Question

three point sized bodies each of mass M are fixed at three corners of light triangular frame of corner length L. About an axis perpendicular to the plane of frame and passing through centre of frame the moment of inertia of three bodies is

Answer 1

Answer:

The answer will be ML^2

Explanation:

According to the problem the mass of the three point each is M which are fixed at a light triangular frame , and the side of the triangle is L

Now we need to find the moment of inertia about  the axis passing through the center of the triangle.

Now we know that of a equilateral triangle the median are cutting at the point which is called the center of gravity or CG

Now we know  cg divides the median into 2:1  ratio

Let the distance from the cg to the points is r

Therefore

Moment of Inertia, I =ΣMr^2

Now as it is a equilateral triangle therefore the angles are 60 °

Now by applying median it will be 30°

if we take a small triangle then the base of that will be l/2 angle will be 30°

Therefore, cos30° = L/2/r

r = L/√3

Therefore,

Moment of Inertia , I = Mr^2 +Mr^2 +Mr^2 = 3Mr^2 = 3M( L/√3)^2 = ML^2