A marble and a cube are placed at the top of a ramp. Startingfrom rest at the sa

Question

A marble and a cube are placed at the top of a ramp. Startingfrom rest at the same height, the marble rolls and the cube slides(no kinetic friction) down the ramp. Determine the ratio of thecenter-of-mass speed of the cube to the center-of-mass speed of themarble at the bottom of the ramp. How do you figure for the cube A marble and a cube are placed at the top of a ramp. Startingfrom rest at the same height, the marble rolls and the cube slides(no kinetic friction) down the ramp. Determine the ratio of thecenter-of-mass speed of the cube to the center-of-mass speed of themarble at the bottom of the ramp. How do you figure for the cube

Explanation / Answer

A marble anda cube are placed at the top of a ramp. Starting from rest
at the same height, the marble rolls and the cube slides (nokinetic friction)
down the ramp. Determine the ratio of the center-of-mass speed ofthe cube
to the center-of-mass speed of the marble at the bottom of theramp.

Apply Conservation of Energy to bothsystems (which begin at rest):

CUBE:
     {Final Total Kinetic Energy} = {Initial Potential Energy}
     ---->   {Translational K.E.}+ {Rotational K.E.} = m*g*h
     ---->   (1/2)*m*v^2 +(0) = m*g*h
     ---->   v = Sqrt[2*g*h ]

MARBLE (recall that = v/R and I = (2/5)*m*R^2):
     {Final Total Kinetic Energy} = {Initial Potential Energy}
     ---->   {Translational K.E.}+ {Rotational K.E.} = m*g*h
     ---->   (1/2)*m*v^2 +(1/2)*I*^2 = m*g*h
     ---->   (1/2)*m*v^2 +(1/2)*I*(v/R)^2 = m*g*h
     ---->   (1/2)*m*v^2 +(1/2)*((2/5)*m*R^2)*(v/R)^2 = m*g*h
     ---->   (1/2)*m*v^2 +(1/5)*m*v^2 = m*g*h
     ---->   (7/10)*m*v^2 = m*g*h
     ---->   v = Sqrt[(10/7)*g*h ]

---->    (v_cube)/(v_marble) =
                = Sqrt[2*g*h]/Sqrt[(10/7)*g*h]
                = Sqrt[14/10]
                = 1.183

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