Derive an expression for kinetic energy of a rotating body.
Answers
Explanation:
Rotational kinetic energy can be expressed as: Erotational=12Iω2 E rotational = 1 2 I ω 2 where ω is the angular velocity and I is the moment of inertia around the axis of rotation. The mechanical work applied during rotation is the torque times the rotation angle: W=τθ W = τ θ .
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Figure shows the cross section of the a body with this paper . Let the body rotate about an axis passing through p and normal to the plane of this page , with an angular velocity . The the body can be assumed to be consisting of a large number of particles of mass m₁ , m₂ , m₃,,,,,, at distance r₁ , r₂ , r₃,,,,
from the axis of rotation , respectively
➝The particles move in circular orbits of radii r₁ , r₂ , r₃ ,,,,,, with respective speed r₁ω₁ , r₂ω₂ , r₃ω₃ ,,,,,, The rotational kinetic energy of the body is equal to the sum of the kinetic energies of the individual particles
➝K = 1/2m₁(r₁ω)² + 1/2m₂(r₂ω)² + m₃(r₃ω)²+,,,,,
➝K = 1/2(m₁r₁²+m₂r₂²+m₃r₃²+,,,,) = (1/2)lω²
➝K=(1/2)Iω²
Where l = ∑m₁r₁² is the moment of inertia about the axis of rotation
