Is Momentum Conserved When a Dumbbell Slides Against a Frictionless Wall?

[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!

 

[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 [tex]|\frac{dx}{dt}| = |\frac{dy}{dt}|[/tex] to solve for y in terms of x. Then go back to [tex]x^2 + y^2 = L^2[/tex] 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.

 

[kyle8921]

I would approach this problem by first understanding the concept of conservation of momentum. This principle states that the total momentum of a closed system remains constant, regardless of any internal changes or external forces acting on the system. In this case, the closed system would be the Earth and the dumbbell.

Looking at the problem, we can see that the dumbbell is initially at rest and then starts sliding out, causing the top weight to move vertically and the bottom weight to move horizontally. This change in motion is due to an external force, which is the frictionless wall. However, the total momentum of the system should still be conserved.

To solve this problem, we can use the conservation of momentum equation, which states that the initial momentum (before the weights start moving) is equal to the final momentum (after the weights are moving). This means that the momentum of the system in the x-direction (horizontal) should be equal to the momentum in the y-direction (vertical).

We can set up the following equations:

Initial momentum in the x-direction: m * 0 = 0
Final momentum in the x-direction: m * v1 = mv1

Initial momentum in the y-direction: m * 0 = 0
Final momentum in the y-direction: m * v2 = mv2

Since the weights are of equal mass, we can set the final momenta equal to each other:

mv1 = mv2

This equation shows that the velocities in both directions should be equal, as you mentioned. However, it is important to note that this is only true after the weights have started moving. Initially, the momentum in both directions is zero. So, we cannot say that v1 = v2 from the beginning.

To find the velocity when the weights are both moving with the same speed, we can use the equation you mentioned: x^2 + y^2 = L^2. This equation represents the Pythagorean theorem, where x and y are the horizontal and vertical distances traveled by the weights, and L is the length of the rod.

We can differentiate this equation with respect to time to get:

2x * dx/dt + 2y * dy/dt = 0

Since the weights are moving at the same speed, we can say that dx/dt = dy/dt = v. Plugging this into the equation, we get:

2x * v + 2y * v = 0