# Uniform Circular Motion Tension

1. Jul 11, 2007

### Kyleman

1. The problem statement, all variables and given/known data
A 100g bead is free to slide along an 80cm long piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40 cm apart. When the pole is rotated about its axis, AB becomes horizontal. Find the tension in the string. Find the speed of the bead at B.

2. Relevant equations
a=v^2/r
F=ma

3. The attempt at a solution
So I figured out that the radius of the the circular motion the bead makes is 30cm. However, that's all I have. I'm not sure where to go from here because I don't know the velocity or acceleration.

2. Jul 11, 2007

### bel

Draw a free body diagram of the bead and use the fact that the only forces are the force of gravity and the centripetal force. From the forces, calculate the acceleration, which will then, by $$a= v^2 / r$$ give you the velocity of the bead, since you already know the radius.

3. Jul 13, 2007

### Kyleman

I thought that because the string is horizontal there is no gravity acting on it. How do I calculate the centripetal force too?

4. Jul 13, 2007

### Staff: Mentor

Why would you think that? Hint: Part of the string is not horizontal.
bel gave you the equation for centripetal acceleration. Combine that with an analysis of the vertical and horizontal force components. Use Newton's 2nd law.

5. Jul 13, 2007

### Kyleman

So would this be the correct formula?: Ft+mg=m v/r

6. Jul 13, 2007

### Kyleman

sorry, Ft - force of the tension.

7. Jul 13, 2007

### Staff: Mentor

No, that is not correct.

Start by drawing a diagram identifying all the forces acting on the bead. Then consider the vertical and horizontal components separately.

8. Jul 13, 2007

### Kyleman

okay, okay

is this correct? haha Fty=mg and Ftx=ma

Thanks for the help by the way.

9. Jul 14, 2007

### Staff: Mentor

You are on the right track, but realize that there are two sections of string pulling the bead. (Draw yourself a picture.)

10. Aug 2, 2008

### KBark15

11. Aug 2, 2008

### KBark15

I got 980 N for the tension and 17.2m/s for the speed not sure if its right but at least its a real #

12. Aug 3, 2008

### Staff: Mentor

No, doesn't look right to me. But as you say, at least they are real numbers!

Show how you got those answers.

13. Jun 7, 2009

### tomotron3000

Hello,
I am looking at the same problem.
I figured out the answers, but there is still something bothering me about the problem.
I understand why Fty=mg
but I don't understand why Ftx=mv^2/r.
Aren't both Ftx and mv^2/r going in the same direction, towards the pole?
Why is Ftx=mv^2/r?

14. Jun 7, 2009

### tomotron3000

I have another question.
Given the question. How would you go about finding the lengths of each side. We know one side is 4, but how do you find the other sides? I just recognized a 3,4,5 side triangle. (maybe because I read that book Fermats Enigma) But had the number been something other then 4 I would not known how to do it.

15. May 1, 2011

### Neale.c

I doing the same problem, and I think I understand what to do. Force in the y direction is mg, and Fx = mv^2/r, and then we can get the tension on BC by adding the two force vectors. The problem I'm running into is where to get velocity to find Fx?

16. May 1, 2011

### Delphi51

Interesting problem! Does it look like this?

tomotron, your right triangle idea is brilliant! It need not be 3,4,5; it just has to fit the pythagorean equation.

17. May 1, 2011

### Staff: Mentor

Don't think of mv^2/r as a force, but as the result of apply Newton's 2nd law in the x direction. The acceleration in that direction is v^2/r. You'll end up solving for the velocity.

The only forces acting are the weight (mg) and the tension in the string. (Careful when considering the string tension.)