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Quick ? about gravitational torque

  1. Nov 12, 2008 #1
    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. jcsd
  3. Nov 12, 2008 #2

    Redbelly98

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    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.
     
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