Dumbbell-shaped flywheel physics problem?
I need some help figuring this problem out.
A 1.0- kg mass is attached to a string wrapped around a shaft of negligible mass and having a 7.0- cm radius. A dumbbell-shaped "flywheel" made from two 0.500- kg masses is attached to one end of the shaft and perpendicular to its axis (see the figure). The mass is released from rest and allowed to fall 1.2 m to the floor. It reaches a speed of 1.3083 m/s just before striking the floor. How far apart are the masses of the dumbbell?
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Not seeing the figure, I assume the masses are on opposite sides of the shaft. Let the distance between masses = 2d, so the radius of the masses = d. Call the falling mass m1 and the total flywheel mass m2. Call the shaft radius r, falling-mass speed v and shaft-and-flywheel rotation rate ω.
KE(flywheel) = PEi - KE(weight)
PEi = m1gh
KE(weight) = m1v^2/2
KE(flywheel) = m1(gh-v^2/2)
KE(flywheel) also = Iω^2/2, where ω = v/r
I = m2*d^2, so KE(flywheel) = m2*d^2ω^2/2
Then d = sqrt(2KE(flywheel)/(m2*ω^2)) = sqrt(2m1(gh-v^2/2)/(m2*ω^2)
and the answer is 2d.
With g = 9.81 I get d = 0.25 m, so flywheel masses are 0.5 m apart.
well, you see, trigonometry calculus mathematical bull$#!t. no one even understands what in the HELL you just said XD
what's your book called? if it's university physics, you better try cramster.com