# Particle in Uniform Motion

1. Sep 27, 2008

### Husker70

1. The problem statement, all variables and given/known data
A light string can support a stationary hanging load of 25.0kg before breaking. A 3.00kg
object attached to the string rotates on a horizontal table in a circle of radius .800m, and
the other end of the string is held fixed. What range of speed can the object have before
the string breaks?

2. Relevant equations
I have drawn a diagram and put these forces together.

3. The attempt at a solution
Sum of Forces in X=n and -mg
Sum of forces in Y= T and C sin theata

I know that the X canels out Fx = N-mg = 0
I get Y forces to be T + C sin theata = v^2/r

Am I starting this correct. Something doesn't seem right here.
Thanks,
Kevin
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 28, 2008

### whyhello112

The only force acting in the direction of the string would be the centripetal force, which is dependent on the velocity. Find the velocity which would cause the string to break. The maximum force for the string can be found with the known property that hanging 25kg will break it.

3. Sep 28, 2008

### Husker70

I do know that the forces in the x direction are Centripital Acceleration and tension
but I'm not sure how to put the equation together. F=ma
So does the Tension = m(v^2/r

Thanks,
Kevin

4. Sep 28, 2008

### Husker70

I get the force to be T=m(v^2/r)
So v = sqrt Tr/M
I get 2.58 m/s but the book says 8.08 m/s?

Not sure what I'm doing wrong
Thanks,
Kevin

5. Sep 28, 2008

### Husker70

I got it duh!! I keep thinking that 25.0kg was the tension. But Tension is F=ma
F = (25.0kg)(9/8m/s2) = 245 N
That is the tension force.

Kevin

6. Sep 28, 2008

### Gear300

In this case...you know what to do from here on?

7. Sep 28, 2008

### Husker70

Thanks for looking Gear 300. I got it. Just one more to go to finish my homework.
It could be a rough one.

Kevin