JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Work, Power & Energy

When a body rolls up an incline without slipping, all of its kinetic energy, both the translational part and the rotational part, is converted into gravitational potential energy at the highest point it reaches. So for either body $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$, and because the contact is rolling, $\omega =v/R$. For the solid sphere $I=\frac{2}{5}m{R}^2$, which makes the left side $\frac{1}{2}m{v}^2+\frac{1}{2}\cdot \frac{2}{5}m{R}^2\cdot \frac{v^2}{R^2}=\frac{7}{10}m{v}^2$, so $h_{sph}=\frac{7v^2}{10g}$. For the solid cylinder $I=\frac{1}{2}m{R}^2$, giving $\frac{1}{2}m{v}^2+\frac{1}{2}\cdot \frac{1}{2}m{R}^2\cdot \frac{v^2}{R^2}=\frac{3}{4}m{v}^2$, so $h_{cyl}=\frac{3v^2}{4g}$. Taking the ratio, the common $v^2/g$ cancels and we are left with $\frac{h_{sph}}{h_{cyl}}=\frac{7/10}{3/4}=\frac{7}{10}\times \frac{4}{3}=\frac{14}{15}$. Notice the sphere climbs a little less high because a smaller share of its energy is tied up in rotation.