A solid uniform cylinder rolls down ramp starting from stationary position at he

Question

A solid uniform cylinder rolls down ramp starting from stationary position at height 1.4m. At bottom of the ramp the cylinder has no potential energy. If mass of cylinder is 2.9kg and radius is .076m what is magnitude of angular momentum of cylinder at bottom of ramp? A solid uniform cylinder rolls down ramp starting from stationary position at height 1.4m. At bottom of the ramp the cylinder has no potential energy. If mass of cylinder is 2.9kg and radius is .076m what is magnitude of angular momentum of cylinder at bottom of ramp?

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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