Question

Four rods of equal mass m each , Ab , BC , CA , DB are placed as shown. The moment of inertia of the system about B perpendicular to the plane ABC is (AB=a)

Answer 1

@JamesJ

Answer 2

Obviously break down the problem rod by rod.

Answer 3

And if you're really stuck, you're just going to integrate rod by rod.

Answer 4

I (BC) = ma^2/6 ryt ?

Answer 5

Its basically a maths ques, you just need to find rod lengths, after doing that you just have to calculate moment of inertia of each rod about B and add all

Answer 6

Yup . i Have Calculated the length of rods BC = a/sqrt(2) , AC = a/sqrt(2) , DB = (sqrt2) * a / sqrt3

Answer 7

I (BC) = ma^2/6 ryt ?

Answer 8

yep

Answer 9

I (DB) = ma^2/18 ?

Answer 10

won' that be (2/9)ma^2

Answer 11

Hw ?