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A solid disc of mass 2 kg and radius 0.5 m is rotating about its own geometrical axis at 2 rps. A constant torque $π N - m$ is applied on the disc. Then angular velocity of the disc after 10 sec will be

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18 $π r a d / s$

24 $π r a d / s$

44 $π r a d / s$

34 $π r a d / s$

answer is C.

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Detailed Solution

$I = \frac{1}{2} \times 2 \times {(0.5)}^{2} = \frac{1}{4} K g - m^{2}$

Angular acceleration, $α = \frac{τ}{I} = \frac{π}{1 / 4} r a d / s^{2}$

$\Rightarrow α = 4 π r a d / s^{2}$

$∴ ω = ω_{0} + α t = (2 \times 2 π + 4 π \times 10) r a d / s = 44 π r a d / s$

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