Engineering Solving Moment Equations with Newton's 2nd Law: Help Needed!

[Pipsqueakalchemist]

Homework Statement

Question and attempt below

Relevant Equations

Newton's 2nd law
Moment equation

So I set up 3 equation for this problem. 1st was the moment equation about point G, 2nd and 3rd were from applying Newton's 2nd law to each of the blocks. I thought once I set those equations up I could solve for alpha (angular acceleration) and then find acceleration of each block but when i plug in the numbers I get the wrong answers. Did I do something wrong?

 

[vela]

You made a sign error in the equation for one of the blocks. Remember if one accelerates up, the other one accelerates down.

 

[Pipsqueakalchemist]

Oh i see thanks. I also had another question. Say I took a different point rather than G, say A, the point of the right rope in contact with the circle. Then the alpha (angular acceleration) would be zero when I set up my moment equation right bc the circle isn’t rotating about point A. Is that correct?

 

[vela]

I'm not entirely sure how the problem works out with that choice. My inclination is to think there would be two angular accelerations, the usual one pertaining to the pulley's rotation about its center of mass and one describing the motion of the center of mass of the pulley about point A. The latter one would be zero.

EDIT:

If you choose A as the origin, the angular momentum of the pulley is given by
$$L_A = Rmv_{\rm cm} + I_G\omega = (mR^2)\omega_{\rm cm} + I_{\rm cm}\omega$$ where $\omega$ is the usual angular velocity about the center of mass and $\omega_{\rm cm}$ describes the motion of the center of mass about A. That term vanishes because the pulley doesn't rotate about A, but if we keep it around a moment, differentiating would give
$$\tau_A = \frac{dL_A}{dt} = (mR^2)\alpha_{\rm cm} + I_G\alpha.$$ The fact that the pulley doesn't rotate about A tells us $\alpha_{\rm cm} = 0$ but it doesn't say anything about the usual rotation of the pulley about its axis.

If you calculate the torque on the pulley about A, taking into account the weight and the reaction force from the support, you should end up with the same net torque as before. So in the end, you get the same equation as when you use G as the origin.

 

[Lnewqban]

... I thought once I set those equations up I could solve for alpha (angular acceleration) and then find acceleration of each block but when i plug in the numbers I get the wrong answers. Did I do something wrong?

I believe that a simpler approach to this problem could help.
The system of multiple weights and strings and pulleys create a unique counter-clockwise torque or moment about the actual axis of rotation named G in the diagram.

That applied torque has to overcome the rotational inertia of the pulley's mass in order to establish an increasing rotational velocity at a constant rate or angular acceleration, increasing the angular momentum of the pulley, as well as the linear momentum of each of masses A and B.

Since we have the values of the radius of gyration and mass of the pulley, we can calculate its moment of inertia (I).
If we divide the applied moment and I, we can calculate the value of the angular acceleration of the pulley, and therefore, the linear accelerations of A and B.

Please, see:
https://en.m.wikipedia.org/wiki/Radius_of_gyration

https://courses.lumenlearning.com/p...mics-of-rotational-motion-rotational-inertia/