Question

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

• A. $0.2kg$$m^2$
• B. $0.7kg$$m^2$
• C. $2kg$$m^2$
• D. $3kg$$m^2$

Answer 1

Answer: (B)

$0.7kg$$m^2$

Initial speed of flywheel ${\nu }_1=60$ $rpm$$=\frac{60}{60}=1$ $rps$

Final speed of flywheel ${\nu }_2=360$ $rpm$ $=\frac{360}{60}=6$ $rps$

Rotational Kinetic energy $E=\frac{1}{2}I{w}^2$ where $w=2\pi \nu$

$\therefore$ $\Delta E=\frac{1}{2}I(w_2^2-w_1^2)=\frac{1}{2}I\times 4{\pi }^2({\nu }_2^2-{\nu }_1^2)$

$\therefore$ $484=\frac{1}{2}I\times 4\left(3.14{)}^2\right(6^2-1^1)$

$⟹$ $I=0.7$ $kg{m}^2$