First slide

Torque

Question

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

Difficult

A

$2 \times 10^{6} N m$
B

$2 \times 10^{- 6} N m$
C

$2 \times 10^{- 3} N m$
D

$12 \times 10^{- 4} N m$

Solution

$Given : Mass M = 2 kg, Radius R = 4 cm Initial angular speed ω_{0} = 3 rpm = 3 \times \frac{2 π}{60} rad / s = \frac{π}{10} rad / s We know that, ω^{2} = ω_{0}^{2} + 2 αθ \Rightarrow 0 = {(\frac{π}{10})}^{2} + 2 \times α \times 2 π \times 2 π \Rightarrow α = \frac{- 1}{800} rad / s^{2} Moment of inertia of a solid cylinder, I = \frac{{MR}^{2}}{2} = \frac{2 \times {(\frac{4}{100})}^{2}}{2} = \frac{16}{10^{4}} Torque τ = Iα = (\frac{16}{10^{4}}) \times (- \frac{1}{800}) = - 2 \times 10^{- 6} Nm$

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