# Circular Motion and centripetal acceleration

1. Feb 21, 2015

### Jimmy87

1. The problem statement, all variables and given/known data
Q.1. Explain the changes in the tension of a piece of string which is being swung vertically with a bucket of water at the end. What would the minimum centripetal acceleration need to be for water to not fall out?

2. Relevant equations
ac = v^2/r

3. The attempt at a solution
The tension in any vertical path will always be greatest at the bottom as the weight force needs to be overcome as well as providing a net force towards the center which is required for circular motion. The centripetal acceleration would need to be 9.81 m/s^2, therefore, at the top gravity will provide all the centripetal force and no tension is required from the string.

From researching the bucket problem online, it seems to be that when the bucket is weightless at the top then the tension at the bottom would be equal to twice the weight of the bucket. However, this assumes uniform circular motion. Can this bucket problem ever be uniform circular motion? Can you ever have water in a bucket having a centripetal acceleration of 9.81 all the time? Surely as it comes down towards the ground the kinetic energy is greater therefore v^2/r cannot be constant?

Last edited: Feb 21, 2015
2. Feb 21, 2015

### Staff: Mentor

It would be better to separate the two problems in separate threads.

What is the equation for the centripetal force opposing the gravitational force mg?

3. Feb 21, 2015

### Jimmy87

Wouldn't it be: Fc = 2mg since the centripetal force at the top is mg and the only way to get that at the bottom (when gravity is opposing the bucket) is to have 2mg.

Can a bucket be swung vertically so that the centripetal force is always mg? Is that physically possible?

4. Feb 21, 2015

### Merlin3189

Since the questions ask you to "explain" and "discuss", I'd say you're right on the money.
I think you've thought about the main issues.

One comment I'm not sure about: "this assumes uniform circular motion" - I don't see why?
At first I thought you were referring to the fact that when you swing the bucket in practice, your hand has to move and the string does not go to a fixed centre of a circular orbit, so maybe the orbit is not a circle? Then I decided you did not mean that and you were thinking the angular speed could not be constant.
As you suggest, I doubt that any such motion could have constant angular speed. I just think about the point when the string is horizontal.

5. Feb 21, 2015

### Staff: Mentor

What is the equation for static equilibrium? Think the condition of weightlessness?

Think about the forces on the mass (m) of water.

6. Feb 21, 2015

### Jimmy87

Thanks for your answer. What I mean by "this assumes uniform circular motion" is that when you read about the minimum speed to keep the water to stay in the bucket they show that the tension in the string is zero at the top and 2mg at the bottom. This therefore gives a centripetal force of mg at the top and bottom and hence uniform circular motion. They say that "if you swing the bucket such that ac is always 9.81 m/s^2........" but I don't see how you can keep a constant centripetal acceleration as surely it inevitably goes slower as it reaches the top. Therefore, surely it is wrong to say that the tension in the string is 2mg at the bottom and mg at the top because this implies a constant centripetal force and hence a uniform circular motion?

7. Feb 22, 2015

### Merlin3189

Ok. I think you are right. I had not followed through, from agreeing with you that v and ω are changing, to realising that as well as the tension in the string changing, the centripetal force must also change in the way you describe and so, again as you say, the tension at the bottom must be greater than 2mg.
And it is easily calculable since the only source of changed speed is the change in PE.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted