At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
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
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
