Question

A light thread with a mass m tied to its end is wound over a uniform solid cylinder of radius R
and mass M and the system is set into motion at t = 0. Obtain the angular velocity of the cylinder.

Answer 1

Answer on Question #40696, Physics, Mechanics | Kinematics | Dynamics

A light thread with a mass $m$ tied to its end is wound over a uniform solid cylinder of radius $R$ and mass $M$ and the system is set into motion at $t = 0$ .

Obtain the angular velocity of the cylinder.

Solution:

Method 1:

Sum of the torques is equal to mass moment of inertia $(I)$ times angular acceleration $\alpha$

$$m g R = \left(\frac {M R ^ {2}}{2} + m R ^ {2}\right) \alpha$$

So,

$$\alpha = \frac {2 m g}{\left(\frac {M}{2} + m\right) R}$$

The angular velocity is

$$\omega = \alpha t = \frac {2 m g t}{\left(\frac {M}{2} + m\right) R} = \frac {g t}{\left(1 + \frac {M}{2 m}\right) R}$$

Method 2:

The equation of motion is

$$m g - T = m a$$ $$M = I \alpha$$

where torque $M = TR$ , momentum of inertia $I = \frac{MR^2}{2}$ , linear acceleration $a = R\alpha$ .

From these equations we obtain:

$$T R = \frac {M R ^ {2}}{2} \alpha$$ $$T = \frac {M R}{2} \alpha$$ $$m g - \frac {M R}{2} \alpha = m R \alpha$$

For angular acceleration

$$\alpha = \frac {2 m g}{\left(\frac {M}{2} + m\right) R}$$

The angular velocity is

$$\omega = \alpha t = \frac {2 m g t}{\left(\frac {M}{2} + m\right) R} = \frac {g t}{\left(1 + \frac {M}{2 m}\right) R}$$

Answer. $\omega = \frac{gt}{\left(1 + \frac{M}{2m}\right)R}$