A flywheel absorbs 24 kJ of energy while increasing its speed from 210 rpm to 214 rpm. What will be its kinetic energy at 220 rpm?

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This question was previously asked in

SJVN ET Mechanical 2019 Official Paper

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Answer 1

Concept:

Kinetic energy of a rotating body is given by,

\( KE = \frac{1}{2} I \omega^2 \)

Since moment of inertia (I) is constant,

\( KE \propto N^2 \)

Given:

A flywheel absorbs 24 kJ of energy when the speed increases from 210 rpm to 214 rpm.

We need to find the kinetic energy at 220 rpm.

Calculation:

Let:

\( KE_1 = k \times (210)^2 = 44100k \)

\( KE_2 = k \times (214)^2 = 45796k \)

Given: \( KE_2 - KE_1 = 24 \) kJ

\( \Rightarrow 45796k - 44100k = 1696k = 24 \Rightarrow k = \frac{24}{1696} = 0.01415 \)

Finding KE at 220 rpm:

\( KE = k \times (220)^2 = 0.01415 \times 48400 = 685.86~\text{kJ} \)