A disc of radius R and mass M is rolling horizontally without slipping with speed $\nu$. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is :

(C) $\frac{1}{2} \frac{v^{2}}{g}$

(B) $\frac{3}{4} \frac{v^{2}}{g}$

(D) $\frac{2}{3} \frac{v^{2}}{g}$

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Work, Power & Energy

A disc of radius R and mass M is rolling horizontally without slipping with speed

ν

. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is :

(A) $\frac{v ^{2}}{g}$
(B) $\frac{3}{4} \frac{v ^{2}}{g}$
(C) $\frac{1}{2} \frac{v ^{2}}{g}$ ✓ Correct
(D) $\frac{2}{3} \frac{v ^{2}}{g}$

Answer: (C)

Solution

On the frictionless incline no torque acts on the disc about its own centre, so its angular velocity — and with it the rotational kinetic energy

\frac{1}{2} I ω^{2}

— stays fixed all the way up. Only the translational kinetic energy is available to convert into gravitational potential energy:

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

. Solving for the maximum height,

h = \frac{v ^{2}}{2 g}

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option

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