I don't know what this is about but the torque providing the tangential acceleration is provided by the falling weight Mg times the radius R.
That torque is equal to I(alpha) where alpha is angular acceleration. I for a solid disc is 1/2mR^2 (I think, might want to check on me). So:
MgR = 1/2mR^2(alpha). therefore alpha = 2Mg/mR. The tangential acceleration =R(alpha) which is 2Mg/m. How you would derive this some other way I don't know.
A homogeneous disk of mass m and of radius r free to rotate without friction about its horizontal axis, is used used as a pulley.
An extensible string of negligible mass is wound around this pulley and a mass M is connected to its free end. When the system is released the center of inertia G of mass M moves downward with an acceleration and the pulley thus aquires an angular acceleration O".
At any date t the velocity of G is equal to that of any point C on the periphery of the pulley (sincethe string ddoesn't slide over the pully). Show that the acceleration a of G is equal to the tangential acceleration of C .
Deduce a relation between a and O"
1 answer
1 answer