Translation Converted To Rotation

Balancing forces on the mass $m$ falling under influence of gravity requires: $mg-T = ma \implies T = mg-ma$ where $a$ is the acceleration of $m$ and $T$ is the tension of the string pulling upwards on $m$ .

This tension force is also what causes the cylinder to spin. It exerts a force perpendicular to the radius of the cylinder, causing a torque: $\tau = TR = I\alpha = \left(\frac12 MR^2\right)\alpha$ Since the string presumably does not slip from the cylinder, the acceleration of the mass (and therefore the string) downwards is $a = R\alpha$ where $\alpha$ is the angular acceleration. Equating the torques and substituting in for $\alpha$ : $(mg-ma)R=TR = \left(\frac12 MR^2\right)\frac{a}{R} = \frac12 MRa.$ Now solving for the acceleration yields: $aR(\frac12 M + m) = mgR \implies a = \frac{mg}{\frac{M}{2} + m} = \frac{2mg}{M+2m}.$ Finally, since this acceleration is the tangential acceleration of the cylinder, the angular acceleration is recovered by dividing by $R$ . The angular velocity at time $t$ is just this angular acceleration times $t$ : $\omega = \frac{2mgt}{R(M+2m)}.$

One can solve this by excluding procedure. Answers 2 and 3 are false because of wrong measurement unit. Answer 4 can not be correct since for M=2m angular velocity would become infinitely and for 2m greater than M even negative! Therefore take no. 1.

Let block is moving with accn = a
Tension = T Now mg-T= ma --eq (1) Torque = I.(linear accn) Now R×T=I.(a/R) eq-----2 Put eq 2 in 1 a = 2mg/(2m+M) eq 3

Use conservation of momentum mgh= 1/2Iw^2 + 1/2mv^2 And put h= 1/2.at^2

Hence we ll get the reqd ans.