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A small thin disk of radius r and mass m is attached rigidly to the face of a second thin disk of radius R and mass M. The centre of the small disk is located at the edge of the large disk. The large disk is mounted on its centre of a frictionless axle. the assembly is rotated through a small angle theetre from its equilibrium position and then released. Show that the speed of the small disk as it passes through the equilibrium position is
v=2{Rg(1cos theetre)/[ M/m+ (r/R)^2+2]}^0.5
and the period is
T=2pi {[(M+2m)R^2+mr^2]/[2mgR]}^0.5
I am stuck in the first part already.
My method was: Since the axis of rotation is through the COM of the larger disk, hence the nett torque is mgRsin theetre.
This torque is equal to I alpha which is equal to (1/2MR^2+mR^2) alpha.
angular aceleration is not constant so i must carry out an integration when using the eqn V^2=U^2+2aS
i have been struggling with this prob. for a long time . pl help me.
v=2{Rg(1cos theetre)/[ M/m+ (r/R)^2+2]}^0.5
and the period is
T=2pi {[(M+2m)R^2+mr^2]/[2mgR]}^0.5
I am stuck in the first part already.
My method was: Since the axis of rotation is through the COM of the larger disk, hence the nett torque is mgRsin theetre.
This torque is equal to I alpha which is equal to (1/2MR^2+mR^2) alpha.
angular aceleration is not constant so i must carry out an integration when using the eqn V^2=U^2+2aS
i have been struggling with this prob. for a long time . pl help me.
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