Question

Show that the rotational kinetic energy of a ball rolling over a horizontal plane is 2/7 of its total kinetic energy.

Answer 1

$\begin{array}{l}\text{Let the radius of the ball be}r\text{and its mass be}m\text{and it is}\\ \text{moving rolling without slipping with a velocity}v\text{in the}\\ \text{horizontal table. Let the ball be uniform solid.}\\ \text{The total KE of the ball about the CM is given by}\\ K{E}_{tot}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\\ \text{for uniform solid ball}I=\frac{2}{5}m{r}^2\\ =>K{E}_{tot}=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{2}{5}m{r}^2{\omega }^2\\ \text{For rolling without slipping condition}v=r\omega \\ K{E}_{tot}=\frac{1}{2}m{v}^2+\frac{1}{5}m{r}^2{\left(\frac{v}{r}\right)}^2=\frac{7}{10}m{v}^2\\ \text{The only rotational KE of the ball is given by}\\ K{E}_{rot}\text{=}\frac{1}{2}I{\omega }^2=\frac{1}{5}m{v}^2\\ \frac{K{E}_{rot}}{K{E}_{tot}}\text{=}\raisebox{1ex}{$\frac{1}{5}m{v}^2$}\!\left/ \!\raisebox{-1ex}{$\frac{7}{10}m{v}^2$}\right.=\frac{2}{7}\end{array}$