# Circular Motion of keys

1. Jul 14, 2011

### fa08ti

Keys with a combined mass of 0.100 kg are attached to a 0.25m long string and swung in a circle in the vertical plane.
a) What is the slowest speed that the keys can swing and still maintain a circular path?
b) What is the tension in the string at the bottom of the circle?

for part a, I used g= v^2/r and rearranged it to v=sqrt(gr)
I got v=2.5 m/s
I'm pretty sure that's right

I think i'd have to use the same equation for the part b?

I'm just stuck on the radius thing for part b
any ideas would be great

2. Jul 14, 2011

### Staff: Mentor

What did you use for 'r'?

Use Newton's 2nd law.

There's only one radius.

3. Jul 14, 2011

### fa08ti

I used 0.25 m...that doesn't seem accurate now

4. Jul 14, 2011

### fa08ti

and isn't the second law F=ma? how would that help since I only know the mass

5. Jul 14, 2011

### Staff: Mentor

Why do you say that?

6. Jul 14, 2011

### fa08ti

well the question said the length of the string is 0.25m. It didn't say the radius was that length. Would I be correct in assuming that the length of the string is the radius?

7. Jul 14, 2011

### Staff: Mentor

You need to identify the forces acting on the keys when they are at the bottom.

What's unclear to me is what speed you're supposed to use at that point. Are you to assume a constant speed as it goes around the circle? (It would naturally pick up speed as it falls.)

8. Jul 14, 2011

### Staff: Mentor

Yes. The keys are at the end of the string, thus the string becomes the radius of the circle. (Someone's hand is the center of that circle, presumably.)

9. Jul 14, 2011

### fa08ti

Wouldn't I just use the answer from part a?

10. Jul 14, 2011

### Staff: Mentor

I wouldn't think so. The minimum speed is attained at the top; you'd need to figure out the speed at the bottom.

Even part a is a bit ambiguous: Do they want the minimum speed at any point or the minimum speed at the bottom to reach the top?

11. Jul 14, 2011

### fa08ti

It would be at any point

12. Jul 14, 2011

### Staff: Mentor

Which is what you solved for with the formula you used for part a. (You need to redo your calculation with the correct radius.) The point of minimum speed is at the top of the circle.

13. Jul 14, 2011

### fa08ti

how do i know what the correct radius is?

14. Jul 14, 2011

### Staff: Mentor

The radius is given; it's the length of the string. (I thought I answered that one.)