Question

TWO THIN DISKS ARE ATTACHED AS SHOWN TO THE RIGHT

Two thin disks are attached as shown to the right. The angle θ = 72.3°, and the masses and radii of the disks are listed in the table below.

Answer 1

first we have to find the length BP

$cos(\pi-\theta) = \frac{AP^2 + AB^2-BP^2}{2AP*AB}$

$cos(107.7^{\circ}) = \frac{16.8^2 + 24.5^2-BP^2}{2*16.8*24.5}$

$-0.304 = \frac{16.8^2 + 24.5^2-BP^2}{2*16.8*24.5}$

$-250.28 = 16.8^2 + 24.5^2-BP^2$

$BP^2 = 16.8^2 + 24.5^2+250.28$

$BP = 33.66cm$

So the moment of inertia can be given as :

$I = \frac{1}{2}*.881*(0.168)^2 + .881*(0.168)^2 + \frac{1}{2}* 0.475*(0.077)^2+ 0.475*(.3366)^2$

$I = 0.0925 Kg.m^2$