Question

A solid spherical ball rolls on a table. Ratio of its rotational kinetic energy to total kinetic energy is

• A. $\frac{1}{2}$
• B. $\frac{1}{6}$
• C. $\frac{7}{10}$
• D. $\frac{2}{7}$

Answer 1

Answer: (D)

$\frac{2}{7}$

Answer 2

Linear K.E. of ball $=\frac{1}{2} mv^2$ and rotational K.E.
of ball $=\frac{1}{2}I\omega^2=\frac{1}{2}\bigg(\frac{2}{5}mr^2\bigg) \omega^2 =\frac{1}{5}mv^2$
Therefore total K.E. =$\frac{1}{2}mv^3+\frac{1}{5}mv^2=\frac{7}{10}mv^2$.
And ratio of rotational K.E. and total K.E.
$\, \, \, \, \, \, =\frac{(1/5)mv^2}{(7/10)mv^2}=\frac{2}{7}$.