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Question
Discuss the interlink between translational, rotational and total kinetic energies of a rigid object rolls without slipping.
Solution
Consider an object of the moment of inertia `I`, rolling uniformly. If the frictional force on the body is large enough, the body rolls without slipping. Following quantities can be related,
v = Linear speed of the centre of mass
R = Radius of the body
ω = Angular speed of rotation of the body, `therefore omega="v"/R` for any particle
M = Mass of the body
K = Radius of gyration of the body `therefore I=MK^2`
Total kinetic energy of rolling = Translational K.E. + Rotational K.E.
∴ `E = 1/2M"v"^2+ 1/2 Iomega^2`
`= 1/2M"v"^2+ 1/2(MK^2)("v"/R)^2`
`= 1/2 M"v"^2 (1 + K^2/R^2)`
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Answer in Brief:
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