First slide

Torque

Question

A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the 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 rev $s^{- 2}$ is

Moderate

A

25 N
B

50 N
C

78.5 N
D

157 N

Solution

$Tr = Iα$
Applying torque equation about center of cylinder
$τ = Iα$
$α$ is the angular acceleration of cylinder and it is given
$α = 2 revolution / s^{2} = 2 \times 2 π = 4 π rad / s^{2}$
$T = \frac{Iα}{r} = \frac{{mr}^{2}}{2} \times \frac{α}{r} = \frac{mrα}{2}$
= $\frac{50 \times 0.5 \times 4 π}{2} N = 157 N$

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