JEE Main 2025 · Physics · Rotational Motion · medium

A uniform solid disc rolls without slipping on a horizontal floor with its centre moving at 4 m s^-1 and then rolls, still without slipping, up an…

Numerical answer

A uniform solid disc rolls without slipping on a horizontal floor with its centre moving at

4\ \mathrm{m\,s^{-1}}

and then rolls, still without slipping, up an incline of angle

30^\circ

. Taking

g=10\ \mathrm{m\,s^{-2}}

, the maximum distance (in cm) it travels along the incline is ______.

Show answer and solution

Answer: 240

Rolling without slipping means no energy is lost to friction, and the total kinetic energy of a solid disc is

\tfrac12 mv^2 + \tfrac12\left(\tfrac12 mR^2\right)\left(\dfrac{v}{R}\right)^2 = \tfrac34 mv^2

. At the highest point all of it is potential energy:

\tfrac34 mv^2 = mgh

, so

h = \dfrac{3v^2}{4g} = \dfrac{3\times16}{40} = 1.2\ \mathrm{m}

. The distance along the incline is

s = \dfrac{h}{\sin30^\circ} = 2.4\ \mathrm{m} = 240\ \mathrm{cm}

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