Moment 8-14-2020

Classical Mechanics Level 3

The following region has an area mass density of 1 1 .

x 2 + y 2 ≤ 1 y ≥ 0 x 2 + ( y − 1 2 ) 2 ≥ 1 4 x^2 + y^2 \leq 1 \\ y \geq 0 \\ x^2 + \Big(y- \frac{1}{2} \Big)^2 \geq \frac{1}{4}

What is the moment of inertia of the mass distribution about an axis perpendicular to the x y xy plane and passing through the origin?


The answer is 0.491.

Steven Chase by Steven Chase , 37, USA

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1 solution

First consider the solid semicircular disc without the cavity. Consider an element of area r d r d α rdrdα of the body. Mass of the element is ρ r d r d α \rho rdrdα . Moment of inertia of the element about the given axis is ρ r 3 d r d α \rho r^3drdα .

So, the moment of inertia of the body is π ρ 4 \dfrac {π\rho }{4}

Now consider the solid disc taken out from the body. It's mass is

π ρ 4 \dfrac {π\rho }{4}

It's moment of inertia about the given axis is

3 2 × π ρ 4 × 1 4 = 3 π ρ 32 \dfrac 32 \times \dfrac {π\rho }{4}\times \dfrac 14=\dfrac {3π\rho }{32}

Hence the moment of inertia of the given body is π ρ 4 − 3 π ρ 32 \dfrac {π\rho }{4}-\dfrac {3π\rho }{32}

= 5 π ρ 32 ≈ 0.49087 =\dfrac {5π\rho }{32}\approx \boxed {0.49087} .

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