Uniform Circular Motion AP - Linville UCM AP.pdf · • Uniform circular motion is motion in a circle at the same speed • Speed is constant, velocity direction changes • the speed

Uniform Circular Motion AP

• Uniform circular

motion is motion

in a circle at the

same speed

• Speed is

constant, velocity

direction

changes

• the speed of an object moving in a circle is given

by

2 r

Tv

circumference

time

r= radius of

circle

T = period, time for 1

complete circle

• Is called the tangential

velocity

2 r

Tv

Moon setting

over Rio de

Janerio

Pictures every

6.5 minutes

showing the

moon is moving

at a uniform

speed

Centripetal Acceleration

• The direction of motion

is changing so the

object is accelerating

t

vv

t

va

if

• If you draw the

vectors tail-tail, v is

from vi to vf you get

f iv v v

• the centripetal acceleration of the object is

r

2vc

a ac = centripetal acceleration

(always to centre of circle)

2

2 2

2

2vac r

2

4

r

T

r

r

Tr

2

2

4 r

T

On data sheet too!

2 ac Equations

•the velocity vector is

always directed along

the tangent of the circle

(at right angles to the

acceleration vector)

•centripetal means

centre-seeking, the

acceleration vector

always points to the

centre of the circle

•Acceleration is radially

inward

Centripetal Force

• inertia tends to keep objects moving in a

straight line, a force is needed to cause a

circular motion (2nd Law of Motion)

• the force causing the motion is called the

centripetal force and it always acts

towards the centre of the circle (parallel to

the acceleration vector)

Centrifuge Training

2

F m a & net acv

r

NOT on data sheet

On data

sheet

2mv F c

r

• an object will move in a circle whenever a

constant magnitude force acts on the

object at right angles to its direction of

motion

• The force and

acceleration are

radially inward

• The velocity is

tangential

• The string pulls ball

towards the centre

of the circle

• 3rd Law of Motion

states that a reaction

force will act

outward (ball pulls

string)

• this is sometimes

called centrifugal

force

Important!!!

• centripetal force is not a fundamental force

like gravity, it is the net force acting on an

object moving in a curved path

• The force will not change the speed of the

object because the force has NO

component parallel to the velocity

vector

Example

• A centrifuge for pilot training is 11 m in

radius. At what speed must it rotate in

order to inflict 7.0 g on a pilot?

Solution

2

27.0 9.81 / 11

c

c

va

r

v a r

v m s m

Frequency

• sometimes speed is given in revolutions per minute (rpm) or revolutions per second (rps)

• this is how many times the object goes around the circle in 1 minute or 1 second

• frequency is the number of cycles per second measured in hertz (Hz)

11 sondsec

Hz

fT

1

2 r

T2v rf

Example

• Find the period of an object that

rotates at 35.0 rps.

sT 0286.0

fT

1

Free Body Diagrams

• A tetherball is attached to a

swivel in the ceiling by a light

cord. When the ball is hit by a

paddle, it swings in a horizontal

circle with constant speed, and

the cord makes a constant

angle with the vertical

direction. Write the expression

for the centripetal force in

terms of the other forces.

Free Body Diagrams

• An object is on a horizontal disc that is

rotating at constant speed. Friction

prevents the rock from sliding. Write the

expression for the centripetal force in

terms of the other forces.

•

Example

• A 3.50 kg object is swung in a 1.50 m radius

horizontal circle at 40.0 rps (40.0 Hz). What

magnitude force acts on the object?

2

2

4 mF m a F net c

T

r

Solution

2

2

4 mF m a F net c

T

r

224F so 1

rmfT

f

or

Effect of radius on speed and forces

• all points on a rotating solid

have the same period, but

different speeds

• because the inner points

have smaller distances to

travel, their speeds are less

• the speed and force depends

on the radius

Rotating

disc

• NASA and other space agencies use this to help launch satellites

• at the equator, the speed is about 1667 km/h, at Edmonton, it is about 900 km/h

Angular displacement & velocity

• The rotational

displacement of a

point, , in radians

• 2 radians = 360o

r

s

Radius

length Arcradians)(in

• Angular velocity,

(lower case omega)

radians/s

• Every point on the

disc has the same

angular velocity

avet

• By convention, the angular displacement

is positive if it is counterclockwise and

negative if it is clockwise.

Example: Adjacent Satellites

Two satellites are put into

an orbit whose radius is

4.23×107m.

If the angular separation of the two

satellites is 2.00 degrees, find the

arc length that separates them.

rad 0349.0deg360

rad 2deg00.2

r

s

Radius

length Arcradians)(in

s = r = 4.23x107 m x 0.0349 rad

s = 1.48 x 106 m

Angular & linear speed

• Linear speed, v = r

• Linear speed is

sometimes called

tangential speed

Period & angular velocity

2 1T

f

Distance (2) radians

in a circle

Angular velocity

Direction of Angular velocity

Rolling Motion

• A rolling wheel has

rotational and

translational velocities

• (a) axle is at rest and

the wheel is rotating

at v = r

Rolling Motion

• A rolling wheel has

rotational and

translational velocities

• (b) the wheel is

moving at v = r,

every point on the

wheel is moving

horizontally (wheel is

skidding)

Rolling Motion

When the rotation and

translational motions

are combined…

Equations of Rotational Motion

2

0 0

1

2t t

0 t

2

0 0

1

2x xx x v t a t

0x x xv v a t

Example

• A diver completes 1.5

rotations in 2.3 s.

Determine the

average angular

speed of the diver

Solution

• =1.5 rotations x 2 rad

= 9.425 radians

4.1 /ave rad st

Example

• A metal cylinder of radius 0.45 m is

spinning at 2000 rpm and a brake is

applied slowing it to 1000 rpm in 10

seconds. What is the angular

acceleration?

Solution

• In one revolution there are 2 radians

• (2000 rev/min) x (2 rad/sec) x (1 min/60

sec) = 209.4 rad/sec

• (1000 rev/min) x (2 rad/sec) x (1 min/60

sec) = 104.7 rad/sec

Solution

• f = i + at

• (104.7 rad/s) = (209.4 rad/s) + a(10 s)

• a = -10.47 rad/s2

Example

• A bullet is fired

through 2 discs

rotating at 99.0 rad/s.

The discs are 0.955

m apart. The angular

displacement

between the holes is

0.260 rad. Calculate

the speed of the

bullet.

Solution

• = 99.0 rad/s.

• d = 0.955 m

• = 0.260 rad.

avet

t = 2.626 x 10-3 s v = 364 m/s

Small Angle Approximation

• For small angles, < 0.5 rad, sin

tan

O

Practice

• P 251: #1, 2