Atwood's machine with angular momentum

[hitemup]

Homework Statement

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An Atwood's machine consists of two masses, m_A and m_B, which are connected by an inelastic cord of negligible mass that passes over a pulley. If the pulley has radius R₀ and moment of inertia I about its axle, determine the acceleration of the masses m_A and m_B.

Homework Equations

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L = mvr
T = dL/ dt

The Attempt at a Solution

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This is an example from my textbook. The solution involves:

angular momentum about the axle = (m_A +m_B)vR₀ + Iv/_R0
torque about the axle = m_BgR₀ - m_AgR₀

Then it uses the equation T = dL/dt and finds accelerations of the masses.

What confuses me is the way it calculates the torque. When we were solving this exact same problem in rotational motion, the equations were like these:

m_B*g - F_TB = m_B*a
F_TA - m_A*g = m_A*a
(F_TB - F_TA )*r = I*a/r

How can angular momentum solution get rid of tensions in the torque equation?

 

[BvU]

The torque equation m_BgR₀ - m_AgR₀ is not correct.
At the pulley you have net torque = T_A - T_B

At m_A you have T_A - m_A g R = m_A a_A
At m_B you have T_B - m_B g R = m_B a_B
since a_B = - a_A the differerence has an extra term ( m_A + m_B ) a_A

Yuo can also check that this gives the right relationship for a massless pulley, where the torque difference at the pulley must be zero.

 

[hitemup]

The torque equation m_BgR₀ - m_AgR₀ is not correct.
At the pulley you have net torque = T_A - T_B

At m_A you have T_A - m_A g R = m_A a_A
At m_B you have T_B - m_B g R = m_B a_B
since a_B = - a_A the differerence has an extra term ( m_A + m_B ) a_A

Yuo can also check that this gives the right relationship for a massless pulley, where the torque difference at the pulley must be zero.

I've uploaded the picture.

 

[BvU]

Can't argue with that, can we :) ?
Book calls that the external torque. I'm using a simpleton approach looking at the upper half and the lower half separately. So my net torque is on the wheel only and it gives the wheel an angular acceleration Iα. Book's external torque gives an angular acceleration to the whole lot: ( (m_A + m_B) R₀² + I ) α

So fortunately we end up doing the same thing to calculate the same α. As we should ;)

 

[BvU]

So, for the record: I was a bit hasty stating "The torque equation is not correct" . Depends on what torque is meant.

And the answer to your original question is now clear, I hope: the book solution can avoid the tensions by considering them internal. In exchange the angular momentum then has to be for the whole system.