physics-class-11-unit-07-chapter-06-moment-of-inertia-theorem-of-perpendicularparallel-axis-l-6-10-zwmecmrpuyy

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Moment of Inertia </primary>
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- Q: what is the analogues of mass (m) in rotational motion?

- Consider rotation about a fixed axis.

- Kinetic energy of the rotating body: $K=\sum_ i k_ i=\frac{1}{2} \sum m_ i\left(r_i \omega\right)^2$

- $K = \frac{1}{2} \omega^2 \sum_ i m_ i r_i^2$

- $\sum_ i m_ i r_i^2$ = moment of inertia

- $k=\frac{1}{2} I \omega^2$

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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/875e4b3a-ad0d-4137-5c10-b16df7bef600/public)

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<footer style = "font-size: 18px"> $\rightarrow$ Moment of Inertia and Parallel and Perpendicular Axes Theorems $\rightarrow$ <span style = "color:blue"> Moment of Inertia </span> $\rightarrow$ Properties of Moment of Inertia $\rightarrow$ Thin Circular Ring </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Properties of Moment of Inertia </primary>
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- (1) does not depend on $\omega$. It depends on mass, shape and size.

- (2) It is a charactisitic of a rigid body and also an axis about which it rotates

- (3) Measure of inertia in rotational motion.

- (4) Nature of rotation about an axis

- (5) [I] =$ML^2$ (kg $m^2$)

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<footer style = "font-size: 18px">Moment of Inertia and Parallel and Perpendicular Axes Theorems $\rightarrow$ Moment of Inertia $\rightarrow$ <span style = "color:blue"> Properties of Moment of Inertia </span> $\rightarrow$ Thin Circular Ring $\rightarrow$ Moment of Inertia of Equatorial Triangle and Rod </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Thin Circular Ring </primary>
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- Radius of the ring is R.

- Total mass is M.

- Moment of inertia of a circular ring about an axis,

- $I = \sum m_ i r_ i^2 = \sum_ i m_ i R^2 = R^2 \sum_ i m_i = M R^2$

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<footer style = "font-size: 18px">Moment of Inertia $\rightarrow$ Properties of Moment of Inertia $\rightarrow$ <span style = "color:blue"> Thin Circular Ring </span> $\rightarrow$ Moment of Inertia of Equatorial Triangle and Rod $\rightarrow$ Radius of Gyration </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Moment of Inertia of Equatorial Triangle and Rod </primary>
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- (1) $I _ {\text{altitude}}= m_1(a)^2+m_2(4)^2+m_3\left(\frac{y}{2}\right)^2$

- $=2\times m(\frac{a}{2})^2 = \frac{m a^2}{2}$

- (2) $I=m\left(\frac{l}{2}\right)^2 + m\left(\frac{l}{2}\right)^2$

- $m \frac{l^2}{2} =m k^2$

- $I=(x) M\left(L^2\right) = Mk^2$

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<footer style = "font-size: 18px">Properties of Moment of Inertia $\rightarrow$ Thin Circular Ring $\rightarrow$ <span style = "color:blue"> Moment of Inertia of Equatorial Triangle and Rod </span> $\rightarrow$ Radius of Gyration $\rightarrow$ Moment of Inertia of Rod </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Radius of Gyration </primary>
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- $M\left(L^2\right) = Mk^2$

- k is called as radius of gyration.

- $\rho$ = Mass per unit length.

- M = $\rho$L

- $ dI  = d m x^2  =(d \times \rho) x^2$

- $I=\int_0^L \rho x^2 d x$

- $\frac{\rho L^3}{3}  = \rho \frac{L^2}{3} =\frac{M L^2}{3}$

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<footer style = "font-size: 18px">Thin Circular Ring $\rightarrow$ Moment of Inertia of Equatorial Triangle and Rod $\rightarrow$ <span style = "color:blue"> Radius of Gyration </span> $\rightarrow$ Moment of Inertia of Rod $\rightarrow$ Moment of Inertia </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Moment of Inertia of Rod </primary>
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- The moment of inertia of a rod which is uniform rod about an axis: $I = \frac{M L^2}{3}$

- The moment of inertia of a rod which is uniform rod about center: $I=\frac{M}{2} \frac{(L/2)^2}{3} \times 2 = \frac{M L^2}{12}$

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<footer style = "font-size: 18px">Moment of Inertia of Equatorial Triangle and Rod $\rightarrow$ Radius of Gyration $\rightarrow$ <span style = "color:blue"> Moment of Inertia of Rod </span> $\rightarrow$ Moment of Inertia $\rightarrow$ Moment of Inertia </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Moment of Inertia </primary>
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- <span style="color:blue">Moment of Inertia of a circular disc</span>

- $\rho$: Mass per unit area

- $d I=(2 \pi r d r \rho) r^2$

- $I=2 \pi \rho \int_{r=0}^R r^3 d r=\frac{M e R^4}{2} \rho$

- $M=\pi R^2 \rho$, I =$\frac{MR^2}{2}$

- Solid cylinder: $I_{solid}$ cylinder about an axis passing through the axis of the cylinder =$\frac{MR^2}{2}$

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<footer style = "font-size: 18px">Radius of Gyration $\rightarrow$ Moment of Inertia of Rod $\rightarrow$ <span style = "color:blue"> Moment of Inertia </span> $\rightarrow$ Moment of Inertia $\rightarrow$ Perpendicular Axis Theorem </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Moment of Inertia </primary>
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- <span style="color:blue">Hollow cylinder</span>

- $\rho$: mass per unit area

- $d z=2 \pi R d l(B) R^2$

- $I=2 \pi R^3\rho \int d l=2 \pi R^3\rho l$

- Mass of the cylinder $=2 \pi R L \rho$
        
- $I=(2 \pi r l\rho) R^2=M R^2$

- <span style="color:blue">MI of a solid sphere</span>

- $I=\int_0^R\left(4 \pi r^2 l r \rho\right) r^2$

- $=4 \pi \frac{8}{5} \rho =\frac{3 M R^2}{5}$

- $M_{\text {sphere}}=\frac{4}{3} \pi R^3 \rho$

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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/a388fc07-7a2e-4925-e989-d48b43952600/public)

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<footer style = "font-size: 18px">Moment of Inertia of Rod $\rightarrow$ Moment of Inertia $\rightarrow$ <span style = "color:blue"> Moment of Inertia </span> $\rightarrow$ Perpendicular Axis Theorem $\rightarrow$ Parallel axis theorem </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Perpendicular Axis Theorem </primary>
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- The MI of a planar body about an axis perpendicular to its plane is equal to the sum of MI's about the two perpendicular axes throughout with the perpendicular axis and lying in the plane.

- $I_z=I_x+I y$

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<footer style = "font-size: 18px">Moment of Inertia $\rightarrow$ Moment of Inertia $\rightarrow$ <span style = "color:blue"> Perpendicular Axis Theorem </span> $\rightarrow$ Parallel axis theorem $\rightarrow$ Circular Ring About Tangent </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Parallel axis theorem </primary>
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- $I_{cm}$ known

- $I_z$ = ?

- $\frac{I}{z}=I_z+M a^2$

- $I_z=\frac{M L^2}{12}$

- $I=\int_0^L I x^2 d x=\frac{\rho L^3}{3} =\rho L \frac{L^2}{3} =\frac{M L^2}{3}$

- $I =\frac{M}{2} \frac{(L /2)^2}{3} \times 2=\frac{M L^2}{12}$

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<footer style = "font-size: 18px">Moment of Inertia $\rightarrow$ Perpendicular Axis Theorem $\rightarrow$ <span style = "color:blue"> Parallel axis theorem </span> $\rightarrow$ Circular Ring About Tangent $\rightarrow$  </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
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- **Example:**

- $2I_{diameter}$= $MR^2$ (Perpendicular axis theorem)

- $I_{diameter}=\frac{MR^2}{2}$

- $I_{Tangent}$ = $I_{Diameter}+MR^2$ (Perpendicular axis theorem)

- $=\frac{M R^2}{2}+M R^2=\frac{M R^2}{3}$

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<footer style = "font-size: 18px">Perpendicular Axis Theorem $\rightarrow$ Parallel axis theorem $\rightarrow$ <span style = "color:blue"> Circular Ring About Tangent </span> $\rightarrow$ Parallel axis theorem $\rightarrow$ Thank You </footer>

### <Success style="font-size:25px"> Moment Of Inertia Theorem Of Perpendicularparallel Axis L-6 </Success>
### <primary style="font-size:24px"> Parallel axis theorem </primary>

- $I_{cm}$ known

- $I_z$ = ?

- $I_ z=I_ z+M_z^2$

- $I_z=\frac{ML^2}{12}$

- $I_ z=I_z+M(\frac{L}{2})^2$

- $=\frac{ML^2}{12}+\frac{ML^2}{4}=\frac{ML^2}{3}$

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<footer style = "font-size: 18px">Parallel axis theorem $\rightarrow$ Circular Ring About Tangent $\rightarrow$ <span style = "color:blue"> Parallel axis theorem </span> $\rightarrow$ Thank You $\rightarrow$ </footer>