A solid uniform cylinder rolls down ramp starting from stationary position at he
ID: 1299193 • Letter: A
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