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JEE Advanced 2026 · Paper 1, Physics question 3

A solid cylinder of radius RR rolls without slipping with a center of mass speed v0=gR3v_0=\sqrt{\dfrac{gR}3} on a horizontal surface with a vertical edge, as shown in the figure. Here, gg is the acceleration due to the gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:

A cylinder (circle) rolling to the right with speed v₀ on top of a hatched block whose right side is a vertical edge.
Choose one option
Answer and worked solution

Official answer: B

Once the contact point reaches the edge, the cylinder pivots about the corner (no slipping) with moment of inertia
Icorner=12MR2+MR2=32MR2.I_{\text{corner}}=\tfrac12MR^2+MR^2=\tfrac32MR^2.

Energy after turning through θ\theta from the vertical:
34Mv2=34Mv02+MgR(1−cos⁡θ)  ⟹  v2=gR3+43gR(1−cos⁡θ).\tfrac34Mv^2=\tfrac34Mv_0^2+MgR(1-\cos\theta)\implies v^2=\frac{gR}3+\frac43gR(1-\cos\theta).

Losing contact. The normal force vanishes when gravity alone supplies the centripetal force: gcos⁡θ=v2Rg\cos\theta=\dfrac{v^2}{R}. Then
cos⁡θ=13+43−43cos⁡θ  ⟹  cos⁡θ=57.\cos\theta=\frac13+\frac43-\frac43\cos\theta\implies\cos\theta=\frac57.

So v2=gRcos⁡θ=5gR7v^2=gR\cos\theta=\dfrac{5gR}7, i.e. v=5gR7v=\sqrt{\dfrac{5gR}{7}}.

Topics:Rolling motionCircular motion dynamicsConservation of mechanical energy
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