Question

four particles each of mass M are placed at the corner of the square of side l the radius of gyration of the system about an Axis perpendicular to the square and passing through the centre is​

Answer 1

Ah it's a baby question

Square of side a

Then radius of gyration as you already know is -

K = root (I /m)

Now moment of inertia about required axis is -

4 * M (a /root 2)^2

a/root2 because half of diagonal

So,

I = 4Ma^2 /2 = 2Ma^2

Now (4M)K^2 = I

So K^2 = I /4M

So K^2 = a^2 /2

K = a /root2

Answered by an IIT JEE ASPIRANT and all India mathematics OLYMPIAD TOPPER in class 10th