Question

A solid cylinder is rolling down on an inclined plane without slipping. The length of the plane is $30m$ and its angle of inclination is ${30}^o$. If the cylinder starts from rest then its velocity at the bottom of the inclined plane is:

• A. $14m{s}^{-1}$
• B. $28m{s}^{-1}$
• C. $7m{s}^{-1}$
• D. $35m{s}^{-1}$

Answer 1

The correct option is A $14m{s}^{-1}$

By work energy theorem$,h=30sin30$
$=15m$
$mgh=\frac{m{V}^2}{2}+\frac{1}{2}\left(\frac{m{R}^2}{2}\right){\left(\frac{V}{R}\right)}^2$ (directly putting up the given physically significant case)
$mgh=\frac{3}{4}m{V}^2$
$\sqrt{\frac{10\times 15\times 4}{3}}=V$
$\sqrt{200}=V=14m/s$