Uniform Circular Motion Physics 12. Uniform Circular Motion object is moving at a constant speed but changing directions acceleration occurs due to direction.

Uniform Circular Motion

Physics 12

Uniform Circular Motion

object is moving at a constant speed but changing directions

acceleration occurs due to direction changes

Circular Motion is Periodic

The motion occurs at a regular time interval called the period (T). The period is the time for one complete cycle.

(sinusoidal curve)

Calculating Speed

Constant speed: v = d/t In a circle, d = circumference and t is the

period

T

rv

2

Example

A mass on a string is swung in a circle of radius 1.0m at a speed of 5.0 m/s. What is the period of the revolution?

1.3 s

Centripetal Acceleration

In order for an object to follow a circular path, a force needs to be applied in order to accelerate the object

Although the magnitude of the velocity may remain constant, the direction of the velocity will be constantly changing

As a result, this force will provide a centripetal acceleration towards the centre of the circular path

Centripetal Acceleration

caused by a centre-seeking (centripetal) force that acts toward the centre of the circular path

Diagrams…

r1 and r2 represent POSITION at two different time v1 and v2 represent VELOCITY at two difference

positions, each tangent to the path.

Now use vector addition** Note that the resultant vector is direction TOWARDS the

CENTRE of the circle.

)( 12

12

vvv

vvv

How can we calculate centripetal acceleration (Derivation)?

r

va

t

v

r

v

v

v

r

tv

tvr

tvdv

v

r

r

c

2

2

Acceleration on a v-t graph = slope

Centripetal Acceleration (derivation on page 552)

Where v is the magnitude of the velocity (or speed) in m/s

r is the radius of the circle (in m) and ac is the centripetal acceleration (m/s2)

r

vac

2

Example

A mass on a string is swung in a circle of radius 0.75m at 7.0 m/s. What is its rate of acceleration?

65 m/s2 (towards centre of circle)

Centripetal Force

Circular motion occurs due to the presence of a centripetal (centre-seeking) force that pulls toward the centre of the path.

This force is UNBALANCED, therefore causing acceleration (Newton’s 2nd Law) directed toward the centre of the circle.

Centripetal Force

The centripetal force works to oppose the object’s inertia.

NOTE: At one time inertia was mistakenly referred to as centrifugal force. There is no such thing as centrifugal force – it is an illusion caused by inertia.

www.physicsclassroom.com/mmedia/circmot/cf.cfm

Notes: Centripetal Force is not really a specific force Any force that causes circular motion is a

centripetal force – it isn’t one type of force. Examples: Mass on a string is really

Tension. Planetary Orbit is really Fg. Car rounding a bend is really Ff.

Centripetal Force

Like the centripetal acceleration, the centripetal force is always directed towards the centre of the circle

The centripetal force can be calculated using Newton’s Second Law of Motion r

mvF

r

va

amF

c

c

2

2

2

2

2

2

2

22

2

2

2

4

4

4

2

T

rmF

T

rmF

r

Tr

m

F

rTr

mF

r

mvF

r

va

amF

c

c

c

c

c

c

NOTE: T is the inverse of f (remember from waves) so you can also manipulate the equation that way as well!

T = 1/f

Problem – horizontal circle

A student attempts to spin a rubber stopper (m=.050kg) in a horizontal circle with a radius of .75m. If the stopper completes 2.5 revolutions every second, determine the following:The centripetal accelerationThe centripetal force

Remember

You only need to look at the forces that are affecting the acceleration.

Gravity does not affect a horizontal circular motion as it acts 90’ to the plane of motion.

The stopper will cover a distance that is 2.5 times the circumference of the circle every second

Determine the circumference

Multiply by 2.5 Use the distance and

time (one second) to calculate the speed of the stopper

smvs

mv

t

dv

md

md

mC

rC

/120.1

12

12

)7.4(5.2

7.4

2

Use the speed and radius to determine the centripetal acceleration

Then use the centripetal acceleration and mass to determine the centripetal force

NF

smxkgF

maF

smxa

m

sma

r

va

smv

c

c

cc

c

c

c

3.9

)/109.1)(050.0(

/109.1

75.

)/12(

/12

22

22

2

2

Try This

A certain string can withstand a maximum tension of 55N before breaking. What is the maximum speed at which we can safely swing a 1.0kg mass on a piece of this string 0.50 m long in a horizontal circle?

5.2m/s

Road Design

You are responsible to determine the speed limit for a turn on the highway. The radius of the turn is 55m and the coefficient of static friction between the tires and the road is 0.90.Find the maximum speed at which a vehicle

can safely navigate the turn If the road is wet and the coefficient drops to

0.50, how does this change the maximum speed

Diagrams

The maximum speed at which a vehicle can safely navigate the turn

smv

grv

r

vg

r

mvmg

r

mvFc

/22

2

2

2

Coefficient drops to 0.50, how does this change the maximum speed

smv

grv

r

vg

r

mvmg

r

mvFc

/16

2

2

2

Try This

A car of mass 1250kg goes around a corner of radius 15.0m. Find the minimum coefficient of friction between the tires and the road required to safely navigate the turn at that speed.

u = 0.979

Page 559

15, 16 ,18

Problem – vertical circle

A student is on a carnival ride that spins in a vertical circle.Determine the minimum speed that the ride

must travel in order to keep the student safe if the radius of the ride is 3.5m.

Determine the maximum force the student experiences during the ride (in terms of number of times the gravitational force)

Vertical Circle

While travelling in a vertical circle, gravity must be considered in the solution

While at the top of the circle, gravity acts towards the centre of the circle and provides some of the centripetal force

While at the bottom of the circle, gravity acts away from the centre of the circle and the force applied to the object must overcome both gravity and provide the centripetal force

Vertical Circle

To determine the minimum velocity required, use the centripetal force equal to the gravitational force (as any slower than this and the student would fall to the ground). This means consider the centripetal force to ONLY be Fg (no Ftens)

To determine the maximum force the student experiences, consider the bottom of the ride when gravity must be overcome

At the top of the circle, set the gravitational force (weight) equal to the centripetal force

Solve for velocity

smv

smv

m

vsm

r

vg

r

mvmg

FF cg

/9.5

/34

5.3/81.9

222

22

2

2

At the bottom of the circle, the net force is equal to the sum of the gravitational force and the centripetal force

Solve for number of times the acceleration due to gravity

mgF

smsmmF

m

smsmmF

r

vgmF

r

mvmgF

FFF

p

p

p

p

p

gpc

2

/81.9/81.9

5.3

)/9.5(/81.9

22

22

2

2

Practice Problems

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