Get Answers to all your Questions

header-bg qa

A wheel is rotating freely with an angular speed \omega on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible . Another wheel of moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is :
Option: 1 \frac{5}{6}
Option: 2 \frac{1}{4}
Option: 3 0
Option: 4 \frac{3}{4}

Answers (1)

best_answer

\begin{aligned} &\text { By anglar momentum conservation }\\ &\omega I+3 I \times 0=4 I \omega^{\prime} \Rightarrow \omega^{\prime}=\frac{\omega}{4}\\ &(\mathrm{KE})_{\mathrm{i}}=\frac{1}{2} \mathrm{I} \omega^{2}\\ &(\mathrm{KE})_{\mathrm{f}}=\frac{1}{2} \times(4 \mathrm{I}) \times\left(\frac{\omega}{4}\right)^{2}=\frac{\mathrm{I} \omega^{2}}{8}\\ &\Delta \mathrm{KE}=\frac{3}{8} \mathrm{I} \omega^{2}\\ &\text { fractional loss }=\frac{\Delta \mathrm{KE}}{\mathrm{KE}_{1}}=\frac{\frac{3}{8} \mathrm{I} \omega^{2}}{\frac{1}{2}{\mathrm{I} \omega^{2}}}=\frac{3}{4} \end{aligned}

Posted by

Deependra Verma

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE