An energy of 484 ~J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm . The moment of inertia of the flywheel is:

Correct answer: A. 0.7 ~kg-m^2 Given, aligned & ω_1=60 rpm=60 × (2 π)/(60)=2 π rad / s & ω_2=360 rpm=360 × (2 π)/(60)=12 π rad / s aligned Change in kinetic energy, aligned Δ K . E & =(1)/(2) I(ω_2^2-ω_1^2) 484 & =(1)/(2) × I(ω_2+ω_1)(ω_2-ω_1) I & =(484 × 2)/(14 π × 10 π)=0.7 ~kg · m^2 aligned (where, the symbols have their usual meanings)

NEET 2022

Exam: NEET
Year: 2022
Subject: Physics
Chapter: Rotational Motion

NEET 2022 · Physics · Rotational Motion

An energy of 484 ~J is spent in increasing the speed of a flywheel...

An energy of

484 \mathrm{~J}

is spent in increasing the speed of a flywheel from

60 \mathrm{rpm}

360 \mathrm{rpm}

. The moment of inertia of the flywheel is:

Options

A $0.7 \mathrm{~kg}-\mathrm{m}^2$
B $3.22 \mathrm{~kg}-\mathrm{m}^2$
C $30.8 \mathrm{~kg}-\mathrm{m}^2$
D $0.07 \mathrm{~kg}-\mathrm{m}^2$

Correct answer

A. $0.7 \mathrm{~kg}-\mathrm{m}^2$

Step-by-step solution

Given,

\begin{aligned} & \omega_1=60 \mathrm{rpm}=60 \times \frac{2 \pi}{60}=2 \pi \mathrm{rad} / \mathrm{s} \\ & \omega_2=360 \mathrm{rpm}=360 \times \frac{2 \pi}{60}=12 \pi \mathrm{rad} / \mathrm{s} \end{aligned}

Change in kinetic energy,

\begin{aligned} \Delta \mathrm{K} . \mathrm{E} & =\frac{1}{2} \mathrm{I}\left(\omega_2^2-\omega_1^2\right) \\ 484 & =\frac{1}{2} \times \mathrm{I}\left(\omega_2+\omega_1\right)\left(\omega_2-\omega_1\right) \\ \Rightarrow \quad \mathrm{I} & =\frac{484 \times 2}{14 \pi \times 10 \pi}=0.7 \mathrm{~kg} \cdot \mathrm{m}^2 \end{aligned}

(where, the symbols have their usual meanings)

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