Question

A marble must be released from a minimum height, h, to just make it through a loop.A marble slides down a track (without rolling), just making it around a loop. The height at which it is dropped is 2.5R (the radius of the loop). Now, the marble rolls down the track. Find the height, h, that a rolling marble must be released from to just pass through the loop. The marble has mass, m, and radius, r.

Answer 1

Case 1 : while sliding

using conservation of energy

initial potential energy = kinetic energy at the top of loop + potential energy at the top of loop

mg (2.5R) = (0.5) m v² + mg(2R)

m v² = mgR

v = sqrt(Rg) eq-1

Case 2 : while rolling

using conservation of energy

initial potential energy = kinetic energy at the top of loop + rotational kinetic energy + potential energy at the top of loop

r = radius of the sphere

mg h = (0.5) m v² + (0.5) I w² + mg(2R)

mg h = (0.5) m v² + (0.5) (0.4) m (r)² (v/r)² + mg(2R)

mg h = (0.7) m v² + mg(2R)

using eq-1

mg h = (0.7) m (sqrt(Rg))² + 2 mgR

mg h = 2.7 mgR

h = 2.7R