Question

Everyone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of diameter Dd=0.273 m. The mass of the hoop part is mh=0.120 kg and the mass of the disk part is md=0.050 kg. You would like to make a boomerang that has the same total moment of inertia, around its center, as the sport disk. The boomerang is to be constructed in the shape of an \"X,\" which can be approximated as two thin uniform rods joined at their midpoints. If the total mass of the boomerang is to be mb = 0.245 kg, what must be the length Lb of the boomerang?

Answer 1

Answer:

The length of the boomerang is 0.364 m

Explanation:

The moment of inertia is:

$I=\frac{1}{2} m_{d} r^{2} +m_{h} r^{2}$

Where

md = 0.05 kg

mh = 0.12 kg

r = d/2 = 0.273/2 = 0.1365 m

$I=\frac{0.05*(0.1365)^{2} }{2} +(0.12)*(0.1365)^{2} =2.7x10^{-3} kgm^{2}$

The length of the boomerang is:

$L_{b} =\sqrt{\frac{12I}{m} } =\sqrt{\frac{12*2.7x10^{-3} }{0.245} } =0.364m$