Question

Moment of inertia of a uniform quarter disc of radius r and mass m about an axis through its centre of mass and perpendicular to its plane is :

Answer 1

The Moment of Inertia Of a Uniform Quarter  disc of Radius r and mass m about axis passing through its center of mass  and perpendicular to its plane is $I_{C} = \dfrac{mr^2}{2}- m\left(\sqrt{2}\times \dfrac{4r}{3 \pi} \right)^{2}$

Explanation:

• Mass Moment of inertia is a product of mass and square of distance from center of mass to reference axis about which moment has to be taken.

        $I_m = mr^2$

• First of all we have to consider a quarter disc of radius r and mass m,  now we can apply parallel axis theorem as:-

        $I_c = I_g + m h^2$

       where $I_c$ - Moment of inertia about center of mass

       $I_G$ - Moment of Inertia of small element about its center of mass

       m - mass of the disc

       r - radius of the disc

• Now we can apply the moment of inertia for a quarter disc by parallel axis theorem as:-

       $\dfrac{mr^2}{2} = I_c + \left(m \dfrac{4r}{3\pi} \times \sqrt {2} \right) ^2$

 we get-

   The moment of inertia of quarter disc about an axis passing through its            center of mass and perpendicular to the pane is -

      $I_C = \dfrac{mr^2}{2} - \left (m {\sqrt{2} \times \dfrac{4r}{3\ip} \right)^2$

you can learn more about Moment of inertia here:

https://brainly.in/question/1443750

 

Parallel Axis theorem for Moment of inertia can be learn from here:

https://brainly.in/question/2765327