Question

A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions $$s^{-2}$$ is :-

A
25 N
B
50 N
C
78.5 N
D
157 N

Solution

The correct option is D 157 N$$I_{Cylinder} = \dfrac{MR^2}{2}$$ $$= \dfrac{50 \times \left (\dfrac{1}{2} \right )^{2}}{2}$$ $$= \dfrac{50 \times 1}{2 \times 4} = \dfrac{50}{8}\ kgm^{2}$$ Torque due to tension $$= TR$$$$z = I \alpha$$ $$TR = \left (\dfrac{50}{8} \right ) \times 2$$ $$T \times \dfrac{1}{2} = \dfrac{50}{4}$$$$\boxed{T = 25\ N}$$  $$\alpha = \dfrac{2 rev}{5^2} = \dfrac{2 \times (2 \pi)}{5^2}$$ rad $$1$$ revolution $$= 2 \pi$$ rad.$$\alpha = 4 \pi\ rad/s^{2}$$ $$z = I \alpha$$ $$T \times \left (\dfrac{1}{2} \right ) = \left (\dfrac{50}{5} \right ) \times 4 \pi = 50 \times \pi = 157.07\ N$$ $$\boxed{T = 157\ N}$$ Physics

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