Question

Obtain expression for kinetic energy of rolling motion

Answer 1

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Answer 2

Let M and R be the mass and radius of the body, V is the translation speed, ω is the angular speed and I is the moment of inertia of the body about an axis passing through the centre of mass.

Kinetic energy of rotation,

ER= $\frac{1}{2}$ MV²

Kinetic energy of translation,

Er = $\frac{1}{2}$Iω²

Thus, the total kinetic energy 'E' of the rolling body is

E = ER + Er

= $\frac{1}{2}$ MV² + $\frac{1}{2}$ Iω²

=$\frac{1}{2}$ MV² + $\frac{1}{2}$ MK²ω² .... (I=MK² and K is the radius of gyration)

= $\frac{1}{2}$ MR²ω² + 1/2 MK²ω² .... (V=Rω)

∴E=$\frac{1}{2}$ Mω²(R²+K²)

∴E=$\frac{1}{2}$ M $\frac{V²}{R²}$ (R²+K²)

E = $\frac{1}{2}$ MV²(1+K²R²)

∴ Hence proved.

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