First slide

Rolling motion

Question

Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the body reaches the ground with maximum velocity?

Moderate

A

Ring
B

Cylinder
C

Sphere
D

All have the same velocity

Solution

We assume conservation of energy of the rolling body, i. e. , there is no loss of energy due to friction etc. The potential energy lost by the body in rolling down the inclined plane (=mgh) must, therefore be equal to kinetic energy gained. Since the bodies start from rest the kinetic energy gained is equal to
the final kinetic energy of the bodies.
$K = \frac{1}{2} {mv}_{CM}^{2} (1 + \frac{k^{2}}{R^{2}})$
where $v_{CM}$ is the final velocity of (the centre of mass) of the body. Equating k andmgh, we have
$mgh = \frac{1}{2} {mv}^{2} (1 + \frac{k^{2}}{R^{2}})$
or $v^{2} = (\frac{2 gh}{1 + k^{2} / R^{2}})$
Note It is independent of the mass of the rolling body. For a ring K²=R²
$v_{ring} = \sqrt{\frac{2 gh}{1 + 1}} = \sqrt{gh}$
$For a solid cylinder, k^{2} = \frac{R^{2}}{2} v_{disc} = \sqrt{\frac{2 gh}{1 + 1 / 2}} = \sqrt{\frac{4}{3} gh}$
Fore a solid sphere, $k^{2} = \frac{2 R^{2}}{5}$
$v_{sphere} = \sqrt{\frac{2 gh}{1 + \frac{2}{5}}} = \sqrt{\frac{10}{7} gh}$
From the results obtained it is clear that among the three bodies, the sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.

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