Question

Find moment of inertia of disc about any tangent parallel to diameter

Answer 1

Answer:

ll perform the calculation in polar coordinates. Refer to the sketch above .Let the mass of the ring and its radius be MM and aa respectively. The mass of an infinitesimal segment adθadθ located at (a,θ)(a,θ) is dM=λadθdM=λadθ, where λ=M2πaλ=M2πa. The moment of inertia of this elementary mass about the tangent is dM(a−acosθ)2dM(a−acos⁡θ)2, the distance of the mass dMdM from the tangent being (a−acosθ)(a−acos⁡θ). Then, the moment of inertia of the circular ring about the tangent is ∫dM(a−acosθ)2=∫2πθ=0λadθ(a−acosθ)2=∫2πθ=0M2πaa(a−acosθ)2dθ=32Ma2∫dM(a−acos⁡θ)2=∫θ=02πλadθ(a−acos⁡θ)2=∫θ=02πM2πaa(a−acos⁡θ)2dθ=32Ma2.

Answer 2

Answer:

You want, for example, the moment of inertia about the y-axis except shifted by a radius. So add mr to the 1/4mr.