Question

A solid cylinder at rest at the top of an inclined plane of height $2.7m$ rolls down without slipping. If the same cylinder has to slide down a frictionless inclined plane and acquire the same velocity as that acquired by the center of mass of the rolling cylinder at the bottom of the inclined plane, the height of the inclined plane in meters should be: $(g=10m/s^2)$

• A. $2.2m$
• B. $1.2m$
• C. $1.6m$
• D. $1.8m$

Answer 1

The correct option is D $1.8m$

loss in gravitational potential energy = gain in rotational kinetic energy +gain in translational kinetic energy

so,

$mg\left(2.7\right)=\frac{1}{2}\times m{V}^2+\frac{1}{2}\times \frac{m}{2}\times V^2$
$\left(2.7\right)g=\frac{3}{4}{V}^2$
$V=\sqrt{3.6g}m/s$
Same velocity is acqurired when inclined plane is frictionless
$\therefore V=\sqrt{2gH}$
$\therefore \sqrt{2gH}=\sqrt{3.6g}$
$H=1.8m$