Question

A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle so that each side has a length of b. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.

Answer 1

The moment of inertia of the wire triangle about an axis perpendicular to the triangle's plane and passing through one of its vertices is (mb^2)/4.

Step-by-step explanation:

The moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices can be calculated using the parallel-axis theorem.

First, we need to find the moment of inertia of each side of the triangle when rotating about the axis passing through its center. The moment of inertia of a thin rod rotated about its center perpendicular to its length is given by I = (mL^2)/12, where m is the mass and L is the length.

Since each side of the triangle is a thin rod with length b, the moment of inertia of each side can be calculated as (mb^2)/12. Finally, using the parallel-axis theorem, we can calculate the moment of inertia of the wire triangle about the axis passing through one of its vertices as 3*(mb^2)/12 = (mb^2)/4.