# Tough problem-conservation of momentum?

1. Apr 6, 2006

### syang9

tough problem--conservation of momentum?

suppose you have a dumbbell standing up vertically against a frictionless wall

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(given, m, L (length of rod))
the weights are of course the same mass. the bottom weight of the dumbbell starts sliding out so the top mass moves vertically, while the bottom moves horizontally. i am asked to find the velocity when the weights are both moving with the same speed. i'm not really clear as to how to approach this problem.. it seems if i apply conservation of momentum,

mv1 = mv2; v1 = v2

which is obviously not true. so does this mean momentum is not conserved? if i take the system to be the earth and the dumbbell, the external forces would sum to zero, wouldn't they?

i also figured that
x^2 + y^2 = L^2; 2x * dx/dt + 2y * dy/dt = 0

but this doesn't seem to be enough. it seems as if the speeds should be the same since it is a rigid body. could anyone explain this to me? or give a hint or two?
any help would be appreciated! thanks!

Last edited: Apr 6, 2006
2. Apr 6, 2006

### Curious3141

First of all, why would you assume that momentum is conserved. There is something that is essential for conservation of momentum which is being clearly violated here, what is it ? In your equation, btw, you didn't include the earth within the system.

The way you differentiated implictly with Cartesian coordinates is fine. Just set $$|\frac{dx}{dt}| = |\frac{dy}{dt}|$$ to solve for y in terms of x. Then go back to $$x^2 + y^2 = L^2$$ to determine the y-value at this instant in terms of L.

From that, find the height thru which the top bell has fallen in terms of L. Use this to determine the decrease in Gravitational Potential Energy of the system. Equate that to the increase in kinetic energy of the system at this point, and solve for v, the speed of the bells.

Last edited: Apr 6, 2006