Question

A light string is wrapped on a cylindrical shell and fraction of length of string is unwrapped. A particle of mass 2M is
attached on another end of string as shown. The system is kept in a vertical plane and cylinder can freely rotate about the axis of cylinder. Particle is released as shown in figure. Assuming thee is no slipping between cylinder and string, find angular velocity and velocity of cylinder just after string is taut.
​

Answer 1

Answer:

τ=Rα

TR=Iα

For  solid cylinder   I=21MR2             and   a=rα   (no slipping)

⟹T=21Ma

Equation of motion,    Ma=Mg−T

2T=Mg−T⟹T=3Mg

Explanation:

hope it will help you out

Answer 2

Answer:

angular velocity = $w= \frac{14*g}{R}$

Explanation:

Torque, $\tau = Iw$ ......(1);

where,  $I$ is the moment of inertia

           $w$ is angular velocity

also, $\tau = F.R$ .....(2);

where, R is the perpendicular distance of the axis of rotation and the acting force

For cylindrical shell =  $I = M.R^2$

at the moment when the string has unwrapped completely, the force that the object will apply on side of the cylinder is, $F=Mg$

At the point of time when the string is taut, equating (1) and (2),

$\implies M.R^2.w=2M.g.7R$

$\implies w=\frac{14g}{R}$

[NOTE: force on the cylinder can be calulated in other ways also e.g. using change in momentum per unit time, but the answer remains the same.]

                                                                                                               #SPJ3