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A string wrapped around a cylinder(M=2.15kg, R=0.131m) on a frictionless thin bearing pulls downward with F=6.769N. Also, the string is on the right side of the cylinder. What is angular acceleration?

I know moment of Inertia=1/2*M*R^{2}for cylinder = 0.01845kg*m^{2}.

I know angular accel=Torque_{net}/I

So the torque of the string pulling down will be negative because the cylinder is rotating cw and the string is vertical so sin90=1 which means Torque_{string}=-6.769N*0.131m=-0.8867N*m

Now for the Torque_{grav}=-MgX_{cm}where cm=center of mass, will the Xcm be equal to the radius and positive because it is to the right of the axle/bearing? Or will there not even be a Torque_{grav}because the axle/bearing is already on the center of mass? Or will Xcm=1 so Torque_{grav}=-Mg?

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# Homework Help: Quick ? about gravitational torque

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