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Four point masses, each of mass 1.9 $kg$ are placed at the corners of a square of side 2.2 $m$. Find the moment of inertia of this system about an axis that is perpendicular to the plane of the square and passes through one of the masses.

Respuesta :

Answer:

I = 36.78 kg m^{2}

Explanation:\

Given data:

side of square = 2.2 m

mass =1.9 kg

The moment of inertia i is the total sum of the moments of inertia of the 4 point masses  and it is given as

[tex]I = I_1+I_2+I_3[/tex]

[tex] = mr^{2} +mr^{2}+m(\sqrt2 r)^{2}[/tex]

       [tex]= 2mr^{2}+2mr^{2}[/tex]

         [tex]=4mr^{2}[/tex]

       = [tex]4*1.9*2.2^{2}[/tex]

I = 36.78 kg m^{2}

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