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Systems of Particles and Rotational Motion — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsSystems of Particles and Rotational Motion3 Marks
Question
A solid sphere of mass m and radius r is impure rolling on a horizontal surface. What fraction of total energy of the sphere is:
  1. Kinetic energy of rotation?
  2. Kinetic energy of translation?
✓

Answer

Mass of sphere = m, Radius of sphere = r
Moment of Inertia $\text{I}=\frac{2}{5}\text{mr}^2$
Total energy $=\text{K}_{\text{R}}+\text{K}_{\text{T}}$
$\text{K}_{\text{tot}}=\frac{1}{2}\omega^2+\frac{1}{2}\text{mv}^2$
$=\frac{1}{2}.\frac{2}{5}\text{mr}^2\Big(\frac{\text{v}^2}{\text{r}^2}\Big)+\frac{1}{2}\text{mv}^2(\text{v}=\text{r}\omega)$
$\text{K}_{\text{tot}}=\frac{1}{2}\Big(\frac{7}{5}\Big)\text{mv}^2$
Fraction of K.E. of rotation $=\frac{\text{k}_{\text{R}}}{\text{K}_{\text{tot}}}$
$=\frac{\frac{1}{2}\Big(\frac{2}{5}\Big)\text{mv}^2}{\frac{1}{2}\Big(\frac{7}{5}\Big)\text{mv}^2}=\frac{2}{7}$
$\therefore$ Fraction of K.E of translation $=\frac{\text{K}_{\text{T}}}{\text{K}_{\text{tot}}}=\frac{5}{7}$

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