# Quick ? about gravitational torque

1. Nov 12, 2008

### bocobuff

1. The problem statement, all variables and given/known data
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*R2 for cylinder = 0.01845kg*m2.
I know angular accel=Torquenet/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 Torquestring=-6.769N*0.131m=-0.8867N*m

Now for the Torquegrav=-MgXcm 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 Torquegrav because the axle/bearing is already on the center of mass? Or will Xcm=1 so Torquegrav=-Mg?

2. Nov 12, 2008

### Redbelly98

Staff Emeritus
That's right. Fgrav can be treated as acting at the center of mass ... which is located at exactly 0 m from the center of mass.