A uniform cylinder of mass `m` and radius `R` starts descending at a moment `t=0` due to gravity, Neglecting the mass of the thread, find
(a) the tension of each thread and the angular acceleration of the cylinder,
(b) the time dependence of the instantaneous power developed by the gravitational force.

Explanation

(a) Let us indicate the forces and their points of application for the cylinder. Choosing the positive direction for x and `varphi` as shown in figure, we write the equation of motion of the cylinder axis and the equation of moments in the C.M. frame relative to that axis i.e. from equation `F_x=mw_c` and `N_z=I_cbeta_z`.
`mg-2T=mw_c`, `2TR=(mR^2)/(2)beta`
As there is no slipping of thread on the cylinder
From these three equations
`T=(mg)/(6)=13N`, `beta=2/5g/R=5xx10^2rad//s^2`
(b) we have `beta=2/3g/R`
So, `w_c=2/3ggt0` or, in vector form `vecw_c=2/3vecg`
`P=vecF*vecv=vecF*(vecw_ct)`
`=mvecg*(2/3vecg t)=2/3mg^2t`