Question

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 :

• A. $0.7\, kg - m ^2$
• B. $3.22\, kg - m ^2$
• C. $30.8\, kg - m ^2$
• D. $0.07\, kg - m ^2$

Answer 1

Answer: (A)

$\omega_{ i }=60\, rpm =60 \times \frac{2 \pi}{60}=2 \pi \, rad / s$
$\omega_{ f }=360 \, rpm =360 \times \frac{2 \pi}{60}=12 \pi \, rad / s$
$\Delta K . E .=\frac{1}{2} I \left(\omega_{ f }^2-\omega_{ i }^2\right)=484$
$\frac{1}{2} I \left(144 \pi^2-4 \pi^2\right)=484$
$I \approx 0.7\, kg - m ^2$