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 passing through any corner 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 M are placed in the three points each is 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