- Find the expression for mass moment of inertia of a sphere.
- Find the mass moment of inertia of a hollow cylinder about its axis passing through the centroid.
- Differentiate area moment of inertia from the mass moment of inertia.
- From a circular plate of diameter 100 mm, a circular part is cut out whose diameter is 50 mm as shown in Figure 11.25. Find the centroid of the remaining part.
The Correct Answer and Explanation is :
1. Mass Moment of Inertia of a Sphere
The mass moment of inertia (I) of a solid sphere of mass ( M ) and radius ( R ) about an axis passing through its center is:
[
I = \frac{2}{5} M R^2
]
Derivation:
- The differential mass element ( dm ) is taken as a thin shell at radius ( r ).
- The moment of inertia of each shell is integrated over the entire sphere volume.
- The result of this integral gives ( I = \frac{2}{5} M R^2 ).
For a hollow sphere, where mass is distributed only on the outer surface:
[
I = \frac{2}{3} M R^2
]
2. Mass Moment of Inertia of a Hollow Cylinder about its Centroidal Axis
For a hollow cylinder (thin-walled) with mass ( M ), inner radius ( R_1 ), and outer radius ( R_2 ), the moment of inertia about the central axis is:
[
I = \frac{1}{2} M (R_1^2 + R_2^2)
]
If it is a thin-walled hollow cylinder, then ( R_1 \approx R_2 = R ), and:
[
I \approx M R^2
]
3. Difference between Area Moment of Inertia and Mass Moment of Inertia
| Property | Mass Moment of Inertia | Area Moment of Inertia |
|---|---|---|
| Definition | Resistance to rotational motion (angular acceleration). | Resistance to bending or deflection. |
| Depends on | Mass distribution about the axis. | Area distribution about the axis. |
| Units | ( \text{kg} \cdot \text{m}^2 ) | ( \text{mm}^4 ) or ( \text{m}^4 ) |
| Used in | Dynamics, angular momentum calculations. | Beam bending, structural analysis. |
| Example Formula | ( I = \int r^2 dm ) | ( I = \int y^2 dA ) |
4. Centroid of the Remaining Circular Plate
We analyze the problem where a circular plate of diameter 100 mm has a 50 mm diameter circular cut-out. The centroid of the remaining shape can be determined using the method of composite areas.
Step 1: Define the System
- Original Plate: A circle of radius ( R = 50 ) mm centered at ( (0,0) ).
- Cut-out Circle: A circle of radius ( r = 25 ) mm, removed at a given location.
Step 2: Apply the Centroid Formula
The centroid of the remaining area is:
[
\bar{x} = \frac{A_1 x_1 – A_2 x_2}{A_1 – A_2}
]
[
\bar{y} = \frac{A_1 y_1 – A_2 y_2}{A_1 – A_2}
]
Where:
- ( A_1 = \pi (50^2) ) (original plate),
- ( A_2 = \pi (25^2) ) (cut-out),
- ( x_1 = 0, y_1 = 0 ) (original center),
- ( x_2 = 25, y_2 = 0 ) (assuming the cut-out is at ( (25,0) )).
Step 3: Compute the Values
[
A_1 = 2500\pi, \quad A_2 = 625\pi
]
[
\bar{x} = \frac{(2500\pi \cdot 0) – (625\pi \cdot 25)}{2500\pi – 625\pi}
]
[
\bar{x} = \frac{-15625\pi}{1875\pi} = -8.33 \text{ mm}
]
[
\bar{y} = 0
]
Final Answer:
The centroid of the remaining shape shifts 8.33 mm left from the original center.