Question

If $484J$ of energy is spent in increasing the speed of a wheel from $60rpm$ to $360rpm$, the $M.I$ of the wheel is

• A. $1.6kg{m}^2$
• B. $0.3kg{m}^2$
• C. $0.7kg{m}^2$
• D. $1.2kg{m}^2$

Answer 1

Answer: (C)

$0.7kg{m}^2$

According to the data given in the question,

$\begin{array}{c}{\eta }_1=60rpm=\frac{60}{60}rps=1rps\\ \therefore w_1=2r{\eta }_1=2\pi \\ {\eta }_2=\frac{360}{60}rpm=6rps\\ w_2=2\pi {\eta }_2=12\pi \end{array}$

484 joule of energy is spent in increasing.

$\begin{array}{c}\frac{1}{2}I{w}_2^2-\frac{1}{2}I{w}_1^2=484\\ \frac{I}{2}\left(w_2+w_1\right)\left(w_2-w_1\right)=484\\ \frac{I}{2}\left(12\pi +2\pi \right)\left(12\pi -2\pi \right)=484\\ I\left(14\pi \right)\left(10\pi \right)=484\\ I=\frac{484\times 2}{140\times {\pi }^2}=\frac{484\times 2}{140\times 9.86}\\ =0.7kg{m}^2\end{array}$

Hence, option C is correct.