Question
State and prove the theorem of 'parallel axes'.
✓
Answer
Statement: The moment of inertia $\left(l_0\right)$ of an object about any axis is the sum of its moment of inertia ( $l_c$ ) about an axis parallel to the given axis, and passing through the centre of mass and the product of the mass of the object and the square of the distance between the two axes. Mathematically, $I _0= I _{ C }+ Mh ^2$
Proof:
1.Consider an object of mass M. Axis MOP is an axis passing through point $O$.
2.Axis $A C B$ is passing through the centre of mass $C$ of the object, parallel to the axis MOP, and at a distance $h$ from it ( $\therefore h = CO$ ).
The theorem of parallel axes
3.Consider a mass element 'dm' located at point D. Perpendicular on OC (produced) from point D is DN.
4. The moment of inertia of the object about the axis $A C B$ is $I _{ C }=\int(D C)^2 dm$, and about the axis MOP, it is $I _0=\int$ (DO) ${ }^2 dm$.
$\text { 5. } I_0=\int(DO)^2 dm$
$=\int\left((DN)^2+(NO)^2\right) dm$
$=\int\left((DN)^2+(NC)^2+2 \cdot NC \cdot CO+(CO)^2\right) dm$
$=\int\left((DC)^2+2 NC \cdot h+h^2\right) dm . . . . . . . . . . . . .(\text { using Pythagoras theorem in } \Delta DNC)$
$=\int(DC)^2 dm+2 h \int NC \cdot dm+h^2 \int dm$
Now, $\int(D C)^2 dm = I _{ C }$ and $\int dm = M$
6.NC is the distance of a point from the centre of mass. Any mass distribution is symmetric about the centre of mass. Thus, from the definition of the centre of mass, $\int NC . dm =0$
$\therefore I_0=I_{C}+M \cdot h^2$
This is the mathematical form of the theorem of parallel axes.
Proof:
1.Consider an object of mass M. Axis MOP is an axis passing through point $O$.
2.Axis $A C B$ is passing through the centre of mass $C$ of the object, parallel to the axis MOP, and at a distance $h$ from it ( $\therefore h = CO$ ).
The theorem of parallel axes
3.Consider a mass element 'dm' located at point D. Perpendicular on OC (produced) from point D is DN.
4. The moment of inertia of the object about the axis $A C B$ is $I _{ C }=\int(D C)^2 dm$, and about the axis MOP, it is $I _0=\int$ (DO) ${ }^2 dm$.
$\text { 5. } I_0=\int(DO)^2 dm$
$=\int\left((DN)^2+(NO)^2\right) dm$
$=\int\left((DN)^2+(NC)^2+2 \cdot NC \cdot CO+(CO)^2\right) dm$
$=\int\left((DC)^2+2 NC \cdot h+h^2\right) dm . . . . . . . . . . . . .(\text { using Pythagoras theorem in } \Delta DNC)$
$=\int(DC)^2 dm+2 h \int NC \cdot dm+h^2 \int dm$
Now, $\int(D C)^2 dm = I _{ C }$ and $\int dm = M$
6.NC is the distance of a point from the centre of mass. Any mass distribution is symmetric about the centre of mass. Thus, from the definition of the centre of mass, $\int NC . dm =0$
$\therefore I_0=I_{C}+M \cdot h^2$
This is the mathematical form of the theorem of parallel axes.
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