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A circular disc of mass m and radius r i...

A circular disc of mass m and radius r is rolling over a horizontal table top with angular velocity `omega`. Prove that the total energy of the disc , `K = (3)/(4) m omega ^(2) r^(2)`.

Text Solution

Verified by Experts

Total kinetic energy of the disc,
K= translational kinetic energy+ rotational kinetic energy
`=(1)/(2) mv^(2)+(1)/(2) Iomega^(2)`
Here v = linear velocity of the disc = `omegar`
`I`= moment of inertia of the disc about the perpendicular axis passing through its centre
`= (1)/(2)mr^(2)`
`:. " " K = (1)/(2) m (omegar)^(2)+(1)/(2).(1)/(2) mr^(2).omega^(2)`
=`(1)/(2) momega^(2)r^(2) +(1)/(4)momega^(2) r^(2) = (3)/(4) m omega^(2)r^(2).`
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Knowledge Check

  • If a particle of mass is moving in a horizontal circle of radius r with a centripetal force (k/r^2) the total energy is

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