Thank you for reporting, we will resolve it shortly
Q.
If a sphere is rolling, then the ratio of its rotational kinetic energy to the total kinetic energy is
Work, Energy and Power
Solution:
Let m be the mass, r the radius of the sphere and let v and $\omega$ be the linear and angular velocities in rolling down.
Thus total kinetic energy = linear kinetic energy + rotational kinetic energy
$=\frac{1}{2}mv^2+\frac{1}{2}l\omega^2$
where I is the moment of inertia ie, $I=\frac{2}{5}mr^2$
Hence, total kinetic energy$=\frac{1}{2}mv^2+\frac{1}{2}\left(\frac{2}{5}mr^2\right)\frac{v^2}{r^2}$
$\frac{1}{2}mv^2+\frac{1}{5}mv^2$
$\frac{7}{10}mv^2$
So, the ratio$=\frac{Rotational \,KE}{Total \,KE}$
=$\frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2}=2:7$