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Q.
A solid cylinder of radius $r$ rolls down on an inclined plane without slipping. The speed of the centre of mass when it reaches the bottom, is
AMUAMU 2003
Solution:
From law of conservation of energy
Translational kinetic energy + rotational
kinetic energy $=$ potential energy of fall
$\therefore \frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=m g h$ ...(i)
where $I$ is moment of inertia, $\omega$ the angular velocity and $v$ the linear velocity.
For a solid cylinder $I=\frac{1}{2} mr ^{2}$ ... (ii)
and $\omega=\frac{v}{r}$ ...(iii)
From Eqs. (i), (ii) and (iii), we get
$\frac{1}{2} m v^{2}+\frac{1}{4} m r^{2} \times \frac{v^{2}}{r^{2}}=m g h$
$\Rightarrow v=\sqrt{\frac{4 g h}{3}}$