Question

An equilateral traingular wire frame is made from 3 rods of equal mass and length l each. The frame is rotated about an axis perpendicular to the plane of the frame and passing through its end. What is the radius of gyration of the frame

Answer 1

Answer: radius of gyration = L(3/2)^½

Explanation: I = MK²

=› K = (I/M)^½

I = (3/2)ML²

If u need the derivation of I = (3/2)ML²

Then ask in comment section.

Answer 2

The radius of gyration of the frame is l/√2.

The MoI of a Rod of mass M and length L that passes through the center of the rod perpendicular to the rod itself will be = 1/12 ml²

The MoI of a Rod of mass M and length L that passes through one end of the rod perpendicular to the rod itself will be = 1/3 ml²

AD = √3/2L

MoI of BC about the axis through perpendicular to the plane of the triangle will be -

= 1/12ml² + m(√3/2l)²

= 1/12ml² + 3/4ml²

= 5/6ml²

According to the principle of superposition -

= 1/3ml² + 1/3ml² + 5/6ml²

= 9/6ml²

= 3/2ml²

Therefore, the radius will be -

k = √ 3/2ml²/3m

= l/√2