Question #87861
A marble and a cube are placed at the top of a ramp. Starting from rest at the same height, the marble rolls without slipping and the cube slides (no kinetic friction) down the ramp. Determine the ratio of the center-of-mass speed of the cube to the center-of-mass speed of the marble at the bottom of the ramp.
1
Expert's answer
2019-04-11T09:40:25-0400

The cube. From the conservation of energy:


mgh=0.5mv2mgh = 0.5mv^2

v=2ghv=\sqrt{2gh}

The marble. From the conservation of energy:


0.5MV2+0.5Iω2=Mgh0.5MV^2 + 0.5 Iω^2 = Mgh

For a sphere:


I=0.4Mr2I=0.4 Mr^2

ω=Vr\omega=\frac{V}{r}

Thus,


0.5MV2+0.5(0.4Mr2)(Vr)2=Mgh0.5MV^2 + 0.5 (0.4 Mr^2)(\dfrac{V}{r})^2 = Mgh

V=10gh7V=\sqrt{\frac{10gh}{7}}

So,


vV=75\frac{v}{V}=\sqrt{\frac{7}{5}}



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