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Complicated Catapult/Conservation of Energy Test Question

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ccmuggs13
#1
Mar29-12, 01:45 AM
P: 4
I just took a physics test today and there was one problem that I cannot figure out if I did correctly. The problem involves a catapult pictured here: http://i39.tinypic.com/a2t76h.jpg
The bar is 2m long and has a mass of 3kg. The rock at the end of it has a mass of 5kg. The spring is .3m from the pivot. As shown, when the spring is unstretched/uncompressed the bar makes an angle of 30 above the horizontal. It is then pushed down so it now makes an angle of 30 below the horizontal. My first question is what is the x for the spring when determining the elastic potential energy? Because I just set up triangles and figured it out to be .3m but my friend used the relationship between linear and angular distance, d=θr, and got x to be larger. But I believe that gives the arc length traveled by the end of the spring as it is compressed rather than how much it is actually compressed but I could be wrong. Here is a picture of what I think it is like: http://i44.tinypic.com/5zm6om.jpg

And so after you figure out what x is, what the problem asks for is the velocity of the rock as it passes back through the original position? (Using conservation of energy) So how would this be done and what is the answer? Because I ended up with a negative value while trying to find the answer meaning the spring did not have enough energy to get the catapult back to the starting position but this is probably wrong because that would be a dumb test question. Sorry this post is so long but please help!!!! I've been going nuts trying to figure out what I did wrong!!!
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K^2
#2
Mar29-12, 02:35 AM
Sci Advisor
P: 2,470
The displacement would depend on where the other end of the spring is attached. If it's right bellow the point where it's attached to arm at equilibrium, your value of .3m is correct. I would go with that given no other information.

Do you happen to remember what the spring constant is? Also, is the equilibrium position equilibrium for the spring, equilibrium for unloaded catapult, or equilibrium for loaded catapult? Each of these would give you different results. In the final case, the speed will be maximum through equilibrium. But in the first two cases the kinetic energy might work out to be negative if the spring coefficient was too low.

The gravitational potential energy as function of angle looks like this.

[tex]U = sin(\theta)\frac{1}{2}mgL + sin(\theta)MgL[/tex]

Kinetic energy as function of velocity.

[tex]T = \frac{1}{6}mv^2 + \frac{1}{2}Mv^2[/tex]

Where L is the length of the arm, m is mass of the arm, M is mass of the boulder, and v is velocity of the boulder. The angle is measured from horizontal.

Assuming equilibrium is equilibrium for the spring, the rest is easy. Add spring potential to gravitational potential at lower point, subtract of gravitational potential at the equilibrium point, and that's your kinetic energy. Solve for v, and you're done. If some other equilibrium condition is meant, you'll have to correct for it.
ccmuggs13
#3
Mar29-12, 02:56 AM
P: 4
Oh yea I completely forgot about the spring constant, it was 2000N/m. And the equilibrium position is equilibrium for the spring.

Also can you use the linear kinetic energy formula for this? Because a large portion of our test involved angular velocity and the problem gave us a hint reminding us of the formula for the moment of inertia of a bar about its end so I assumed we needed to use angular velocity and kenetic energy and such.

ccmuggs13
#4
Mar29-12, 03:01 AM
P: 4
Complicated Catapult/Conservation of Energy Test Question

Oh I think I see that you did take angular motion into account as opposed to just linear which is why you have the kinetic energy of the bar start with 1/6 as opposed to 1/2

But unless I am doing my calculations wrong I am still getting a negative kinetic energy... Please check the calculations. This is a very dumb question if the spring really does not have enough energy to return to equilibrium
azizlwl
#5
Mar29-12, 03:37 AM
P: 963
I think x=2(.3)tan30 since Cos30=-Cos30
K^2
#6
Mar29-12, 05:16 AM
Sci Advisor
P: 2,470
If you remembered all the numbers correctly, and the equilibrium is equilibrium for the spring, you are absolutely correct. Here is a plot of kinetic energy as function of angle. The points where the curve crosses zero are turning points. The angles are in radians. So the arm will get just a little past 0, and then start coming down again, never reaching the equilibrium point. To get the arm to swing past equilibrium, you must pull the arm down past -60 before firing.
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