MCQ
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 kinetic energy is :
- A$\frac{2}{5}$
- ✓$\frac{5}{2}$
- C$\frac{3}{4}$
- D$\frac{4}{3}$
✓
Answer
Correct option: B.
$\frac{5}{2}$
(B) $\frac{5}{2}$
Sol. $\frac{\text { Lincar KE }}{\text { Rotational K.E }}=\frac{\frac{1}{2} mv _{ cn }^2}{\frac{1}{2} I \omega^2}$
$
\frac{mv_{cm}^2}{\frac{2}{5} mR^2 \omega^2}=\frac{5}{2} \quad(V=\omega R)
$
Sol. $\frac{\text { Lincar KE }}{\text { Rotational K.E }}=\frac{\frac{1}{2} mv _{ cn }^2}{\frac{1}{2} I \omega^2}$
$
\frac{mv_{cm}^2}{\frac{2}{5} mR^2 \omega^2}=\frac{5}{2} \quad(V=\omega R)
$
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