# Homework Help: Object on a string

1. Oct 9, 2007

### bulldog23

1. The problem statement, all variables and given/known data

The string in the Figure is L = 109.0 cm long and the distance d to the fixed peg P is 74.1 cm. When the ball is released from rest in the position shown, it will swing along the dashed arc. How fast will it be going when it reaches the lowest point in its swing?How fast will it be going when it reaches the highest point in its swing?

I am totally lost as to how to even start this problem. I drew a free body diagram, but I do not know what equations I need to use. Could someone please walk me through this problem?

2. Oct 9, 2007

### Dick

You want to use energy conservation equations. Do you know any of those that might apply?

3. Oct 9, 2007

### bulldog23

I have no idea which ones would apply.

4. Oct 9, 2007

### Dick

You aren't trying. KE+PE=constant. KE=(1/2)*m*v^2. PE=mgh. Take it from there.

5. Oct 9, 2007

### bulldog23

I've never taken physics before, so this is my first time ever doing this stuff. So are these the only equations that apply?

6. Oct 9, 2007

### Dick

They aren't the only ones that apply. But they are the only ones that you need.

7. Oct 9, 2007

### bulldog23

I don't understand how you take the length of the string and incorporate that into those equations to get a speed. Could you please explain that

8. Oct 9, 2007

### Dick

Do you know what the equations that I wrote mean? Have you seen them before? Do you know conservation of energy?

9. Oct 9, 2007

### bulldog23

I understand what those equations mean, I just don't know how to apply them to this problem. I have seen these equations before.

10. Oct 9, 2007

### Dick

Ok, then put h=0 to be the bottom of the arc. What the total energy of the ball at it's initial position?

11. Oct 9, 2007

### bulldog23

At it's initial position it would just be the PE=mgh, but what is the mass?

12. Oct 9, 2007

### Dick

m is m. It will cancel if you are patient. What's h in terms of your diagram?

13. Oct 9, 2007

### bulldog23

h would be the length L of the string. 109 cm

14. Oct 9, 2007

### Dick

Ok. So total energy is mgL. Now what are KE and PE at the bottom of the arc? Answer both to earn points. Remember their total is the same energy that you started with, mgL.

15. Oct 9, 2007

### bulldog23

KE would be mgL, and PE would be 0. Is that right?

16. Oct 9, 2007

### Dick

Right, but KE is also (1/2)mv^2. What's v at the bottom of the arc? Get that and you have half of the points.

Last edited: Oct 9, 2007
17. Oct 9, 2007

### bulldog23

I am not sure what the velocity would be.

18. Oct 9, 2007

### Dick

KE=(1/2)mv^2. KE=mgL. I think we've agreed on that. Can you solve for v in terms of L?

19. Oct 9, 2007

### bulldog23

so v would be the square root of g*L/.5?

20. Oct 9, 2007

### Dick

Yessssssss. Now what's PE at the top of the circle of radius r. What's KE at the top of the circle? The total is still mgL. Energy is conserved. I'm hoping you'll get this soon. It's bedtime, help me out.

21. Oct 9, 2007

### bulldog23

sorry man, I'm tryin...So then PE would be mg*2r?

22. Oct 9, 2007

### Dick

Great! You are helping! KE is still (1/2)*mv^2 but this is a different v than before, it's v at the top of the circle, not v at the bottom. Can you put it all together?

23. Oct 9, 2007

### bulldog23

alright so lets go back to the speed at the bottom of the circle. I calculated the speed to be 46.22 m/s because I used the KE=square root of g*L/.5 Is this right?

24. Oct 9, 2007

### Dick

Your decimal point is in the wrong place. L is 109cm, not 109m.

25. Oct 9, 2007

### bulldog23

oh yeah duh! that makes sense. So to get the radius I just take 109-74.1, which makes r=34.9. Then the speed would be square root of g*r/.5???