# Find the speed of the bucket

1. Jun 20, 2008

### 0338jw

1. The problem statement, all variables and given/known data
A bucket of mass 2.00kg is whirled in a vertical circle of radius 1.10 m. At the lowest point of its motion the tension in the rope supporting the bucket is 25.0 N. a) Find the speed of the bucket. b) How fast must the bucket move at the top of the circle so that the rope does not go slack?

2. Relevant equations

$$\Sigma$$Fy= Ft - Fg =m*$$a_{c}$$
Ac=v^2/r

3. The attempt at a solution
I have solved for the net force in the Y direction and I get 5.4N. Is this the net centripetal force that i can use to plug back into F=m *ac for the centripetal acceleration? I get 2.7m/s when i do this. I know this question has been posted before, and I'm sorry but I don't quite get it after reading and rereading bits and pieces of information by others posted in the forum. all help is greatly appreciated!

2. Jun 20, 2008

### 0338jw

or when solving for speed do I plug the net force back into F=m(v^2/r)? When I do this and solve for the speed I get 1.72m/s. I'm not sure what I'm doing wrong and which is the correct equation to use.

3. Jun 20, 2008

### Staff: Mentor

Yes.
That's the same thing, so I don't see how you got two different answers. Show exactly what you did.

4. Jun 20, 2008

### Staff: Mentor

I see what you did here: You solved for the centripetal acceleration. (That should be 2.7 m/s^2--not m/s.) You need to solve for the speed, which you did in your second post.

5. Jun 20, 2008

### 0338jw

Ohh okay! So now I have the acceleration as 2.7 m/s/s and the speed as 1.72 m/s. Does the centripetal acceleration remain the same for when the bucket is at the top of the circle? I know the force of tension is 0 and that the bucket must be swinging fast enough to overcome the weight of the bucket and water to keep it from falling down. I'm not sure how to write this in equations though. the sum of the forces in the Y direction is FT + Fg =m*ac? Again, sorry if I'm not seeing something obvious I'm taking a couple summer classes and we're flying through some important stuff pretty quickly.

6. Jun 20, 2008

### Staff: Mentor

No.
Right! You solve it exactly like you solved the first part.

7. Jun 20, 2008

### Fanaticus

Wait, centripetal acceleration is v^2/r. At the top of the circle neither v nor r change. Its circular motion.

8. Jun 20, 2008

### Staff: Mentor

The speed changes. (Don't confuse this with uniform circular motion, where the speed stays constant.)

9. Jun 20, 2008

### 0338jw

thank you for the help! When I solved for velocity I got 3.28 m/s. It should be right, but I'm still not very confident in my work yet. Again, thanks for the help!

10. Jun 20, 2008

### Staff: Mentor

Sounds good to me!