JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and Rotational Equilibrium is :
- A \(\frac{3}{4}\)
- B \(\frac{4}{3}\)
- C \(\frac{5}{2}\)
- D \(\frac{2}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{KE}_{(T)}=\frac{1}{2} m v^2 \\ & \mathrm{KE}_{(R)}=\frac{1}{2} \cdot \frac{2}{5} m R^2 \cdot \frac{v^2}{R^2}=\frac{1}{2} m v^2\left(\frac{2}{5}\right) \end{aligned}\) So, \(\frac{\mathrm{KE}_{(T)}}{\mathrm{KE}_{(R)}}=\frac{5}{2}\)
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