- #1

jzhu

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## Homework Statement

An Atwood machine consists of two masses, M and m, which are connected by an inelastic cord of negligible mass that passes over a pulley. If the pulley has radius R and moment of inertia I about its axle, determine the acceleration of the masses M and m.

## Homework Equations

torque = dL/dt

L(angular momentum) = R x v

L = Iω

## The Attempt at a Solution

This is the solution in the book. I have no idea why tension of the string is ignored.

L = (m + M)vR + I(v/r) <- this part makes sense to me

torque = mgR - MgR <- this part does not make sense to me

Following the Atwood machine, shouldn't the forces for both m and M be something like

ΣF = F(tension) - mg = ma (differing signs depending on which is going down, of course)[/B]

**Instead, the book has it as ΣF = mg**

So then since torque is equal to RF, they get that torque = mgR

But my idea is that torque in this case is equal to RF(tension of m) - RF(tension of M)

which would mean that

torque = R(ma + mg) - R(mg - ma)

After this, we just plug torque into the t = dL/dt equation, and with the factored out velocity that we get from the total angular momentum, get acceleration from dv/dt and simple algebra reveals that the answer is a = (m-M)g/(m+M)+I/R^2.

Any help would be appreciated!

So then since torque is equal to RF, they get that torque = mgR

But my idea is that torque in this case is equal to RF(tension of m) - RF(tension of M)

which would mean that

torque = R(ma + mg) - R(mg - ma)

After this, we just plug torque into the t = dL/dt equation, and with the factored out velocity that we get from the total angular momentum, get acceleration from dv/dt and simple algebra reveals that the answer is a = (m-M)g/(m+M)+I/R^2.

Any help would be appreciated!