# Homework Help: Rotation in a Vertical Plane - tension on a bucket

1. Feb 26, 2012

### testme

1. The problem statement, all variables and given/known data
A 4kg bucket of water is rotated in a vertical circle of radius 2m at the minimum rate required to keep the water in the bucket. Find the tension in the handle of the bucket at a) the bottom of the circle and b) the top of the circle.

mass: 4kg

2. Relevant equations

Fnet = ma
Fnet = mv^2/r

3. The attempt at a solution

a) Well, I'm not sure if I'm doing this right but since we're talking about the tension at the bottom of the circle I took the velocity to be 0 (I think I'm wrong but I'm not sure what else to do).

Fnet = mv^2/r
Fg + T = mv^2/r
-39.2 + T = 4(0)^2/2
-39.2 + T = 0
T = 39.2 N

b) I'm having trouble with this one, so far I have

Fnet = mv^2/r
Fg + T = mv^2/r
-39.2 + T = 4v^2/2
-39.2 + T = 2v^2

Sorry I know it's not much that I've done but I'm out of ideas.

Last edited: Feb 26, 2012
2. Feb 26, 2012

### LawrenceC

If you have the minimum speed where the water does not come out of the bucket, the water then is 'weightless'. If the water is weightless, what can you say about the water and the bucket together?

Then what happens at the bottom? You have the same two forces but now they combine differently.

3. Feb 26, 2012

### testme

Hmm, I'm not really getting what you're trying to say, why is the water then 'weightless' and wouldn't that mean the water and the bucket together have the combined weight of the bucket only?

At the bottom you'd have the same forced but the tension is upwards while gravity is downwards right?

I'm still not sure of how I'm supposed to calculate the tention force though since we don't have a velocity

4. Feb 26, 2012

### LawrenceC

I'm trying to give you hints without telling you how to do it.

When an object revolves in circular motion like this pail with water in it, there is a force that keeps it in the curved path. That force is mv^2/r. The acceleration part of what makes up the force is v^2/r. If this acceleration just equals gravity, 9.81 m/sec^2, weightlessness occurs. The force outward away from the center of rotation just equals the weight when in the vertical position.

At the bottom, the outward force and weight are in the same direction.

5. Feb 26, 2012

### testme

The centripetal force, I believe. So at the top we're saying it only equals gravity? Does that mean there's no tension force at the top?

Wouldn't the two forces be in opposite directions at the bottom?

Sorry, it's just my first question like this.

Here is a picture of the FBD I drew - they're just rough drawings

Last edited by a moderator: May 5, 2017
6. Feb 26, 2012

### LawrenceC

"The centripetal force, I believe. So at the top we're saying it only equals gravity? Does that mean there's no tension force at the top?"

Yes, you are correct.

You have left out a force in each sketch. It is mv^2/r pointing away from the center of rotation in each case. At the top it points up. At the bottom it points down. So what does this do to the tension at the bottom?

7. Feb 26, 2012

### testme

What my teacher told us was that the centripetal force is the combined force that will be towards the center. Isn't that what mv^2/r is? That's what I'm kind of confused about.

8. Feb 26, 2012

### LawrenceC

At the bottom the tension is the sum of the weight and mv^2/r providing the speed is the same at the top as it is at the bottom.

"What my teacher told us was that the centripetal force is the combined force that will be towards the center. Isn't that what mv^2/r is?"

Centripital force is the force that keeps the pail moving in circular motion. So it is the sum of mv^2/r and the weight.

9. Feb 26, 2012

### testme

Okay, so I think I finally got it.

Since we know that the T at the top is 0.

Fnet = mv^2/r
Fg = mv^2/r

One sig fig.. (and using 9.8 m/s^2 for gravity)

40 = 4v^2/2
40 = 2v^2
20 = v^2
v = 4 m/s

Then at the bottom

Fnet = mv^2/r
Fg + T = mv^2/r

Assuming up is positive

-40 + T = 4(4)^2/2
-40 + T = 2(16)
T = 72 N

Thanks a lot for the help.