The moment of inertia of a circular area of radius r with respect to diameter axis isa)πr2 ÷ 4b)πr2 ÷ 2c)πr4 ÷ 4d)None of theseCorrect answer is option 'C'. Can you explain this answer?

Question

The moment of inertia of a circular area of radius r with respect to diameter axis isa)&pi;r2 &divide; 4b)&pi;r2 &divide; 2c)&pi;r4 &divide; 4d)None of theseCorrect answer is option 'C'. Can you explain this answer?

Answer

I = (1/4)mr^2, where m is the mass of the circular area.

Explanation:

The moment of inertia of a 2D object depends on its mass distribution and the axis of rotation. For a circular area with uniform mass distribution, the moment of inertia with respect to the diameter axis can be found using the formula:

I = (1/4)mr^2

where m is the mass of the circular area and r is the radius. This formula assumes that the circular area is thin and has negligible thickness.

To derive this formula, we can divide the circular area into infinitesimally small concentric rings, each with mass dm and radius r'. The moment of inertia of each ring with respect to the diameter axis is:

dI = (dm)(r')^2

The total moment of inertia of the circular area can be obtained by integrating over all the rings:

I = ∫dI = ∫(dm)(r')^2

Since the mass density is uniform, dm = (m/πr^2)dA, where dA is the infinitesimal area of the ring. Substituting this into the integral and using the limits of integration from 0 to r, we get:

I = ∫(m/πr^2)dA(r')^2

I = (m/πr^2)∫dA(r')^2

I = (m/πr^2)∫2πr'dr'

I = (m/πr^2)(2π/3)(r')^3 |_0^r

I = (1/4)mr^2

Therefore, the moment of inertia of a circular area of radius r with respect to diameter axis is (1/4)mr^2.

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