PrayogX › Physics › System of Particles and Rotational Motion

A solid cylinder rolls off a vertical edge: rotation about the corner, energy with the parallel-axis inertia and the speed when it loses contact

A steel cylinder rolls at v0 = sqrt(gR/3) to the edge of a block and turns about the corner. The corner's push shrinks as the centre swings down and reaches zero at cos(theta) = 5/7 - the run freezes there and measures v = sqrt(5gR/7), option B - then the cylinder flies off on a parabola. Change v0 and watch where it leaves.

Sub-topics covered

• Rolling without slipping: the contact point is at rest, so reaching the edge gives no impulse
• Moment of inertia about the corner by parallel axes: I = I_cm + mR^2 = 3/2 mR^2
• Energy about a fixed pivot: 3/4 m v^2 = 3/4 m v0^2 + mgR(1 - cos theta)
• Radial equation at the corner: N = mg cos theta - m v^2/R; contact lost when N = 0
• Leaving at once when v0^2 >= gR

What it teaches

• At the edge the rolling cylinder's contact point is already at rest, so it simply starts turning about the corner with omega = v0/R
• About the corner the kinetic energy is (1/2)(3/2 mR^2)(v/R)^2 = 3/4 m v^2
• Energy: 3/4 v^2 = 3/4 v0^2 + gR(1 - cos theta)
• The corner's push N = mg cos theta - m v^2/R falls as theta grows; contact is lost when v^2 = gR cos theta
• cos theta = (3 v0^2/gR + 4)/7 = 5/7 for v0^2 = gR/3, so v = sqrt(5gR/7)

Simulation ID

• ADV-2026-P1-PHY-Q03 · revision 1

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