Science, asked by anjuangelnaiju, 8 months ago

three point masses m 2m 3m are located at three vertices of an equilateral triangle of side l .the moment of inertia of the system os particles about an axis perpendicular to their plane and equidistant from vertices is a) 2ml^2 b) 3ml^2 c) 2√3 ml^2 d) 6ml^2

Answers

Answered by midhun1midhun1
10

Explanation:

Moment of inertia of particle 1 is 0 about an axis along the altitude of the triangle passing through m1

Moment of inertia of m2 about an axis along the altitude of the triangle passing through m1 = m2d2=m2(a/2)2

Moment of inertia of m3 about an axis along the altitude of the triangle passing through m1 = m3d2=m3(a/2)2

Total Moment of inertia about an axis along the altitude of the triangle passing through m1  = m2(a/2)2+m3(a/2)2=(m2+m3)(a)2/4

Answered by swethassynergy
1

The moment of inertia of the system of particles about an axis perpendicular to their plane and equidistant from vertices is 2ml^{2} and option (a) is correct.

Explanation:

Given:

Three point masses m, 2m  and 3m are located at three vertices of an equilateral triangle of side l.

To Find:

The moment of inertia of the system of particles about an axis perpendicular to their plane and equidistant from vertices.

Solution:

As given, three point masses m, 2m and  3m are located at three vertices of an equilateral triangle of side l.

The point masses  located at three vertices of an equilateral triangle are = m,2m amd 3m.

The  length of side of equilateral triangle = l

The axis is equidistant from the vertices .

Therefore, it passes through the circumcenter of the triangle and is perpendicular to the plane of the rectangle.

Distance of each point mass from the axis,r =\frac{l}{\sqrt{3} }.

Moment of inertia = =mass\times r^{2}

Moment of inertia of  m point mass+ Moment of inertia of  2m point mass+Moment of inertia of 3 m point mass =m\times (\frac{l}{\sqrt{3} } )^{2} +2m\times (\frac{l}{\sqrt{3} } )^{2}+ 3m\times (\frac{l}{\sqrt{3} } )^{2}

=\frac{ml^{2} }{3} +\frac{2ml^{2} }{3}+\frac{3ml^{2} }{3}

=\frac{6ml^{2} }{3}

=2ml^{2}

Thus,the moment of inertia of the system of particles about an axis perpendicular to their plane and equidistant from vertices is 2ml^{2} and option (a) is correct.

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