At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
The moment of inertia about the y-axis of the thin semicircular region of constant density is given below.
[tex]\rm I_y = \dfrac{1}{8} \times \pi r^4[/tex]
What is rotational inertia?
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
[tex]\rm I_x = \int y^2 dA\\\\I_y = \int x^2 dA[/tex]
x = r cos θ
y = r sin θ
dA = r dr dθ
Then the moment of inertia about the x-axis will be
[tex]\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[/tex]
On integration, we have
[tex]\rm I_x = \dfrac{1}{8} \times \pi r^4[/tex]
Then the moment of inertia about the y-axis will be
[tex]\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[/tex]
On integration, we have
[tex]\rm I_y = \dfrac{1}{8} \times \pi r^4[/tex]
Then the moment of inertia about O will be
[tex]\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[/tex]
More about the rotational inertia link is given below.
https://brainly.com/question/22513079
#SPJ4
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.