A 10.0-kg block resting on a horizontal, frictionless surface is attached to a 6
ID: 1998010 • Letter: A
Question
A 10.0-kg block resting on a horizontal, frictionless surface is attached to a 6.00-kg weight by a thin, light cord that passes over a solid, cylindrical pulley. The pulley rotates on frictionless bearings. The weight is released from rest, and the cord does not slip on the pulley. At the instant the 6.00-kg weight has fallen 1.25 m, the speed of the 10.0-kg block is 0.450 m/s. The effective radius of the pulley is 0.075 m. a) Calculate the moment of inertia of the pulley. b) If the solid cylindrical pulley were replaced by a hollow cylindrical pulley of the same mass and radius, would the speed of the 6.00-kg block after falling 1.25 m be greater than, less than, or equal to 0.450 m/s? Briefly explain your reasoning.Explanation / Answer
(A) Applying energy conservation,
PEi + KEi = PEf + KEf
(6 x 9.81 x 1.25) = (6 v^2 /2 + 10 v^2 /2 + I w^2 / 2)
{ w = v/R }
73.575 = 8 v^2 + (I) (v/0.075)^2 / 2
73.757 = (8 x 0.450^2) + (I x 0.450^2 / 0.075^2) / 2
I = 71.955 / 18 = 4 kg m^2
(B) if hollow cylinder is used then I will increase.
tne 73.575 = 8 v^2 + (0.5 I v^2 / R^2)
v = sqrt[ 73.575 / (8 + 0.5I/R^2) ]
as I increases then denominator term will also increase.
hence v will decrease.
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