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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?

Respuesta :

Answer:

The length of the boomerang is 0.364 m

Explanation:

The moment of inertia is:

[tex]I=\frac{1}{2} m_{d} r^{2} +m_{h} r^{2}[/tex]

Where

md = 0.05 kg

mh = 0.12 kg

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

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

The length of the boomerang is:

[tex]L_{b} =\sqrt{\frac{12I}{m} } =\sqrt{\frac{12*2.7x10^{-3} }{0.245} } =0.364m[/tex]

The length of the boomerang(L_b) is; L_b = 0.364 m

We are given;

Mass of hoop part; m_h = 0.12 kg

Mass of disk part; m_d = 0.05 kg

Total mass of boomerang; m_b = 0.245 kg

Diameter of both hoop and disk; d = 0.273 m

Radius; r = d/2 = 0.273/2

r = 0.1365 m

Total moment of inertia for the hoop and disk system is given by the formula;

I = (½m_d + m_h)r²

I = ((½ × 0.05) + 0.12)0.1365²

I = 0.145 × 0.01863225

I = 0.0027 kg.m²

Now, moment of inertia of the boomerang is given by;

I = (1/12)m_b × (L_b)²

Since moment of inertia of boomerang is same as the sports disk, then we have;

L_b = √(12 × 0.0027/0.245)

L_b = 0.364 m

Read more about moment of inertia at; https://brainly.com/question/2254624

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