A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop it after revolutions is [2019]

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Solution

Q.
A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop it after $2 π$ revolutions is [2019]

1 $2 \times 10^{6} Nm$

2 $2 \times 10^{- 6} Nm$

3 $2 \times 10^{- 3} Nm$

4 $12 \times 10^{- 4} Nm$

Ans.
(2)

Given: Mass M = 2 kg, Radius R = 4 cm

Initial angular speed

$ω_{0} = 3 rpm = 3 \times \frac{2 π}{60} rad/s = \frac{π}{10} rad/s$

We know that, $ω^{2} = ω_{0}^{2} + 2 α θ$

$\Rightarrow 0 = {(\frac{π}{10})}^{2} + 2 \times α \times 2 π \times 2 π \Rightarrow α = \frac{- 1}{800} {rad/s}^{2}$

Moment of inertia of a solid cylinder,

$I = \frac{M R^{2}}{2} = \frac{2 \times {(\frac{4}{100})}^{2}}{2} = \frac{16}{10^{4}}$

Torque $τ = I α = (\frac{16}{10^{4}}) \times (\frac{- 1}{800}) = - 2 \times 10^{- 6} Nm$

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