Pumpkin Swing: Calculating Speed at θ=30°, L=5m, g=9.8m/s2, m=10kg

[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.

 

[cookiemonster]

Why don't you try an energy argument?

potential + kinetic energy = constant

cookiemonster

 

[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 don't 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

 

[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

 

[Drakon25th]

Originally posted by cookiemonster
Your approach is good, but your expression for potential energy is not correct. Try to see if you can't include the angle in there.

cookiemonster

can't or can?

 

[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

 

[Drakon25th]

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

 

[cookiemonster]

What triangle did you use?

cookiemonster

 

[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

 

[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

 

[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

 

[cookiemonster]

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

cookiemonster

 

[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?

 

[cookiemonster]

Yup. Good job.

cookiemonster

 

[Drakon25th]

thanks for all the help