A 1.0-kg mass is attached to a string wrapped around a shaft of negligible mass
ID: 1438000 • Letter: A
Question
A 1.0-kg mass is attached to a string wrapped around a shaft of negligible mass and having a 8.0-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. The mass is released from rest and allowed to fall 1.4 m to the floor. It reaches a speed of 1.5973 m/s just before striking the floor. How far apart are the masses of the dumbbell ? please include some extra steps so that I can see where the numbers are coming from.
Explanation / Answer
By law of conservation of energy,
KEfly = PEi - KEweight ------------(1)
PEi = Mgh ------------(2)
M= Falling mass = 1.0kg
KEweight = Mv^2/2 ------------(3)
KEfly = M(gh-v^2/2) = I^2/2 ------------(4)
where = v/r and I = m*r^2
m= mass of flywheel = 2*0.5*1.0kg
Hence
KEfly = m*r^2^2/2 ------------(5)
From (1),(2), (3), (4) and (5)
r= sqrt(2KEfly/(m2*^2)) = sqrt(2M(gh-v^2/2)/(m*^2) = sqrt(2*1.0(9.8*1.4-1.5973^2/2)/(1.0*(1.5973/0.08)^2) = 0.25
Distance between the masses of the dumbbell = 2r = 2*0.25 = 0.5m
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