
The surface density (mass/area) of a circular disc of radius ‘a’ depends on the distance from the center as $\rho \left( r \right)=A+Br$. Find its moment of inertia about an axis perpendicular to the plane of disc and passing through its center.
Answer
555.3k+ views
Hint:
The moment of inertia can be found by the relation $dI=dM\cdot {{r}^{2}}$, $r$ is the distance taken from the center of the disc to the axis about which the moment of inertia is being calculated. Using this relation, find the moment of inertia of the circular disc with the help of its surface density.
Complete answer:
Given the surface density (mass/area) of the circular disc of radius ‘a’ depends on the distance from the center as $\rho \left( r \right)=A+Br$
We are having the relation to find the moment of inertia as follows
$dI=dM\cdot {{r}^{2}}$ Where $r$ is the radius
This radius is nothing but the distance from the center of the disc to the axis perpendicular to the disc about which moment of inertia needs to be calculated.
Given that this radius is 'a’
We know that mass is equal to the product of volume and density
For a small elemental ring of width $dr$
We have $dM=\rho (r)\cdot 2\pi rdr$
Put this in the above formula and integrate on both sides
$dI=(A+Br)\cdot 2\pi rdr\cdot {{r}^{2}}$
$dI=(A2\pi {{r}^{3}}+B2\pi {{r}^{4}})dr$
Now integrate on both sides
We obtain $I=\left( \dfrac{A\pi {{r}^{4}}}{2}+\dfrac{B2\pi {{r}^{5}}}{5} \right)_{0}^{a}$
$I=\left( \dfrac{A\pi {{a}^{4}}}{2}+\dfrac{B2\pi {{a}^{5}}}{5} \right)$
Hence we obtain the moment of inertia of the circular disc as $I=\left( \dfrac{A\pi {{a}^{4}}}{2}+\dfrac{B2\pi {{a}^{5}}}{5} \right)$
Note:
We need to take care while substituting the value of $r$ or integrating with $r$ because it is not the actual radius of the disc; it is the distance from the center of the disc to the axis about which we are calculating the moment of inertia. Also it is the simplest way to find the moment of inertia, by taking an elemental area and obtaining its moment of inertia first and then integrating it to find the moment of inertia for the whole object instead calculating for the bulk at a time.
The moment of inertia can be found by the relation $dI=dM\cdot {{r}^{2}}$, $r$ is the distance taken from the center of the disc to the axis about which the moment of inertia is being calculated. Using this relation, find the moment of inertia of the circular disc with the help of its surface density.
Complete answer:
Given the surface density (mass/area) of the circular disc of radius ‘a’ depends on the distance from the center as $\rho \left( r \right)=A+Br$
We are having the relation to find the moment of inertia as follows
$dI=dM\cdot {{r}^{2}}$ Where $r$ is the radius
This radius is nothing but the distance from the center of the disc to the axis perpendicular to the disc about which moment of inertia needs to be calculated.
Given that this radius is 'a’
We know that mass is equal to the product of volume and density
For a small elemental ring of width $dr$
We have $dM=\rho (r)\cdot 2\pi rdr$
Put this in the above formula and integrate on both sides
$dI=(A+Br)\cdot 2\pi rdr\cdot {{r}^{2}}$
$dI=(A2\pi {{r}^{3}}+B2\pi {{r}^{4}})dr$
Now integrate on both sides
We obtain $I=\left( \dfrac{A\pi {{r}^{4}}}{2}+\dfrac{B2\pi {{r}^{5}}}{5} \right)_{0}^{a}$
$I=\left( \dfrac{A\pi {{a}^{4}}}{2}+\dfrac{B2\pi {{a}^{5}}}{5} \right)$
Hence we obtain the moment of inertia of the circular disc as $I=\left( \dfrac{A\pi {{a}^{4}}}{2}+\dfrac{B2\pi {{a}^{5}}}{5} \right)$
Note:
We need to take care while substituting the value of $r$ or integrating with $r$ because it is not the actual radius of the disc; it is the distance from the center of the disc to the axis about which we are calculating the moment of inertia. Also it is the simplest way to find the moment of inertia, by taking an elemental area and obtaining its moment of inertia first and then integrating it to find the moment of inertia for the whole object instead calculating for the bulk at a time.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

