Three point sized bodies each of mass M are fixed at three corners of a light triangular frame of side 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:
A
ML2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
3ML22
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
√3ML2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3ML2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is AML2 Height of the triangle is, h=√32L
The center (centroid) of the frame is at 23h=23√32L=1√3L from each of the vertex.
MI of the system about center of the frame is: I=3×(M(1√3L)2)=ML2