving in a circle, this is called the centripetal force.The centripetal force points toward the center of the circle.

Centripetal Force

In order to make an object revolve about an axis, the net force on the object must pull it toward the center of the circle.This force is called a centripetal (center seeking) force. Fc

Centripetal Force

Centripetal force on an object depends on:The object’s mass - more mass means more force.The object’s velocity - more speed means more force.And…

Centripetal Force

The centripetal force on an object also depends on:

The object’s distance from the axis (radius).

If linear velocity is held constant, more distance requires less force.If rotational velocity is held constant, more distance requires more force.

22

ωmrrvmmaF t

cc

tvcF

s)speed(rad/ angularpath(m) circular of radiusr

speed(m/s) tangentialv(kg) massm

(N) force lcentripeta

t

ω

Fc

Centripetal Force

TORQUE

Torque

Torque

Is the measure of how effectively a force causes rotation.

angleθdistancerforceFtorqueτ

sinθFr τ

Torques in equilibrium

When the torques associated with two masses balance each other.

rFτ:where0τ-τ

g

21

Torques in equilibrium

Two people with the same mass. Which picture shows torques in equilibrium?

A B

Torques in equilibrium

A – the distances are not the same in B so the torques will not balance.

A B

PROBLEM 1

Alfred weighs 400 N. He sits on one end of a seesaw 1.5 m from the fulcrum. Ann weighs 200 N. How far from the fulcrum must she sit to balance the seesaw?

?

PROBLEM 1

F1 = 400 Nr1 = 1.5 mF2 = 200 Nr2 = ?

1 = 2 F1r1 = F2r2

400(1.5) = 200(r2)r2 = 3 m

1.5 m3 m

GRAVITATIONAL MOTION

Newton’s Law of Universal Gravitation

states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Gravity is a consequence of mass. The force of gravity between two objects is directly proportional to the product of the objects’ masses. The force of gravity between two objects is inversely proportional to the square of the distance between the two objects’ centers of mass.

Newton’s Law of Universal Gravitation

Universal Gravitational Constant:

Mass of Earth:

Radius of Earth:

Universal Constants

ACCELERATION DUE TO GRAVITY

g

2

2

2

RGM

RGMmmg

RGMmF

Planet of Radiusobject of mass

Planet the of MassConstant nalGravitatio Universal

nGravitatio of Force

RmMGF

So . . . g = ?Now work it out. Now work it out. Work that problem out. Get that problem right.

The mass of planet Jupiter is 1.9 x 1027 kg and that of the sun is 1.99 x 1030 kg. The mean distance of Jupiter from the sun is 7.8 x 1011m. Calculate the gravitational force which the sun exerts on Jupiter.

Universal Gravitation #1

mr

kgNmG

kgm

kgm

11

2211

302

271

108.7

/1067.6

1099.1

109.1

NN

mkgkgkgNm

rmmGF

23

24

22302711

211

30272211

221

101454104145104145

108710991109110676

...

).(

).)(.() /.(

Where will the ball go?

What will happen to the Moon?

http://www.mrwaynesclass.com/teacher/circular/CarAndCurve.htm