A uniform, solid cylinder with mass M and radius 2R restson a horizontal tableto

Question

A uniform, solid cylinder with mass M and radius 2R restson a horizontal tabletop. A string is attached by a yoke to africtionless axle through the center of the cylinder so that thecylinder can rotate about the axle. The string runs over adisk-shaped pulley with mass M and radius R that ismounted on a frictionless axle through its center. A block ofmass M is suspended from the free end of the string (thefigure ). The string doesn't slip over the pulley surface, and thecylinder rolls without slipping on the tabletop. Find the magnitude of the acceleration of the block after thesystem is released from rest. Express your answer in terms of the variables M, R, andappropriate constants. ablock=

Explanation / Answer

This is a Moment of Inertia/Energy Problem. . Find KE of a rotating disk Moment of Inertia: m r^2/2 KE = KE.translation + KE.rotation KE = 1/2 mv^2 + 1/2 I ^2 = v/r KE = 1/2 m v^2 + 1/2 (m r^2/2)(v/r)^2 KE = 1/2 mv^2 + 1/4 mv^2 KE.disk = 3/4 mv^2 . Add up the KEs and set equal to PE KE.2rdisk + KE.pulley + KE.block = PE.block 3/4 Mv^2 + 3/4Mv^2 + 1/2 Mv^2 = Mgd 2M v^2 = Mgd v^2 =1/2gd, but v^2 = 2ad 2ad = 1/2gd a =1/4 g . Don't worry about M (since it is the same for all objects) andR (which can be incorporated back into velocity) ******Addendum********* I realize that I made a small error above. The pulleyhas no translational energy -- just rotational. So 3/4 Mv^2 + 1/4Mv^2 + 1/2 Mv^2 = Mgd 1.5 v^2 = gd a = 1/3 g . Find KE of a rotating disk Moment of Inertia: m r^2/2 KE = KE.translation + KE.rotation KE = 1/2 mv^2 + 1/2 I ^2 = v/r KE = 1/2 m v^2 + 1/2 (m r^2/2)(v/r)^2 KE = 1/2 mv^2 + 1/4 mv^2 KE.disk = 3/4 mv^2 . Add up the KEs and set equal to PE KE.2rdisk + KE.pulley + KE.block = PE.block 3/4 Mv^2 + 3/4Mv^2 + 1/2 Mv^2 = Mgd 2M v^2 = Mgd v^2 =1/2gd, but v^2 = 2ad 2ad = 1/2gd a =1/4 g . Don't worry about M (since it is the same for all objects) andR (which can be incorporated back into velocity) ******Addendum********* I realize that I made a small error above. The pulleyhas no translational energy -- just rotational. So 3/4 Mv^2 + 1/4Mv^2 + 1/2 Mv^2 = Mgd 1.5 v^2 = gd a = 1/3 g

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