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A plank is placed on a solid cylinder which rolls

A plank is placed on a solid cylinder which rolls on a horizontal surface.The two are of equal mass.There is no slipping in any of the surfaces in contact.Find the ratio of kinetic energy of plank and cylinder.
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ratio=\frac{0.5m(2v)^2}{0.5mv^2+0.5\frac{mr^2v^2}{2r^2}}=8:3

 

 

Kinetic energy of plank: \frac{1}{2} (m)((2v)^2)

Kinetic energy of cylinder is \frac{1}{2} (m)((v)^2) +\frac{1}{2}IW^2

where W is angular velocity

So kinetic energy of cylinder is \frac{1}{2} (m)((v)^2) +\frac{1}{2}\frac{mR^2}{2}(\frac{V}{R})^2=\frac{3}{4}(mv^2)

   ratio=\frac{0.5m(2v)^2}{0.5mv^2+0.5\frac{mr^2v^2}{2r^2}}=8:3???????

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