Question

A rectangular lamina of uniform density is situated with opposite corners at (0,0) and (15,4). calculate its radii of gyration about the y and x axes.

Answer 1

First, you have to find the moment of inertia along the x and y axes. Constant density is denoted as k.


$I_{x}= \int\limits^15_0\int\limits^4_0 {k y^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

$I_{y}= \int\limits^15_0\int\limits^4_0 {k x^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

Then, the radii of gyration for

x = √[I_x/m]
y = [I_y/m]

where m = k(15-4)² = 121k. Then,

x = y = [4880.33k/121k] = 40.33

I hope I was able to help you. Have a good day.