Corner Tip-Off Lab — a rolling cylinder leaves a vertical edge — JEE Advanced 2026 Paper 1 Q.3

01The question

Q.3 A solid cylinder of radius R rolls without slipping with a center of mass speed v₀ = √gR/3 on a horizontal surface with a vertical edge, as shown in the figure. Here, g 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)0
(B)√5gR/7
(C)√gR/15
(D)√3gR/7

02The experiment

Answer— Press START: the run measures it at the moment contact is lost.

Ready.

Speed

Forces

Launch speed v₀√(gR/3)

03Detailed solution

Answer: — v = √(5gR/7) ≈ 0.845 √(gR).

Reaching the edge — no jolt. In rolling without slipping the contact point is momentarily at rest. When that point arrives at the corner, the corner simply becomes the pivot: nothing hits anything, so there is no impulse. The cylinder carries on with ω = v₀/R, now turning about the corner. (Check: angular momentum about the corner before, I_cmω + mv₀R = (3/2)mRv₀, equals I_cornerω after.)
Turning about the corner. The centre moves on a circle of radius R about the corner. Let θ be the angle of the line corner → centre from the vertical. The speed of the centre is v = Rω, and the moment of inertia about the corner is
I_corner = ½mR² + mR² = (3/2)mR², KE = ½I_cornerω² = ¾mv².
Energy. Only gravity does work (the corner does not move). The centre drops by R(1 − cos θ):
¾mv² = ¾mv₀² + mgR(1 − cos θ).
The corner's push. Along the radius towards the corner, gravity's component minus the corner's push N provides the centripetal force:
mg cos θ − N = mv²/R ⇒ N = mg cos θ − mv²/R.
As θ grows, cos θ falls and v rises, so N falls. The experiment's N arrow shows exactly this.
Losing contact: N = 0. Then v² = gR cos θ. Put it in the energy equation with v₀² = gR/3:
¾gR cos θ = ¾·gR/3 + gR − gR cos θ ⇒ (7/4) cos θ = 5/4 ⇒ cos θ = 5/7 (θ ≈ 44.4°).

v² = gR cos θ = 5gR/7 ⇒ v = √(5gR/7) — option (B).
Why it does not leave at once. At the edge (θ = 0) the corner must supply N = mg − mv₀²/R = (2/3)mg > 0. It stays on until θ ≈ 44.4°. In general cos θ = (3v₀²/gR + 4)/7. If v₀² ≥ gR, the cylinder leaves right at the edge with v = v₀: try it with the v₀ slider.

The centre moves on the dashed circle about the corner. N points from the corner to the centre. Friction f acts along the tangent and keeps the cylinder from slipping. The needed amount is f = (1/3)mg sin θ, from the torque about the centre. As N → 0 the ratio f/N grows without limit, so a real corner would let the cylinder slip slightly before θ = 44.4°. The question, and this experiment, take the ideal grip that it states: "rotation around the corner".

Every option

Option	v²/gR	What it would mean
(A) 0	0	The cylinder would have to stop. But it starts with v₀² = gR/3, and gravity only adds energy as the centre falls. Impossible.
(B) √(5gR/7)	5/7 ≈ 0.714	Energy about the corner with N = 0: cos θ = 5/7. This is what the experiment measures. Correct.
(C) √(gR/15)	1/15 ≈ 0.067	Less than v₀² = gR/3: the cylinder would have slowed down while falling. Impossible.
(D) √(3gR/7)	3/7 ≈ 0.429	It would need cos θ = 3/7. There, energy gives v² = (23/21)gR, so N = mg(3/7 − 23/21) = −(2/3)mg < 0: contact is already lost. It is what you get by subtracting the initial kinetic energy instead of adding it.

Understand the concept

A body turning about a fixed point stays in contact only while the support pushes, N ≥ 0. Contact ends when the pull of gravity towards the pivot (mg cos θ) is no more than the centripetal force the motion needs (mv²/R). Energy decides v; the contact condition decides where. For a rolling body, the kinetic energy about the contact point is ½I_contactω². That includes both the translation and the spin: ¾mv² for a solid cylinder, mv² for a ring, (7/10)mv² for a solid sphere.

Key takeaways

At the edge, a rolling body pivots about the corner with no impulse: its contact point is already at rest.
Use I about the pivot, (3/2)mR² for a solid cylinder, so KE = ¾mv².
Leaving the surface means N = 0, i.e. v² = gR cos θ.
Solid cylinder: cos θ = (3v₀²/gR + 4)/7. For v₀² = gR/3 this gives 5/7, so v = √(5gR/7).
If v₀² ≥ gR, it leaves at the edge itself.

Verification: independent derivation, a Newton–Euler integration that does not use the energy equation, and the page's own engine run by tests/verify_p1phyq03.py. The official key (B) and the MathonGo solution (B) agree. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded.

04Energy against grip

Blue: v²/gR the cylinder has at each angle (energy). Green: the most the corner can hold, cos θ (where N = 0). While blue is below green, the corner pushes. Where they cross, contact is lost. Both curves come from the model and follow the v₀ slider.

How to use this simulation

Press ▶ START EXPERIMENT. The cylinder rolls to the edge and turns about the corner.
Watch the green N arrow at the corner shrink, and the readouts for v, v²/gR and θ.
The run freezes when contact is lost: N = 0 at θ = 44.4°, v² = 5gR/7. Then the cylinder flies off on a parabola.
Drag v₀ to try other launch speeds, and press Question value to return to √(gR/3). Use Speed and Forces to slow it down or hide the arrows.
Replay runs it again. Reset returns to the start. Classroom fills the screen for a projector.