There is a hanging mass m1 and a mass sliding on the table m2. They are connecte

Question

There is a hanging mass m1 and a mass sliding on the table m2. They are connected by a rope passing over a pulley. The pulley is a wheel with mass M, inner radius R1 and outer radius R2, and assume the spokes have a negligible mass which we will ignore. The coefficient of kinetic friction between m2 and the table is k. The system is released and the sliding block has a velocity of v1 toward the pulley when it passes a reference point on the table and a larger velocity of v2 toward the pulley when it passes a second reference point a distance d closer to the pulley. Use energy methods to express the speed v2 in terms of m1, m2, M, R1, R2, uk, v1, and d. The moment of inertia of a hollow cylinder is given by I = 1/2M(R12 + R22)

Explanation / Answer

using work energy theorem,

work done by gravity   + work done by friction = change in KE

m2 gd + ( - uk m1g d) = [ (m1 v2^2 /2 ) + ( m2 v2^2 /2 ) + ( M (R1^2 + R2^2) /2 (V2/R2)^2 / 2) ] - [ (m1 v1^2 /2 ) + ( m2 v1^2 /2 ) + ( M (R1^2 + R2^2) /2 (V1/R2)^2 / 2)

2m2gd - 2ukm1gd   + m1 v1^2 + m2 v1^2 + ( M(R1^2 + R2^2) v1^2 / 2 R2^2 ) = ( m1 + m2 + ( M(R1^2 + R2^2) / 2R2^2 )) v2^2

v2 = sqrt [ (2m2gd - 2ukm1gd   + m1 v1^2 + m2 v1^2 + ( M(R1^2 + R2^2) v1^2 / 2 R2^2 ) ) / ( m1 + m2 + ( M(R1^2 + R2^2) / 2R2^2 ))]

v2 =sqrt [ ( 4m2gdR2^2 - 4ukm1gdR2^2 + 2m1v1^2 R2^2 + 2m2 v1^2 R2^2 + M v1^2 R1^2 + M v1^2 R2^2 ) / (2m1R2^2 + 2 m2 R2^2 + M R1^2 + M R2^2 ) ]

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