Moment Of Inertia Of a Hollow Sphere

Moment of inertia is a physical quantity that quantifies an object's rotational inertia. It is similar to mass in linear motion and plays a crucial role in rotational dynamics. Moment of inertia is essential in various engineering applications, such as designing rotating machinery, understanding the stability of structures, and analyzing the motion of rigid bodies in physics and engineering problems. Understanding moments of inertia is crucial for solving problems in rotational mechanics and dynamics, providing insights into how objects behave when subjected to rotational forces and torques.

1.0Moment Of Inertia

• The measure of the property by which a body revolving about an axis opposes any change in its rotational motion is known as the Moment of Inertia.
• The moment of inertia of a particle concerning an axis of rotation is equal to the product of its mass and the square of its distance from the rotational axis.

$\mathrm{I}=\mathrm{m}{r}^2$, r = perpendicular distance from the axis of rotation

• Moment of Inertia of a system of particles

$\mathrm{I}=m_1{r}_1^2+m_2{r}_2^2+m_3{r}_3^2+\dots \dots \dots \dots =\sum \mathrm{m}{r}^2$

• The moment of Inertia depends on the following:

1. Mass of the body
2. Mass distribution of the body
3. Position of the axis of rotation.

• Moment of inertia does not depend on:-

1. Angular velocity
2. Angular acceleration
3. Torque
4. Angular momentum

• SI Unit-kg-m²
• CGS Unit-g-cm²
• Dimensional Formula - [M¹L²T⁰]

2.0Moment of Inertia For Continuous Mass Distribution

Choose an appropriate element (of mass dm) on the body at a particular distance r from the axis .Then r² dm integrated over the appropriate limits to cover the whole body gives a moment of Inertia.

$I_{AB}=\int r^{2}dm$

Note: Moments of inertia of two or more than two bodies can be added or subtracted only when all moments of inertia are written on the same axis.

3.0Moment of Inertia of a Hollow Sphere

• Moment of Inertia of Hollow Sphere about its diametric axis $=\frac{2}{3}\mathbf{M}{R}^2$

4.0Moment of Inertia of a Solid Sphere

• Moment of Inertia of a solid Sphere about its diametric axis = $\frac{2}{5}\mathbf{M}{R}^2$

Illustration: If I₁,I₂,I₃ is the MOI of  the solid sphere, hollow sphere, and ring, each having the same mass and radius, the statement that best holds true concerns their respective moments of inertia about their geometric axes

1. I₁>I₂>I₃       
2. (2) I₃>I₂>I₁    
3. (3) I₂>I₁>I₃    
4. (4) I₂>I₃I₁

Solution:(2)   I₃>I₂>I₁ 

$I_1=\frac{2}{5}\mathrm{M}{R}^2$  ; 

$I_2=\frac{2}{3}\mathrm{M}{R}^2$  ;                

$I_3=\mathrm{M}{R}^2$

5.0Derive Moment of Inertia of a Hollow Sphere

Mass of element: $\mathrm{dm}=\frac{M}{4\pi R^2}\times 2\pi R\mathrm{Cos}\theta \cdot \mathrm{Rd}\theta \Rightarrow dm=\frac{1}{2}\mathrm{M}\mathrm{Cos}\theta \mathrm{d}\theta$

$\mathrm{dl}=\mathrm{dm}(R\mathrm{Cos}\theta {)}^2$

$\mathrm{dl}=\left(\frac{1}{2}\mathrm{M}\mathrm{Cos}\theta \mathrm{d}\theta \right)(R\mathrm{Cos}\theta {)}^2$

$=\frac{1}{2}{\mathrm{MR}}^2{\mathrm{Cos}}^3\theta \mathrm{d}\theta$

Moment Of Inertia of Hollow:

$\text{ Sphere }=\mathrm{I}=\int dI={\int }_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}\frac{1}{2}{\mathrm{MR}}^2{\mathrm{Cos}}^3\theta d\theta$

$\frac{1}{2}{\mathrm{MR}}^2{\int }_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}{\cos }^3\theta d\theta$, on solving integration we get

$\mathrm{Ml}=\frac{2}{3}{\mathrm{MR}}^2$

6.0Derivation of Moment of Inertia of a Solid Sphere 

Mass of element: $\mathrm{dm}=\frac{M}{\frac{4}{3}\pi R^3}\times 4\pi {\mathrm{r}}^2\mathrm{dr}$

$\mathrm{dm}=\frac{3M}{R^3}{r}^2\mathrm{dr}$

Moment of Inertia of the element is, $\mathrm{d}=\frac{2}{3}\mathrm{dm}{\mathrm{r}}^2$

$\mathrm{dl}=\frac{23M}{3R^3}{\mathrm{r}}^2\mathrm{dr}\cdot {\mathrm{r}}^2$

$\mathrm{dl}=\frac{2M}{R^3}{\mathrm{r}}^4\mathrm{dr}$

Moment of inertia of Solid Sphere:

$I\mathrm{I}=\int \mathrm{d}\mathbf{d}={\int }_0^R\frac{2M}{R^3}{\mathrm{r}}^4\mathrm{dr}=\frac{2M}{R^3}{\int }_0^R{\mathrm{r}}^4\mathrm{dr}=\frac{2M}{R^3}\times \frac{R^5}{5}=\frac{2}{5}{\mathrm{MR}}^2$

$\mathrm{I}=\frac{2}{5}{\mathrm{MR}}^2$

7.0Formula of M.I of a Hollow Sphere and Solid Sphere

• Moment of Inertia of a Hollow Sphere = $\frac{2}{3}{\mathrm{MR}}^2$
• Moment of Inertia of a Solid Sphere = $=\frac{2}{5}{\mathrm{MR}}^2$

8.0Solved Problems

Question 1. Solid and hollow spheres of the same mass have the same M.I. about their geometrical axes. What would the ratio of their radii be?

Solution:

$I_{\text{Solid Sphere }}=I_{\text{Hollow Sphere }}$

$\frac{2}{5}{\mathrm{Mr}}_1^2=\frac{2}{3}M{r}_2^2$

$\Rightarrow \frac{r_1}{r_2}=\frac{\sqrt{5}}{\sqrt{3}}$

Question 2. Moment of Inertia of a sphere about its diameter is $\frac{2}{5}{\mathbf{MR}}^2$.What is its moment of inertia about an axis perpendicular to its two diameter and passing through their point of intersection?

Solution: Moment of Inertia of a sphere about its diameter is $\frac{2}{5}{\mathbf{MR}}^2$.

Two diameter always intersect at the centre, however a sphere in symmetric from all directions, so taking any other diameter axis will have moment $\frac{2}{5}{\mathbf{MR}}^2$. Because the axis in question is also the diameter of the sphere.

Question 3. A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. Find the ratio of their kinetic energies of $\frac{E_{\text{Sphere }}}{E_{\text{Cylinder }}}$?

Solution:    $\frac{E_{\text{Sphere }}}{E_{\text{Cylinder }}}=\frac{\frac{1}{2}{I}_s{\omega }_s^2}{\frac{1}{2}{I}_C{\omega }_C^2}=\frac{I_s{\omega }_s^2}{I_c{\omega }_c^2}$

$I_S=\frac{2}{5}m{R}^2,I_C=\frac{1}{2}m{R}^2,{\omega }_C=2{\omega }_S$

$\frac{E_{\text{Sphere }}}{E_{\text{Cylinder }}}=\frac{\frac{2}{5}m{R}^2\times {\omega }_S^2}{\frac{1}{2}m{R}^2\times {\left(2{\omega }_S\right)}^2}$

$or\frac{E_{\text{Sphere }}}{E_{\text{Cylinder }}}=\frac{2}{5}\times \frac{1}{2}=\frac{1}{5}$

Question 4. What is the moment of inertia of a solid sphere of density() and radius R relative to its diameter?

Solution: Moment of Inertia of a Solid Sphere $=\frac{2}{5}{\mathrm{MR}}^2$

$\mathrm{M}=\mathrm{V}\times \rho =\frac{4}{3}\pi R^3\rho$

$\text{ M.I }=\frac{2}{5}\times \frac{4}{3}\pi R^3\rho \times R^2$

$\text{ M.I }=\frac{2}{5}\times \frac{4}{3}\times \frac{22}{7}\times R^3\rho \times R^2$

$\text{ M.I }=\frac{176}{105}{R}^5\rho$

Question 5. Calculate the kinetic energy of a hollow sphere with a mass of 3 kg, a radius of 1 m, and an angular velocity of 10 rad/s, rotating about an axis that passes through its diameter.

Solution:   Kinetic energy of Hollow sphere = $\frac{1}{2}I{\omega }^2$

$\Rightarrow \frac{1}{2}\times \frac{2}{3}M{R}^2\times {\omega }^2$

$\Rightarrow \frac{1}{2}\times \frac{2}{3}\times 3\times (1{)}^2\times (10{)}^2=100\mathrm{J}$