Angular acceleration of meter stick after removing one mass

[Nm]

Two masses (mA= 2 kg, mB= 4 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks. The system is then hung from a string, so that it stays horizontal. Now, if mass B was removed, and no additional force was supplied, calculate the size of the angular acceleration of the meter stick at that instant.

I calculated the center of mass to be at the 50 cm mark, but I don't know how to get the angular acceleration. Do I just use +rF/mr^2 = angular acceleration? I got the wrong answer form that. This seems like a simple problem, but maybe I'm thinking too much..

 

[Tide]

I think what you're looking for is the relationship between torque and angular acceleration. Here it is: torque = moment of inertia X angular acceleration.

 

[Nm]

I used that relationship to get inertia of a point mass to equal mr^2 and the net torque to equal rFsin90. From the two equations I got angular acceleration to equal F/mr. Am I really far off because I don't know the force...?

 

[Tide]

The force is the weight of the suspended object!

 

[Nm]

If the weight is mg, the angular acceleration would be 9.8m/s^2/(2kg)(0.75m)?

 

[Tide]

Don't forget the force holding the meter stick in place!