Question

A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights $h_{\text{sph}}$ and $h_{\text{cyl}}$ on the incline. The ratio $\frac{h_{\text{sph}}}{h_{\text{cyl}}}$ is given by

• A. $\frac{2}{\sqrt{5}}$
• B. $\frac{14}{15}$
• C. $1$
• D. $\frac{4}{5}$

Answer 1

The correct option is B $\frac{14}{15}$

When a spherical/circular body of radius $r$ rolls without slipping, its total kinetic energy is
$K_{\text{total}}=K_{\text{translation}}+K_{\text{rotation}}$
$=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}m{v}^2+\frac{1}{2}I.\frac{v^2}{r^2}$ $[\because \omega =\frac{v}{r}]$
Let $v$ be the linear velocity and $R$ be the radius for both solid sphere and solid cylinder.
$\therefore$ Kinetic energy of the given solid sphere will be
$K_{\text{sph}}=\frac{1}{2}m{v}^2+\frac{1}{2}{I}_{\text{sph}}\frac{v^2}{R^2}$
$=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{R}^2\times \frac{v^2}{R^2}$
$=\frac{7}{10}m{v}^2...\left(i\right)$
Similarly, kinetic energy of the given solid cylinder will be $K_{\text{cyl}}=\frac{1}{2}m{v}^2+\frac{1}{2}{I}_{\text{cyl}}\frac{v^2}{R^2}$
$=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{m{R}^2}{2}\times \frac{v^2}{R^2}=\frac{3}{4}m{v}^2...\left(ii\right)$
Now, from the conservation of mechanical energy,
$mgh=K_{\text{total}}$
$\therefore$ For solid sphere, using Eq. $\left(i\right)$
$mg{h}_{\text{sph}}=\frac{7}{10}m{v}^2...\left(iii\right)$
Similarly, for solid cylinder, using Eq. $\left(ii\right)$
$mg{h}_{\text{cyl}}=\frac{3}{4}m{v}^2...\left(iv\right)$
Taking the ratio of Eqs. $\left(iii\right)$ and $\left(iv\right)$, we get
$\frac{mg{h}_{\text{sph}}}{mg{h}_{\text{cyl}}}=\frac{\frac{7}{10}m{v}^2}{\frac{3}{4}m{v}^2}\Rightarrow \frac{h_{\text{sph}}}{h_{\text{cyl}}}=\frac{7}{10}\times \frac{4}{3}=\frac{14}{15}$