Get the answers you need from a community of experts on IDNLearn.com. Get prompt and accurate answers to your questions from our community of knowledgeable experts.
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 value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide clear and concise answers, so visit us again soon.
