A 5.0 kg \rm kg, 52-cm \rm cm-diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released.

Question

a)What is the magnitude of the cylinder's initial angular acceleration

My work:
I found the moment of inertia from the parallel axis theorem:
I=I_cm+Md^2
I=(1/2)MR^2+Md^2
(plugging in the #'s you get...
I=0.507kgm^2
torque=Ix angular acceleration
I_grav=-Mgx_cm

don't know what my center of mass is?
How do I find the torque so I can isolate for angular acceleration?

Answer 1

Torque = Ix = rF
T = -mgx_cm
The x_cm is the distance it was moved. Since it is on the edge this would just be -r. Therefore,
Ix = mgr
x= (mgr/I)
x= [(5kg)(9.8 m/s^2)(0.26 m)] / (0.507 kgm^2)

x= 25.128 rad/s^2