Physics · Rotational Dynamics
JEE Main 2026 — 28 January, Evening Shift — Question 50
A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is J.
Answer: 30
Numerical answer — enter this value.
Step-by-step solution
\mathrm{mg} \times 3=\frac{1}{2} \cdot \frac{\mathrm{mR}^{2}}{2} \omega^{2}+\frac{1}{2} \mathrm{mv}^{2}
\end{gathered}$$
$$\begin{gathered}
\& \mathrm{v}=\omega \mathrm{R}
\end{gathered}$$
From equation (i) \& (ii)
$\mathrm{g} \times 3=\frac{3}{4} \cdot \mathrm{v}^{2}$
K.E. of flywheel $=\frac{1}{2} \times \frac{\mathrm{mR}^{2}}{2} \times \omega^{2}=\frac{1}{4} \mathrm{mv}^{2}$
$=\frac{1}{4} \times 3 \times 40=30$ Joule
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2026
- Subject
- Physics
- Chapter
- Rotational Dynamics
- Topic
- Work-Energy Theorem in General Motion