Dear Pranay

Plz see the pic for solution.

All the best.

AKASH GOYAL

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Approved

net torque on the shaft is given...

now we can find net angular accleration of the system by

         T=I(alfa)                         (T -torque,alfa - angura accleration and I is total moment of inertial of system )

 alfa = T/I .........1

moment of inertial of any of cylender about the axis of rotation is I₁

 I₁ = I(about its own axis) + I(about axis of rotation)

I₁ = ma² /2 + md²                     ( a is radius of cylender and r is perpendicular distance from axis of rotation)

  I

Approved

I posted the comlete solution but i dont know why remaining part is not appearing ...

here is the remaining solution remaining solution :

                  I_{1 = I (about axis of rotation) + I(abot its own axis)}

                  I_{1 = mr² + ma² /2                          ( Icom =ma² /2 )}

  total moment of inertia is 4I₁

         I = 4I_{1 = 4mr² + 2ma²}     

for constant torque we can use this eq

                     W =W_{0 + (alfa)t                             (alfa is angular accleration)}

                t=W/alfa                         (w_{0  is 0 )}

                t = WI/T

               t = 2m(a_{² + 2r² )W/T}