# Catapult velocity help

1. May 25, 2013

### kervyn

For grade 12 physics we have built a catapult and it is now our job to determine a few things about it. Its a fixed arm rotating around a lever. It's energy comes from a band stretched when the catapult is wound back. (band is attached to base and onto non-throwing portion of arm) I am just curious as whether or not the law of conservation of energy will help find the theoretical speed of the catapult's arm.

I know the elastic potential energy in the band, if i solve for Velocity will that correspond to the system?

0.5kx^2=0.5mv^2

I have an excel spread sheet solving for Velocity but based on my video evidence the theoretical does not match up with actual velocity. I'm aware its not fully efficient. Are my calculations relevant to the problem?

Thanks for any help.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 25, 2013

### Mandelbroth

The blunt answer is yes, though the extent of how much it will correspond depends on how the system works. I think we need more information to figure this out. Could you attach a diagram?

3. May 25, 2013

### kervyn

here is a photo of it.

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• ###### Screen Shot 2013-05-25 at 7.58.57 PM.png
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4. May 26, 2013

### haruspex

Typically, the tension doesn't start from 0. Even in the 'relaxed' state of the mechanism, the elastic material should be already under considerable tension. This makes it much more efficient. The energy stored becomes k{(x+δx)2-x2}/2, or about k x δx, instead of k δx2/2. However, from the photograph this might not apply to yours.
Some of the energy will go into the KE of the arm etc. You should be able to calculate this.
The projectile may launch before the mechanism has reached its most relaxed state. It might be hard to judge if that is the case.

5. May 26, 2013

### consciousness

One of your hurdles is to find the rotational kinetic energy of the arm at the time of release. However it is possible to calculate the angular velocity of the arm(ω) in terms of the velocity of the projectile(v) at this instant by a very simple relation v=rω where r is the distance between the pivot and the point where the projectile is kept.

Also I would advise you to properly oil the pivot so that minimum energy is lost due to friction as the band moves.

The apparatus looks quite light so be careful that it doesn't move backwards and spoils the calculation.

I agree that most probably the projectile wont release when the band is at natural length. This is probably the biggest hurdle but the angle of release can be found experimentally. Eg. from using a video camera preferably with slow motion recording.