
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
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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.
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