A body slides down a smooth inclined plane of inclination θ and reaches the bottom with velocity 'V'. If the same body is a ring which rolls down the same inclined plane then linear velocity

Question

A body slides down a smooth inclined plane of inclination θ and reaches the bottom with velocity 'V'. If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of the plane is

TestHub · Accepted Answer

Let the height of the inclined plane be $h$ .

For the body sliding down the smooth inclined plane, the conservation of mechanical energy gives:

$m g h = \frac{1}{2} m V^{2}$

$V = 2 g h$

For the ring rolling down the same inclined plane, the total kinetic energy at the bottom is the sum of translational and rotational kinetic energy:

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

For a ring, the moment of inertia is $I = m R^{2}$ and for pure rolling, $ω = \frac{v}{R}$ . Substituting these values:

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} (m R^{2}) (\frac{v}{R})^{2}$

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} = m v^{2}$

$v = g h$

Comparing the two velocities:

$v = \frac{2 g h}{2} = \frac{V}{2}$

Physics - Rotation Question with Solution | TestHub

Options:

Doubts & Discussion