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First off, I swear this is not homework. I'm doing some practice problems because I got an exam coming up. I'm stuck on this one: alt text

I figured I would use energy conservation for this problem. So since the thing is not moving initially, I tried doing

$mgh=1/2 I\omega^2+1/2 mv^2$, but that doesnt give me the right answer. Any ideas?

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+1 For being a good old-fashioned quantitative question! –  Noldorin Nov 14 '10 at 19:53
    
I'm all for "quantitativity," but I hope these sorts of super-specific homework-like questions remain a minority on the site. –  David Z Nov 15 '10 at 4:38
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2 Answers 2

up vote 6 down vote accepted

The basic setup is correct, conservation of energy might be the quickest way to go.

$$(m_1 -m_2) g h = \frac 1 2 I \omega^2 + \frac 1 2 (m_1+m_2) v^2, I=\frac{MR^2}{2}, \omega = v/R$$ gives me one of your options as the result.

The two $m$ in your formula seem to refer to different quantities.

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Good, concise answer. Just to clarify, MR^2/2 is the moment of inertia of a uniform disc around an axis passing through its center orthogonally to the plane of the circle. –  Noldorin Nov 14 '10 at 20:22
    
you are indeed correct sir. Only question is, why do you subtract m1-m2 for potential energy? –  maq Nov 14 '10 at 20:27
    
@fprime: It's really m_1gh - m_2gh with the common factor taken out. There are two tensions on the rope due to the gravitational forces, in opposite directions. –  Noldorin Nov 14 '10 at 20:57
    
@fprime: You identified the tricky part here (which was obvious to me because I worked from a Lagrangian and only later said that "yes, that was a waste of time, let's do it using conservation of energy"). Noldorin explains this correctly in the force picture, but in the energy picture I'd say "I wrote down the `change of energy' between the before and after states, not the absolute energies". The absolute energy formulation is what the other other answer points out, for equivalent results. –  Thomas Themel Nov 15 '10 at 7:59
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Don't forget the potential energy of m2 which is at height 2h when m1 hits the ground.

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How would I use that? –  maq Nov 14 '10 at 20:05
    
You add it at the right hand side of the equation. –  Vagelford Nov 14 '10 at 20:49
    
Thomas has taken that term from the right side of the equation to the left side. The subtraction gives the (m1-m2)gh expression (I am assuming that m is the total mass m1+m2 in your expression). –  Vagelford Nov 14 '10 at 20:52
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