In the arrangement shown in figure the mass of the uniform solid cylindrical pulley of radius `R` is equal to `m` and the masses of two bodies are equal to `m_(1)` and `m_(2)`. The thread slipping and the friction in the axle of the pulley are supposed to be absent. Find the angular acceleration of the cylinder and the ratio of tensions `(T_(1))/(T_(2))` of the vertical sections of the thread in the process of motion.

Explanation

Let `alpha=` angular acceleration of the pulley and `a=` linear acceleration of two bodies
Equations of motion are
for mass `m_(1)` `T_(1)-m_(1)g=m_(1)a` ..(i)
for mass `m_(2)` `m_(2)g-T_(2)=m_(2)a` ..(ii)
for pulley `alpha=((T_(2)-T_(1))R)/((1)/(2)mR^(2))`
For no slipping condition `a=Ralpha` ..(iv)
Solving these equation, we get `alpha=(2(m_(2)-m_(1))g)/((2m_(1)+2m_(2)+m)R)`
and `(T_(1))/(T_(2))=(m_(1)(m+4m_(2)))/(m_(2)(m+4m_(1)))`