Question

A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be

• A. $\frac{2}{R}\sqrt{g h}$
• B. $\frac{2}{R}\sqrt{\frac{g h}{2}}$
• C. $\frac{2}{R}\sqrt{\frac{g h}{3}} \, $
• D. $\frac{1}{2 R}\sqrt{g h}$

Answer 1

Answer: (C)

Loss in PE = Gain in KE
$mgh=\frac{1}{2}mv^{2}+\frac{1}{2}T\left(\omega \right)^{2} \, \left(v = R \omega \right)$
$mgh=\frac{1}{2}m\left(R \omega \right)^{2}+\frac{1}{2}\left(\frac{m R^{2}}{2}\right)\left(\omega \right)^{2}$
$gh=\frac{3}{4}R^{2}\omega ^{2}$
$\Rightarrow \, \, \omega ^{2}=\frac{\text{4} g h}{3 R^{2}} \, \Rightarrow \omega =\frac{2}{R}\sqrt{\frac{g h}{3}}$