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Oksi-84 [34.3K]
2 years ago
15

Find the moment of inertia about the y-axis of the thin semicirular region of constant density

Mathematics
1 answer:
arlik [135]2 years ago
3 0

The moment of inertia about the y-axis of the thin semicircular region of constant density is given below.

\rm I_y = \dfrac{1}{8} \times \pi r^4

<h3>What is rotational inertia?</h3>

Any item that can be turned has rotational inertia as a quality. It's a scalar value that indicates how complex it is to adjust an object's rotational velocity around a certain axis.

Then the moment of inertia about the y-axis of the thin semicircular region of constant density will be

\rm I_x = \int y^2 dA\\\\I_y = \int x^2 dA

x = r cos θ

y = r sin θ

dA = r dr dθ

Then the moment of inertia about the x-axis will be

\rm I_x = \int _0^r \int _0^{\pi}  (r\sin \theta )^2  \ r \  dr \  d\theta\\\\\rm I_x = \int _0^r \int _0^{\pi}  r^3 \sin ^2\theta  \  dr \  d\theta

On integration, we have

\rm I_x = \dfrac{1}{8} \times \pi r^4

Then the moment of inertia about the y-axis will be

\rm I_y = \int _0^r \int _0^{\pi}  (r\cos\theta )^2  \ r \  dr \  d\theta\\\\\rm I_y = \int _0^r \int _0^{\pi}  r^3 \cos ^2\theta  \  dr \  d\theta

On integration, we have

\rm I_y = \dfrac{1}{8} \times \pi r^4

Then the moment of inertia about O will be

\rm I_o = I_x + I_y\\\\I_o = \dfrac{1}{8} \times \pi r^4 + \dfrac{1}{8} \times \pi r^4\\\\I_o = \dfrac{1}{4} \times \pi r^4

More about the rotational inertia link is given below.

brainly.com/question/22513079

#SPJ4

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