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.
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!