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

First of all, let us sketch free body diagram of each body. Since the cylinder is rotating and massive, the tension will be different in both the sections of threads. From Newton's law in projection form for the bodies `m_1` and `m_2` and noting that `w_1=w_2=w=betaR`, (as no thread slipping), we have `(m_1gtm_2)`
`m_1g-T_1=m_1w=m_1betaR`
and `T_2-m_2g=m_2w` (1)
Now from the equation of rotational dynamics of a solid about stationary axis of rotation. i.e.
`N_z=Ibeta_z`, for the cylinder.
or, `(T_1-T_2)R=Ibeta=mR^2beta//2` (2)
Similtaneous solution of the above equations yields:
`beta=((m_1-m_2)g)/(R(m_1+m_2+m/2))` and `(T_1)/(T_2)=(m_1(m+4m_2))/(m_2(m+4m_1))`