Example Problem in the book, why is tension ignored?

[jzhu]

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!

 

[Doc Al]

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

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

The book is treating the two masses and pulley as a single system. The tensions in the string would be internal to the system, so they can be ignored.

You can certainly break up the problem and treat the two masses and pulley separately, getting three equations. Then you can solve for the acceleration that way.

 

[jzhu]

Hm, I just solved it using t = Ia with the 2 forces. So does does mean that dl/dt can only be used from an inertial frame of reference? Or do dl/dt = ia?

 

[Doc Al]

Hm, I just solved it using t = Ia with the 2 forces.

If you mean the two tensions, then that's fine. ΣT = Iα

So does does mean that dl/dt can only be used from an inertial frame of reference? Or do dl/dt = ia?

If you let L = the angular momentum of the pulley only, then dL/dt will equal Iα.

 

[jzhu]

Ok, thanks!