A cylinder of mass m rolls down an inclined plane from a height h above a horizontal level. If the velocity of the axis of the cylinder is v when it reaches the lower level, then

Concept:

Since there is no energy loss from the system, the total energy at the initial position will be equal to the total energy at the final position.

At the initial position only potential energy is present and at the final position, only the kinetic energy is present so,

The potential energy at height h will be equal to the total kinetic energy at the lower level.

P.E. = K.E._{translational} + K.E._rotational

Calculation:

Given:

Height of inclined plane = h, Velocity of axis of cylinder at lower level = v

P.E. = K.E.translational + K.E.rotational

mgh = \(\frac{1}{2}\)mv² + \(\frac{1}{2}\)Iω² 

mgh = \(\frac{1}{2}\)mv² + \(\frac{1}{2} \times \frac{{m{r^2}}}{2} \times \frac{{{v^2}}}{{{r^2}}}\)  (∵ v = ωr)

\(v = \frac{{\sqrt {4gh} }}{{\sqrt 3 }}\)

Hence, ​\(v < \sqrt {\left( {2gh} \right)}\) will be correct.