# Having trouble with a centripetal force problem

jap129

## Homework Statement

A 2.0 kg stone is whirled in a horizontal circle on the end of a 2.1 m long rope. The stone travels around the circle 50 times each minute. What centripetal force is exerted by the rope on the stone?

## The Attempt at a Solution

I've tried to work put the variables into the formula: m*v^2/r. But it didn't work. I've tried anything I could figure, and this is the only problem I have been stumped on from the last 50 problems I've done.

## The Attempt at a Solution

jegues
I've tried to work put the variables into the formula: m*v^2/r. But it didn't work.

Can you show us your work? I'm fairly certain it should work...

Homework Helper
The stone travels around the circle 50 times each minute.

so then how many oscillations are in one second?

jap129
Can you show us your work? I'm fairly certain it should work...

50/60= 0.83

2*0.83^2/1.05 = 1.3 <That answer did not look correct to me. I probably did something wrong. But don't remember what.

jegues
2*0.83^2/1.05 = 1.3 <That answer did not look correct to me. I probably did something wrong. But don't remember what.

Where did you get that radius from? Isn't the rope 2.1m?

jap129
Where did you get that radius from? Isn't the rope 2.1m?

Woops. I was looking at a different problem when I wrote down the radius. So...
2*0.83^2/2.1 = 0.65
The answer is still wrong though.

jegues
You can compute its circumference and with it solve for a linear velocity. Then think about the relationship between linear and centripetal velocity.

BoldKnight399
Have you thought about the Fc=mw^2R equation? That might help.

jegues
Have you thought about the Fc=mw^2R equation? That might help.

I want you to solve for this equation yourself using your understanding of the relationship between linear and angular velocity.

I'd rather you understand how the equations came to be, rather than simply memorizing them.

Last edited:
jap129
I believe I figured it out, the answer is 120 N. It makes sense now. Thanks for the help. :)