JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is \(\frac{x}{5}\). The value of \(x\) is _______.
- A \(2\)
- B \(5\)
- C \(4\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\frac{\frac{1}{2} \mathrm{I} \omega^2}{\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} m v^2}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3} m R^2\right) \omega^2}{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^2\right) \omega^2+\frac{1}{2} \mathrm{~m}(\mathrm{R} \omega)^2}…
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