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A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom -
A
$\sqrt{2gh}$
B
$\sqrt{\dfrac{3}{4}gh}$
C
$\sqrt{\dfrac{4}{3}gh}$
D
$\sqrt{4gh}$
Detailed Solution
In rolling without slipping, friction does no work, so mechanical energy is conserved. The loss in potential energy becomes translational plus rotational kinetic energy:
$Mgh = \dfrac{1}{2}Mv^2 + \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}Mv^2\left(1 + \dfrac{K^2}{R^2}\right)$ (using $I = MK^2$ and $v = R\omega$)
$\Rightarrow v = \sqrt{\dfrac{2gh}{1 + K^2/R^2}}$
For a solid cylinder $I = \dfrac{1}{2}MR^2$, so $\dfrac{K^2}{R^2} = \dfrac{1}{2}$
$v = \sqrt{\dfrac{2gh}{1 + \frac{1}{2}}} = \sqrt{\dfrac{4gh}{3}}$
The speed does not depend on the length L of the incline, only on the height h.
$Mgh = \dfrac{1}{2}Mv^2 + \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}Mv^2\left(1 + \dfrac{K^2}{R^2}\right)$ (using $I = MK^2$ and $v = R\omega$)
$\Rightarrow v = \sqrt{\dfrac{2gh}{1 + K^2/R^2}}$
For a solid cylinder $I = \dfrac{1}{2}MR^2$, so $\dfrac{K^2}{R^2} = \dfrac{1}{2}$
$v = \sqrt{\dfrac{2gh}{1 + \frac{1}{2}}} = \sqrt{\dfrac{4gh}{3}}$
The speed does not depend on the length L of the incline, only on the height h.