Question

A solid sphere of mass 0.5 kg and diameter 1 m rolls without sliding with a constant velocity of 5 m/s, the ratio of the rotational K.E. to the total kinetic energy of the sphere is :

• A. $\frac{7}{10}$
• B. $\frac{4}{9}$
• C. $\frac{2}{7}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{2}{7}$

$\begin{array}{c}\text{Given-}\ast \text{mass of Solid sphere}=0.5kg\\ \ast \text{Diameter}=1m\text{then Radius}r=0.5m\\ \text{* velocity}=5m/s\end{array}$

$\begin{array}{c}\text{Kinetic ener}gy\text{of sphere}kE-\\ k{E}_{\text{Total}}=\text{Translation}\left(kE\right)+\text{Rotational}\left(KE\right)\\ \text{Translation}\left(KE\right)=\frac{1}{2}m{v}^2\\ \text{Rotational}\left(KE\right)=\frac{1}{2}I{\omega }^2\\ \text{moment of Inertia of Sphere (I)}\\ I=\frac{2}{5}m{R}^2\end{array}$

$\begin{array}{c}\text{Therefore Rotational}\left(KE\right)=\frac{1}{2}\times \frac{2}{5}m{R}^2{\omega }^2=\frac{1}{5}m{v}^2\\ \text{Sinu}(v=R\omega )\text{in Rotational motion}\\ \Rightarrow k{E}_{\text{Total}}=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2\\ \frac{KE\text{Rotational}}{kE\text{total}}=\frac{\frac{1}{5}m{v}^2}{\frac{7}{10}m{v}^2}=\frac{2}{7}\end{array}$