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Secondly, I am Swedish but I will try to use the correct scientific terms in explaining the problem and my proposed solution.

## Homework Statement

There is a string of negligible mass which "hangs" over a pulley attached to a frictionless horizontal axis. On each end of the string there is a attached a weight. One of the weights (i've chosen the one on the right hand side if drawn the "normal way") is twice the mass of the weight of the other.

The system starts out at rest.

[pulley] = Radius R, and Moment of Inertia I = (1/5)*m*R².

[weights] = A = m and B = 2m.

[string tension] = left side (A): T ; right side (B): S.

[gravity] = As always g denoted the pull of the earth (9.82 m/s² here in Sweden).

When the weights are released from rest (and left to fall freely), what will be the tension on the string on the two sides of the pulley?

## Homework Equations

In my attempted solution below I used:

Newtons II equation where the sum of the forces are equal to the mass times the acceleration (F = m*a).

I also used Eulers II law for momentum (torque) which in this case would be that the momentum is equal to the moment of inertia times the angular acceleration (M = I*alpha)

I also used that alpha = R * a (along the x-axis (which I set to be positive upward))

## The Attempt at a Solution

So, as listed above I will call the momentum about the z-axis of the pulley M, the angular acceleration alpha. I called the "left" pulley A and the "right" pulley B.

I set the x-axis to the positive in the upward direction. And I assumed that the string was not elastic in any way.

First I used newtons second law to describe the forces acting on pulley A to be

F = T - mg = m * a

and respectively for B:

F = S - 2mg = m * a

Then I used Eulers II with the moments origo in the center of the pulley to be

M = RT - RS = I * alpha (and since alpha = R * a) = T-S = I * a.

I then simply solved the equations using both NII and EII to get my expressions for what the tensions on S and T are.

I got that T =m(a((R²/5)-1)+2g)

and S = m(a(1-(R²/5))+g)

It should be added I am at the moment of typing a little tired. Also I should probably have made a picture, but I hope my explanation of the problem will be enough.

Usually I would be satisfied with my answer but the "hint" to the question was that I could use that acceleration (dv/dt) could be written as v*dv/ds (using the chain rule)! Why and how could that be used? I am now worried that I have made some very basic assumption or something wrong. Anyways help much appreciated!