# Homework Help: Need help on uniform circular motion

1. Nov 27, 2007

### surebud

1. The problem statement, all variables and given/known data
A 100 g bead is free to slide along an 80 cm 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

I know this question has been posted before but i am not sure if i understand it properly

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

3. The attempt at a solution
I have broken down the F in its components Fy and Fx. I calculated the angle to be 53 degrees. I used newtons second law Ftx(cos53) = mg and Fty(sin53) = mg to get the tensions for the 2 string components. For the string AB there is only a horizontal component Ftx(cos0) = mg. If i sum these up, did I correctly calculate the net force of tension exerted onto the strings?

2. Nov 28, 2007

### Shooting Star

You should formulate the problem properly. Which is higher, A or C? If you've done some calculations, you can show them here so that we can understand from where to help you.

3. Nov 28, 2007

### surebud

the string makes a triangle. A to C follows the Y axis; A to B the X axis; and C to B the hypotenuse. I have calculated the leangth on the of the triangle sides and determined that the angle at point B is 53 degrees. Since the only force given in the question is gravity, I am working with it to determine the uniform circular motion using newtons second law. The calculations i just spoke off have led me to the equations in my first message. I have come to the conclusion that the force of tension on the string must be at least the magnititude of gravity. Given the vectors of the strings I got the equations above (first message). I have no idea if this is correct though. It makes sense to me but unfortunatly that doesnt make it right :)

4. Nov 28, 2007

### Shooting Star

Ftx(cos 0) = mg? Why? Also, you have written "Ftx(cos53) = mg and Fty(sin53) = mg"?

Your tension along AB + Ftx(cos 53) = mv^2/AB.

Resolve all the tensions along vert and horzntl directions and equate to weight and centrpetal force respectively. That should do it.

5. Nov 28, 2007

### surebud

this might sound stupid but i am not sure how to resolve for the tensions, from what i understand that u have written, I should add the sum of Ftx(cos53) with the tension of the radius. But how can i solve this equation without velocity. I have been trying to figure this out for some reason i cant put it together, i dont know if i am making it clear as to what i am missing but i seem to be stuck here. I see what u have written but it seems to me that i am still missing a parameter. If the force of tension on AB is
Ft = m(v^2/r) and adding this equation with the horzontal component
Ftx(cos53) = m(v^2/r) than dont i just cancel out all the parameters?? Isnt there also tensoion on the string between CB?

6. Nov 30, 2007

### Shooting Star

I thought you were denoting the tension in CB by Ft and its horizontal and vert components by Ftx and Fty. But you do need the velo to solve for the tensions.

Let me just use with different notation.

Let T1 be the tension along BA and T2 the tension along BC. let r = length AB.

mv^2/r = T2*cos B + T1 – equating horz forces
mg = T2*sin B. – equating vert forces.

T2 can be found and hence T1, if the velo is given.

(I have a "feeling" that there is only one value of v which will result in the given configuration, but I'm missing it. I'm a bit too busy at present to pursue it further right now, But I'll post a definiive yes or no as soon as I'm a bit free.)

Last edited: Nov 30, 2007
7. Nov 30, 2007

### surebud

I appreciate the help!!