Rotational Motion

JEE Main level, single correct, medium

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Question 7 of 280

An object of uniform density rolls up the curved path with the initial velocity v0v_0 as shown in the figure. If the maximum height attained by an object is 7v0210g\dfrac{7v_0^2}{10g} (gg = acceleration due to gravity), the object is a _______.

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Answer: (D) solid sphere

By the law of conservation of mechanical energy, the total initial kinetic energy of the rolling object is converted into potential energy at the maximum height.

Total initial kinetic energy for an object rolling without slipping is given by:
K=Ktranslational+KrotationalK = K_{\text{translational}} + K_{\text{rotational}}
K=12mv02+12Iω2K = \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}I\omega^2

Substituting ω=v0R\omega = \dfrac{v_0}{R} and I=mk2I = mk^2 (where kk is the radius of gyration):
K=12mv02+12(mk2)(v0R)2=12mv02(1+k2R2)K = \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}(mk^2)\left(\dfrac{v_0}{R}\right)^2 = \dfrac{1}{2}mv_0^2\left(1 + \dfrac{k^2}{R^2}\right)

The potential energy at the maximum height hh is:
U=mghU = mgh

Given h=7v0210gh = \dfrac{7v_0^2}{10g}, we have:
U=mg(7v0210g)=710mv02U = mg\left(\dfrac{7v_0^2}{10g}\right) = \dfrac{7}{10}mv_0^2

Equating kinetic and potential energy (K=UK = U):
12mv02(1+k2R2)=710mv02\dfrac{1}{2}mv_0^2\left(1 + \dfrac{k^2}{R^2}\right) = \dfrac{7}{10}mv_0^2

1+k2R2=1410=751 + \dfrac{k^2}{R^2} = \dfrac{14}{10} = \dfrac{7}{5}

k2R2=75−1=25\dfrac{k^2}{R^2} = \dfrac{7}{5} - 1 = \dfrac{2}{5}

The moment of inertia is I=25mR2I = \dfrac{2}{5}mR^2, which corresponds to a solid sphere.

Answer: solid sphere

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