Question

A solid cylinder of mass M and radius R rolls from rest down a plane inclined at an angle q to the horizontal . The velocity of the centre of mass of the cylinder after it has rolled down a distance d is –

• A. $\sqrt{\frac{2}{3}gd\tan\theta}$
• B. $\sqrt{gd\tan\theta}$
• C. $\sqrt{\frac{3}{4}gd\sin\theta}$
• D. $\sqrt{\frac{4}{3}gd\sin\theta}$

Answer 1

Answer: (D)

$\sqrt{\frac{4}{3}gd\sin\theta}$

Answer 2

v = $\sqrt{2ad}$ = $\sqrt{2\frac{g\sin\theta}{1 + \frac{K^{2}}{R^{2}}}.d}$

Here $\frac{K^{2}}{R}$= $\frac{1}{2}$

\ v = $\sqrt{\frac{4gd\sin\theta}{3}}$