A solid uniform cylinder rolls down a ramp, starting from a stationary position

Question

A solid uniform cylinder rolls down a ramp, starting from a stationary position at height 1.4m. At the bottom of the ramp, the cylinder has no potential energy. If the mass of the cylinder is 2.9kg and it's radius is 0.076m, what is the magnitude of the angular momentum of the cylinder at the bottom of the ramp with respect to the cylinder's center of mass? A solid uniform cylinder rolls down a ramp, starting from a stationary position at height 1.4m. At the bottom of the ramp, the cylinder has no potential energy. If the mass of the cylinder is 2.9kg and it's radius is 0.076m, what is the magnitude of the angular momentum of the cylinder at the bottom of the ramp with respect to the cylinder's center of mass?

Explanation / Answer

First by conservation of energy

PE = KEr + KEt

KEr = .5Iw2 and I for a cylinder is .5mv2 -- Also v = w/r, so

KEr = .5(.5mv2)(v2/r2) which simplifies to .25mv2

Thus mgh = .5mv2 + .25mv2 (mass cancels)

(9.8)(1.4) = .75v2

v = 4.27 m/s

w = v/r = 4.27/.076 = 56.3 rad/s

Finally for momentum L = Iw

L = (.5)(2.9)(.076)2(56.3)

L = .471 kg m2/s

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