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Eddi Din [679]
3 years ago
13

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

bout the y and x axes.
Mathematics
1 answer:
Sphinxa [80]3 years ago
7 0
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.
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Step-by-step explanation:

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