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 hsph and hcyl on the incline. The ratio \frac{h_{sph}}{h

Engineering

Physics

Conservation of Energy

Rolling

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_sph and h_cyl on the incline. The ratio $\frac{h_{sph}}{h_{cyl}}$ is given by :

$\frac{4}{5}$

$\frac{2}{\sqrt{5}}$

$\frac{14}{15}$

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For solid sphere

$\frac{1}{2} {mv}^{2} + \frac{1}{2} . \frac{2}{5} {mR}^{2} . \frac{v^{2}}{R^{2}} = {mgh}_{sph}$

For solid cylinder

$\frac{1}{2} {mv}^{2} + \frac{1}{2} . \frac{2}{5} {mR}^{2} . \frac{v^{2}}{R^{2}} = {mgh}_{cyl}$

$⇒ \frac{h_{sph}}{h_{cyl}} = \frac{7 / 5}{3 / 2} = \frac{14}{15}$