Circular MotionCircular Motion

For a car going around a curve at constantspeed, the free-body diagram is:

where Fw is the weight of the car, FN is thenormal (perpendicular) force, and Fcp is thecentripetal force.

Fw

FN

Fcp

ΣFy = 0 ΣFX = Fcp FN = Fw

The vector sum of the forces in a vertical direction equals zero which makes sense because there is no vertical acceleration.

The vector sum of the forces in a horizontal direction is greater than zero which makes sense because the car is accelerating in a horizontal plane.

Fcp is the net force which by Newton’s 2nd

law causes the car to accelerate toward the

center of the circle.

The centripetal force is given by:

Fcp = mv2/r

where m is the mass in kg, v is the velocity in

m/s, and r is the radius of the circle in m.

An object undergoing circular motion is constantly

accelerating because its instantaneous velocity is

always changing direction.

The instantaneous velocity is always tangent

to the circle.

The centripetal acceleration is always directed

toward the center of the circle is given by:

ac = v2/r

As is the case with projectile motion, Newton’s

three laws can be found in circular motion.

ac

Newton’s 1st law is found when the object wantsto continue its uniform motion in a straight line.

An object does want to accelerate but can be made to do so with a net force.

Newton’s 2nd law is found by the net force and theacceleration acting in the same direction, towardthe center of the circle.

Newton’s 3rd law is also found but this is where acolossal misinterpretation is made.

Remember that Newton’s 3rd law requires twodifferent objects.

When you go around a corner in a car, who or what is accelerating?

That’s right, you and the car are accelerating

by a net force which is the centripetal force.

Both you and the car have mass (hopefully not the same mass) and because of inertia both of you want a state of zero acceleration.

The centripetal force is the force that the road exerts on the car (and you) via friction.

The reaction to this force is the car exerting a force on the road which is called the centrifugal force.

When the driver takes an especially sharp turn while you are the passenger, many people say that centrifugal force pushes them into the door.

This is where your “physical” intuition can

fool you.

The centrifugal force acts on the road, not

you!

In actuality, it is nothing more than Newton’s

1st law kicking into action.

Acceleration and Velocity

Any net force produces an acceleration which is

in the same direction as the net force.

When the acceleration is parallel to the velocity, the speed will increase.

When the acceleration is in the opposite direction of the velocity, the speed will decrease.

If the acceleration is perpendicular to the

velocity, the speed remains the same but

the velocity changes in direction.

Circular Motion ProblemCircular Motion Problem

A 0.017 kg rubber stopper is attached to a0.89 m length of string. The stopper is

swung ina horizontal circle making one revolution

in1.2 s.

(a)Draw a free body diagram for the rubber

stopper and identify the origin of each force.

Fcp is the centripetal force and represents theforce the string exerts on the rubber stopper.

Fw is the weight of the stopper which represents

the earth pulling down on the stopper.

Remember that in a force diagram you onlyinclude the forces that act on the object!

Fw

Fcp

(b) What is the speed of the stopper?

vave = Δx/Δt = C/Δt = 2πr/Δt

vave = 2 × 3.14 × 0.89 m/1.2 s = 4.7 m/s

(c) What is the instantaneous velocity?

v = 4.7 m/s tangent to the circle.v

(d) What is the instantaneous acceleration?

a = v2/r = (4.7 m/s)2/0.89 m = 24 m/s2

a = 24 m/s2 toward the center of the circle.

(e) Determine both centripetal and centrifugal force.

Fcp = ma = 0.017 kg × 24 m/s2 = 0.41 N

Fcf = Fcp = 0.41 N

(f) What happens if the string breaks?

When the string breaks, Fcp = 0, the stopper

will be in free fall and will attempt to travel in a

straight line with a velocity of 4.7 m/s while

accelerating vertically.

Vertical Circular MotionVertical Circular Motion

When an object moves in a vertical circle, the

weight of an object can not be ignored.

Free-body diagram at the top of a circle:

Fw

T

Fnet = Fc

Fnet = Fw + TFc = Fw + T

Free-body diagram at the bottom of a circle:

Fw

T

Vertical Circle ProblemVertical Circle Problem

A 0.147 kg ball at the end of a 1.20 m longstring is swung in a vertical circle. If the

massof the string is neglected:

(a) What is the minimum speed the ball must

have at the top of the circle so that it continues to uniformly accelerate towards the center?

m = 0.147 kg r = 1.20 m

Fw

At the top, the minimum speed occurs when T = 0,that is when the weight of the object supplies the centripetal force.

Fnet = Fc = Fw

Fc = Fw

mv2

r= mg

v = (rg)1/2 =(1.20 m × 9.80 m/s2)1/2=3.43 m/s

(b) What is the tension in the cord at the bottom

of its path if the ball is moving at three times

the speed at the top?

Fw

T

At the bottom, the tension is a maximumbecause it must support the weight of the objectand also provide the centripetal force.

Fnet = Fc = T – Fw

Fc = T- Fw = mv2/r – mg

T = Fc + Fw

T = 0.147 kg × (10.3 m/s)2 + 0.147 kg × 9.80 m/s2

T = 17.0 N

Another Circular Motion Problem Another Circular Motion Problem

A boy weighing 294 N sits 5.0 m from the axis

of rotation on a merry-go-round. Themerry-go-round spins at 4.7 rev/min.

(a) What is the frictional force required to keep

the boy on the merry-go-round.

Fw

FNFc

Fw = 294 N r = 5.0 m v = 4.7 rev/min

Ff = Fc =mv2

r

Ff =294 N/9.80 m/s2×

v =rev

4.7 ×1 min2 × π × 5.0 m

1 rev ×1 min60 s

v = 2.5 m/s

(2.5 m/s)2

5.0 m= 38 N

(b) Determine the coefficient of friction.

µ=Ff

FN

=38 N

294 N= 0.13

More Circular Motion Thoughts More Circular Motion Thoughts

In the case of a merry-go-round, friction must

provide the centripetal force.

You must always pay attention to units!

Because 1 N = kg•m/s2, all the units must becompatible to cancel out.

One complete revolution is equal to thecircumference of the circle, 2πr.

An interesting tidbit is that the coefficient of

friction (µ) is independent of mass in thisproblem.

How do you know? I’m glad you asked!

Aren’t you glad you asked?

µ =Ff

FN

=Fc

Fw

mv2/rmg= =

v2

rg

Wrap Up QuestionsWrap Up Questions

If the average velocity of an object is zeroduring a time interval, what can be said

for thedisplacement of that object for the same

timeinterval?

The displacement equals zero becausedisplacement is used to define velocity.

If the displacements of a particle are known attwo distinct points and the time to travel thatdisplacement is known, can you determine theinstantaneous velocity?

The instantaneous velocity can not bedetermined unless you are told that the object istraveling at a constant velocity, in which case the two velocities would be the same.

You could determine the instantaneous velocityfrom a Displacement vs Time graph.

If the speed of an object is constant, can it beaccelerating?

It could accelerate by following a curved path,either going around in circular motion or turninga corner in a car.

If the velocity of an object is constant, can it beaccelerating?

No, because acceleration is defined as the rateat which the velocity changes.