Question

# Three identical thin rods each of mass $$m$$ and length $$L$$ are joined together to form an equilateral triangular frame. The moment of inertia of frame about an axis perpendicular to the plane of frame and passing through a corner is:

A
2mL23
B
3mL22
C
4mL23
D
3mL24

Solution

## The correct option is A $$\dfrac{3mL^{2}}{2}$$Sum of MI of the two rods, at vertex of which the axis passes, is $$2\times mL^2/3=2mL^2/3$$For the third rod,  the distance between center of the rod and the axis is $$d=\sqrt3L/2$$Using parallel axis theorem, its MI is $$mL^2/12+3mL^2/4=5mL^2/6$$So total MI of the system is $$2mL^2/3+5mL^2/6=3mL^2/2$$Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More