# Homework Help: Help with circular motion

1. Sep 30, 2014

### Jimmy87

1. The problem statement, all variables and given/known data
Assessing the required forces at the top and bottom of a loop de loop. I understand that you need a larger force at the bottom of the loop because you have to counter gravity. My question is, when you are undergoing uniform circular motion and your centripetal acceleration is constant then would your centripetal force also be constant?

If you have a loop de loop with a centripetal acceleration of 9.81m/s^2 then at the bottom the reaction force is 2mg which provides the centripetal force but at the top the weight provides the centripetal force which is only mg as the reaction force would be zero. But if v^2/r is constant then how can the centripetal force be different at the top and bottom?

2. Relevant equations
F = mv^2/r.

3. The attempt at a solution
Have listed my thoughts in the top section.

2. Sep 30, 2014

### nasu

Why do you assume constant speed? What is the effect of gravity on the motion around the loop?

3. Sep 30, 2014

### Jimmy87

Sorry, I meant to say that if we are purely considering uniform circular motion on a loop de loop where we have a centripetal acceleration of 9.81. Gravity acts against the centripetal force at the bottom but provides it at the top. I thought the reaction force provides the centripetal force at the bottom but then I don't get how the centripetal force can be greater at the bottom if the acceleration is constant? Or is the centripetal force not the whole force but just the net bit towards the centre?

4. Sep 30, 2014

### nasu

How can you have uniform circular motion on a vertical loop? Do you consider some engine attached to the object moving around the loop?
The usual situation assumes a ball sliding freely around the loop. When it moves upwards it slows down, have a minimum velocity at the top and then accelerates again when it goes down.
So neither speed nor centripetal acceleration is constant for this motion.

5. Sep 30, 2014

### Jimmy87

Thanks for the reply. Ok forget that I see what you mean and take your point. What happens if you take a rope and a bucket with water in it. If you knew the radius and the mass then you could swing it with a constant velocity (v) such that the centripetal acceleration is 9.81 all the time. At the bottom the normal force from the bucket on the water would be 2mg and at the top it would be zero. Here the centripetal acceleration would be constant but wouldn't the force be greatest at the bottom?

Last edited: Sep 30, 2014
6. Sep 30, 2014

### Jimmy87

Also, if the rollercoaster does indeed move at different speeds, would this strictly be non-uniform circular motion?

7. Sep 30, 2014

### nasu

Ok, i see that you are really focused on constant speed. Then indeed the centripetal acceleration will be constant as well. And so will be the centripetal force. IF you can arrange this.
If you want the centripetal force to be equal with the weight of the ball, the reaction at the top will be zero. This is actully the limiting case for speed at the top. Any speed less than this will requir negative reaction force.

8. Sep 30, 2014

### Jimmy87

Thanks nasu. So, in my example, am I correct in saying that the reaction force of the bucket at the bottom of the swing will be 2mg but the centripetal force at the bottom will be mg? What I am trying to get at - is the centripetal force just the net force towards the centre as appose to the total force towards the centre? And would this mean that you should never label a FBD with the centripetal force as it is a net force only?

9. Sep 30, 2014

### Jimmy87

Also is it only true that the magnitude of the centripetal acceleration and force is constant since the direction is changing?

10. Sep 30, 2014

### CWatters

Correct. Centripetal force is the force required to cause the bucket to move in a circular path. It doesn't matter what other forces act on the bucket as long as the net force is mv^2/r.

There are two forces acting on the bucket.

1) The tension in the rope T which will vary
2) Gravity

In order to move in a circle the two components must sum to mv^2/r

Lets call the net force acting towards the center F

At the top

F = T + mg (they act in same direction)
For circular motion F must equal mv^2/r so
mv^2/r = T + mg
Rearrange to give a general equation for the tension..
T = mv^2/r - mg

If you then arrange for the velocity and radius to be such that mv^2/r = mg then T=0

At the bottom

F = T - mg (they act in opposite directions)
For circular motion F must = mv^2/r so
mv^2/r = T - mg
Rearrange to give a general equation for the tension..
T = mv^2/r + mg

If you then arrange for the velocity and radius to be such that mv^2/r = mg then T=2mg
So the tension in the string varies top to bottom but in both cases the net force F = mv^2/r

11. Sep 30, 2014

### Jimmy87

Perfect, thank you very much, that makes complete sense. So in the situation I described where you try to keep the velocity at a particular constant value such that the centripetal acceleration is always 9.81 then the magnitude of the centripetal force will be mg at every point around the circle? Is that right? Also, since the centripetal force is a net force is it wrong to label it in a FBD?

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