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A rope is wound around a hollow cylinder of mass 3kg and radius 40cm is free to rotate about its geometrical axis. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N ?

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$25 {rads}^{- 2}$

$2.5 {rads}^{- 2}$

$0.25 {rads}^{- 2}$

$5 {rads}^{- 2}$

answer is B.

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Detailed Solution

Here, torque $(τ) = l α \to (l)$

where I = moment of inertia, here I = MR².

$α$ = angular acceleration.

$τ = R \times F$

Then, from (1)

$α = \frac{τ}{I} = \frac{R F}{M R^{2}} = \frac{F}{M R} = \frac{30}{3 \times 0.4} = 25 rad / s^{2}$

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