Split Disk Moment

Consider a solid, uniform circular disk of mass 1 1 and radius 1 1 , with its center at ( x , y ) = ( 0 , 1 ) (x,y) = (0,1 ) . The disk resides within the x y xy plane.

Each point on the disk can be represented in polar coordinates as ( r , θ ) (r,\theta) , where r r is the distance from the origin (not from the center of the disk) and θ \theta is the angle with respect to the positive x x axis.

Let us move each infinitesimal bit of mass on the disk from ( r , θ ) (r,\theta) to ( r , θ + δ ) (r,\theta + \delta) , where δ = π 4 \delta = -\frac{\pi}{4} for ( 0 θ π 2 ) (0 \leq \theta \leq \frac{\pi}{2}) and δ = + π 4 \delta = +\frac{\pi}{4} for ( π 2 < θ π ) (\frac{\pi}{2} < \theta \leq \pi) . The resulting mass arrangement is as shown in the diagram.

What is the moment of inertia of this split disk with respect to the y y axis?

Note: The mass of each infinitesimal portion is preserved as it is moved.


The answer is 1.175.

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1 solution

Otto Bretscher
Dec 19, 2018

Let's denote the right half of the "nutmeg" by D D and proceed in polar coordinates. The moment of inertia we seek is 2 D x 2 d m = 2 π D x 2 d A = 2 π π 4 π 4 0 2 sin ( θ + π 4 ) r 3 cos 2 θ d r d θ = 3 4 + 4 3 π 1.1744 2\int\int_D x^2 dm=\frac{2}{\pi}\int\int_D x^2 dA=\frac{2}{\pi}\int _{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{2\sin(\theta+\frac{\pi}{4})}r^3\cos^2\theta \ dr \ d\theta=\frac{3}{4}+\frac{4}{3\pi}\approx \boxed{1.1744}

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