This is one question, please answer ALL parts. ICM = M2R2%3Dso that the final speed isVCM, hollowVg(ho - h) :(9.80 m/s2)(14 m 3 m) =m/s.%3D(B) Which cylinder takes less time to reach the bottom of the ramp?At any height h, the results already obtained show that the solid cylinder is moving at a speed4g(ho -- h)VCM, solid= V3and the speed of the hollow cylinder at that same height isVCM, hollowV glh, - h).= VSo the solidO v cylinder moves faster at each height, and takes less time to reach the bottom of theramp.(C) What is the ratio of the final speed of the solid cylinder to that of the hollow cylinder?From the results already obtained, the ratio of the final speeds is4gh/3VCM, solidVCM, hollowghFINALIZENotice that the final speeds depended only on the shape of the object and the way that the mass was distributed noton the mass or radius of eitho ANALYZE(A) How fast is each cylinder moving when it has reached a height of 3 m above the ground?Each cylinder starts out with the total mechanical energy consisting entirely of gravitational potential energy:Emech = Mgho%3Dwhere M is the mass of the cylinder. At a height h above the ground, the kinetic energy has increased at the expenseof the potential energy, with the total mechanical energy the same, so that2EmechMvCM+ Mgh.%3D2Equate the second expression for Emech to its initial value given by the first equation, and use the relation betweenthe rotational and translational speeds of the rolling cylinderV CMto obtainon(e)Mon- Mg(ho – h)22which can be rearranged into29(ho - h)+ Icm/(MR)²VCM =Next consider the two cases. For a solid cylinder of mass M = M,, the moment of inertia isso that the final speed at height h is4g(ho - h)m/s.= V%3DVCM, solid3

The solution of this problem is as follows,

ICM = M2R2 %3D so that the final speed is VCM, hollow V g(h, - h) = v (9.80 m/s2)(14 m - 3 m) = m/s. %3D (B) Which cylinder takes less time to reach the bottom of the ramp? At any heighth, the results already obtained show that the solid cylinder is moving at a speed 49(ho - h) VCM, solid V and the speed of the hollow cylinder at that same height is VCM, hollow = Vg(h, – h). So the solid O v cylinder moves faster at each height, and takes less time to reach the bottom of the ramp.

ICM = M2R2 %3D so that the final speed is VCM, hollow V g(h, - h) = v (9.80 m/s2)(14 m - 3 m) = m/s. %3D (B) Which cylinder takes less time to reach the bottom of the ramp? At any heighth, the results already obtained show that the solid cylinder is moving at a speed 49(ho - h) VCM, solid V and the speed of the hollow cylinder at that same height is VCM, hollow = Vg(h, – h). So the solid O v cylinder moves faster at each height, and takes less time to reach the bottom of the ramp.