Question

Three balls of mass 1kg, 2kg and 3kg respectively are arranged at the centre of equilateral triangle of side 1m. What will be the moment of inertia of system sbout an axis through the centroid and perpendicular to the plane of the triangle

Answer 1

Answer:

Sorry buddy i don't know the answer....xplanation:

Answer 2

The moment of inertia of system about an axis through the centroid and perpendicular to the plane of the triangle is $2\ kg\ m^2$ .

We know , distance between vertex  and center in equilateral triangle is given by :

$D=\dfrac{a}{\sqrt{3}}$

Where , a is side of equilateral triangle .

Now ,  Moment of inertia is given by :

$I=mR^2$  { here , R is the distance between balls and center }

Therefore moment of inertia of system is given by summation of all single

moment of inertia .

Therefore ,

$I=1\times( \dfrac{1}{\sqrt3})^2+2\times( \dfrac{1}{\sqrt3})^2+3\times( \dfrac{1}{\sqrt3})^2\\\\I=2\ kg\ m^2$

Hence , this is the required solution .

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Moment of inertia

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