# Centripetal Force Help

1. Nov 28, 2009

### PinkEraser

1. The problem statement, all variables and given/known data
A ball of mass M is attached to a string of length R and negligible mass. The ball moves clockwise in a vertical circle. When the ball is at point P, the string is horizontal. point Q is at the bottom of the circle and point Z is at the top of the circle. Air resistance is negligible. Express all algebraic answers in terms of the given quantities and fundamental constants.

a.What are the forces exerted on the ball at point P and Q,respectively

b. Derive an expression for Vmin, the minumum speed the ball can have at point Z without leaving the circular path.

c. The maximum tension the string can have without breaking is Tmax. Derive an expression for Vmax, the maximum speed the ball can have at point Q without breaking the string

d.Suppose the string breaks at the instant the ball is at point P. Describe the motion of the ball immediately after the string breaks

2. Relevant equations

Fc = mac
ac = V^2/r

3. The attempt at a solution

a. At point P, Tension is to the right, Gravity is pusing the mass down, and the centripetal force is going to the center of the circle so it is also going to the right

At point Q which is at the bottom of the circle, Tension is now going up as well as the centripetal force. Gravity is pushing down

b. At point Z, all forces are going down.
So Fc = mac
T + mg = m(Vmin)^2 / R
Vmin^2 = R(T+mg) / m
Vmin = radical R(T + mg) /m

c. Same thing, just with maximum tension at pt Q
Fc = mac
Tmax - mg = m(Vmax)^2 / R
Vmax = radical R(Tmax - mg) / m

d. It goes up with the same speed as it was going in a circle. Gravity is the only force pushing down on it

Is this correct? I'm not sure where the centripetal force is at in point P, Q, and Z

2. Nov 28, 2009

### PinkEraser

Can someone verify if I'm wrong or right?

3. Nov 28, 2009

### ideasrule

Centripetal force is not a separate force. It's just a name for anything that acts towards the center; in this case, it's tension and sometimes gravity.

4. Nov 28, 2009

### ideasrule

What's the minimum possible value for T?

5. Nov 28, 2009

### PinkEraser

Since there's no force acting on the Tension, does Tension = 0?

Last edited: Nov 28, 2009
6. Nov 28, 2009

### pgardn

Tension is one of the forces along with gravity that is supplying the Centripetal Force keeping the mass moving in a circle at Point Z. If you set T = 0 as you stated, this means that only the force of gravity is keeping the mass moving in its circular path at point Z. So use the equation you have and set T = O and find Vmin.

Then realize that at this speed it is actually in free fall but still moving in a circle at point Z.

Does this make any sense?

7. Nov 28, 2009

### PinkEraser

and Vmax is correct?

and then from what you said, the motion when it breaks at point P, it is in free-fall

8. Nov 28, 2009

### pgardn

9. Nov 28, 2009

### pgardn

Is P on the left or right hand side of the circular path? Since I read above it says the mass is moving clockwise, this means if P is on the right hand side of the circular path and the string breaks it will move direclty down with an initial velocity of V and only be subject to gravity so yes, free fall. If P is on the left hand side of the circle as you look at the page then the mass will be given an initial velocity straight up and be in free fall.

10. Nov 28, 2009

### pgardn

I might add you need to be a little careful when doing these problems about what Force is supplying Centripetal Force. In your problem it was tension and gravity. Sometimes it is the normal force of a seat on a ride, sometimes it is friction, sometimes it the electrical force. Centripetal force, or the force that keeps an object in a circular path has to be supplied by some other type of force or forces you have already discussed. And notice the position in the path that you are looking at really changes which of the forces are actually contributing to the centripetal force or opposing it.

It should make sense that the tension should be the largest on the bottom part of the path because it must "beat out" gravity to supply enough force to hold the object in the circular path.