Angular Acceleration and Linear Acceleration of a Pulley

[EchoTheCat]

Homework Statement

There are two blocks, m1 and m2, that are hanging from a rope which passes over a pulley.
The mass of the pulley is 0.15 kg.
The radius of the pulley is 0.055 m.
The mass of m1 is 0.165 kg.
The mass of m2 is 0.18 kg.
The linear acceleration is m1 is 0.3504 m/s/s downward, and the linear acceleration of m2 is 0.3504 m/s/s upward.
What is the magnitude of the angular acceleration of the pulley?

Homework Equations

The tension on mass 1 is m1g-m1a.
The tension on mass 2 is m2a+m2g.
T1-T2 = 1/2 mp * a * r.

The Attempt at a Solution

m1g-m1a - (m2a+m2g) = 1/2 * mp * a * r
0.165 (9.81 - 0.3504) - 0.18(0.3505+9.81) = 1/2 * 0.15 * a * 0.055
a = 64.98 rad/s/s

 

[EchoTheCat]

Ok, so I was able to solve for
(m1g+m1a)R – (m2g-m2a)R = -(1/2mr2)a, so, a = 6.37 rad/s/s.

 

[tnich]

Ok, so I was able to solve for
(m1g+m1a)R – (m2g-m2a)R = -(1/2mr2)a, so, a = 6.37 rad/s/s.

I get the same numerical result, but I can't get there using your equation. I am not sure how to parse -(1/2mr2)a. Do you mean $-(\frac 1 2 m_p r^2) α$? It looks like you might be using the moment of inertia of the pulley multiplied by the angular acceleration to get the torque. And that could work if you knew that the pulley was a perfect disc. However, when I plug in the numbers I don't get the 6.37 $\frac {rad} {s^2}$.
Here is another approach: the angular acceleration α must be equal to the linear acceleration of the cable divided by the radius of the pulley.

 

[tnich]

I get the same numerical result, but I can't get there using your equation. I am not sure how to parse -(1/2mr2)a. Do you mean $-(\frac 1 2 m_p r^2) α$? It looks like you might be using the moment of inertia of the pulley multiplied by the angular acceleration to get the torque. And that could work if you knew that the pulley was a perfect disc. However, when I plug in the numbers I don't get the 6.37 $\frac {rad} {s^2}$.
Here is another approach: the angular acceleration α must be equal to the linear acceleration of the cable divided by the radius of the pulley.

I see I made a mistake in my calculations, so my numbers now agree with yours. I think you have the right answer. Now that you have solved the problem that way, try the approach I outlined above. I think you will see that you get the same answer and you don't have to assume anything about the distribution of mass in the pulley.

 

[gneill]

The mass of m1 is 0.165 kg.
The mass of m2 is 0.18 kg.
The linear acceleration is m1 is 0.3504 m/s/s downward, and the linear acceleration of m2 is 0.3504 m/s/s upward.

How is it that the lighter mass is accelerating downwards?

Edit: You should be aware that there is a simple relationship between angular acceleration and the linear acceleration of the pulley's rim that doesn't involve any involvement of the masses, torques, etc. Do you know what it is?