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A 1.0- kg mass is attached to a string wrapped around a shaft of negligible mass

ID: 2255171 • Letter: A

Question

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 1.3 m to the floor. It reaches a speed of 1.5971 m/s just before striking the floor. How far apart are the masses of the dumbbell?

Can someone please help me on this the answer is not .5m or .44m so please don't comment with those answers!

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 1.3 m to the floor. It reaches a speed of 1.5971 m/s just before striking the floor. How far apart are the masses of the dumbbell? Can someone please help me on this the answer is not .5m or .44m so please don't comment with those answers!

Explanation / Answer

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 ?.
KE(flywheel) = PEi - KE(weight)
PEi = m1gh
KE(weight) = m1v^2/2
KE(flywheel) = m1(gh-v^2/2)
KE(flywheel) also = I?^2/2, where ? = v/r
I = m2*d^2, so KE(flywheel) = m2*d^2?^2/2
Then d = sqrt(2KE(flywheel)/(m2*?^2)) = sqrt(2m1(gh-v^2/2)/(m2*?^2)
and the answer is 2d.