JEE Main 2026 Apr 5 Shift 2, Physics Q6: Rolling Motion

JEE Main2026Apr 5, Shift 2Physics

An object of uniform density rolls up the curved path with the initial velocity $v_{0}$ as shown in the figure. If the maximum height attained by an object is $\frac{7 v _{0}^{2}}{10 g}$ ( $g$ = acceleration due to gravity), the object is a _______.

A
solid cylinder
B
ring
C
disc
D
solid sphere

Solution

Energy conservation (the body rolls without slipping, so both translational and rotational KE convert to PE):

$m g h = \frac{1}{2} m v_{0}^{2} + \frac{1}{2} I ω^{2}$ , with $ω = v_{0} / R$ .

Substituting $h = \frac{7 v _{0}^{2}}{10 g}$ :

$\frac{7 m v _{0}^{2}}{10} = \frac{m v _{0}^{2}}{2} + \frac{I v _{0}^{2}}{2 R ^{2}}$

$\Rightarrow \frac{m v _{0}^{2}}{5} = \frac{I v _{0}^{2}}{2 R ^{2}} \Rightarrow I = \frac{2 m R ^{2}}{5}$ .

This matches the moment of inertia of a solid sphere about its diameter.

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JEE Main 2026 Apr 5 Shift 2, Physics Q6: Rolling Motion

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