Rotational Mechanics · Rolling Cylinder at an Edge

See the cylinder leave the corner at a lower height

A solid cylinder rolls to a vertical edge with speed v0 = √(gR/3). After it catches the corner, it rotates about that corner. The contact breaks only when the normal reaction becomes zero, and that happens after the center has dropped below its original level.

2026 Paper 1 Q3 · rolling cylinder break-off
Break-off point is below the original COM level · cosθ = 5/7
Observation stage
Motion speed 1.0×
Angular momentum about the corner
At the instant the cylinder catches the corner, angular momentum about the corner is conserved.
ExpressionIP = Icm + mR²
MeaningIP = 3mR²/2
Original COM leveldashed line before falling
Break-off pointlower by 2R/7
Normal reactionvanishes at cosθ = 5/7

Derivation path

01 / CORNER PIVOT

Use angular momentum about the corner

The corner impulse passes through the corner, so angular momentum about that point is conserved at impact.

IP = 3mR²/2,   Ω₀ = v₀/R
02 / ENERGY

The center drops while rotating

When the radius to the center has turned by θ from the vertical, the center has fallen by R(1 − cosθ).

v² = (gR/3)(5 − 4cosθ)
03 / CONTACT BREAKS

Set the normal reaction to zero

Along the radius, contact is lost when gravity alone supplies the needed centripetal acceleration.

N = 0 ⇒ v²/R = gcosθ ⇒ cosθ = 5/7
Speed of the center of mass at break-off
v² = gRcosθ = 5gR/7   ⇒   v = √(5gR/7)
Correct option: (B)