# The Pumpkin Swing

1. Feb 24, 2004

### physicsgivesmeCs

How fast is the pumpkin moving at the bottom if theta = 30 degrees, L = 5 m, g = 9.8 m/s^2, m = 10 kg.

L is the length of the string holding the pumpkin. M is the mass of the pumpkin. It's like a pendulum, the pumpkin is positioned at 30 degrees from the vertical and let go. Assume no air resistance/friction.

Last edited: Feb 24, 2004
2. Feb 24, 2004

### cookiemonster

Why don't you try an energy argument?

potential + kinetic energy = constant

cookiemonster

3. Feb 24, 2004

### Drakon25th

could you please explain what you mean a little? I am lost with that problem, too.

This is what I'm thinking, but i dont know what to do with the angle and all and if i'm doing this correctly or not:

TE = PE + KE
PE = KE
mgh = mv^2
sqrt ((mgh)/m) = v
sqrt (gh)= v
sqrt (9.8m/s^2 * 5.0m) = v
v = 7 m/s

Please help me if you can, thanks in advance

Last edited by a moderator: Feb 25, 2004
4. Feb 25, 2004

### cookiemonster

Your approach is good, but your expression for potential energy is not correct.

Let the position at theta = 0 (hanging straight down) have a potential of 0. So now the question is, "If the pendulum is at an angle of 30 degrees, how high is the bob?" Why don't you draw a triangle?

cookiemonster

5. Feb 25, 2004

### Drakon25th

can't or can?

6. Feb 25, 2004

### cookiemonster

I hadn't read your post carefully enough to realize you needed help with the angle. I edited my post to include some helpful information.

And just for a fun little exercise in English, the sentences with "can" and "can't" mean exactly the same thing!

cookiemonster

Last edited: Feb 25, 2004
7. Feb 25, 2004

### Drakon25th

if i did it right, i got 1.25m high from the bottom point, is that correct?

8. Feb 25, 2004

### cookiemonster

What triangle did you use?

cookiemonster

9. Feb 25, 2004

### Drakon25th

Hold on, i am about to post up the picture i drew

http://mysite.verizon.net/vze3ss2y/sitebuildercontent/sitebuilderpictures/pumpkin.jpg

i used the big triangle to find the hypotenuse of the small triangle (circled in green) then i used the laws of sine to determine the height of da pumpkin from the bottom point where PE = 0

Last edited by a moderator: Feb 25, 2004
10. Feb 25, 2004

### cookiemonster

Well, I'm not quite sure I understand your triangle. I'll describe the one you're going to want to use:

We're going to let the string and bob form the hypotenuse. So the hypotenuse will always have length 5. Additionally, we want to get a value for y, so the other two sides will be in the up-down direction and left-right direction. Since we want the up-down and left-right dimensions to form a triangle with the string, put the up-down side directly below the place where the string is hanging and the left-right side right to the side of the bob. I assume the angle is of the hypotenuse off of the up-down side, so the angle on the top of this triangle will be 30. Now you can calculate the y height.

Just remember that this is the height from the top, whereas you're looking for the height from the bottom.

Edit: In response to your post and diagram (I understand it now), you're assigning the value of 5 to the wrong side. Your approach can work, but it's more difficult than it needs to be.

cookiemonster

11. Feb 25, 2004

### Drakon25th

could you please post a picture of what you're trying to say?

edit: wait i think i know what you're talking about now

Last edited by a moderator: Feb 25, 2004
12. Feb 25, 2004

### cookiemonster

Hope it works... I don't have a webspace of my own (that I know how to use).

cookiemonster

#### Attached Files:

• ###### new bitmap image.jpg
File size:
1.9 KB
Views:
108
13. Feb 25, 2004

### Drakon25th

thanks for the picture, it is very helpful.

Ok, so when i solve for the adjacent side (up-down), i get 4.33m from the top, meaning the pendulum is .66m from the bottom. So when i solve the problem to find the velocity, for height in PE = mgh, i use .66m to get:

PE = KE
mgh = .5mv^2
gh = .5v^2
sqrt (gh/.5) = v
sqrt ((9.8m/s^2 * .66m)/.5) = v
v = 3.6 m/s

Is that correct?

14. Feb 25, 2004

### cookiemonster

Yup. Good job.

cookiemonster

15. Feb 25, 2004

### Drakon25th

thanks for all the help

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook