Find the polar moment of inertia of a hollow circular section of external diameter ‘D’ and internal diameter ‘d’ 4.Define principal axes and principal moment of inertia.5.Locate the centroid and calculate the moment of inertia about centroidal axes of a semicircular lamina of radius 2m. 6.A semicircular area having a radius of 100 mm is located in the XY-plane such that its diameter coincides with Y-axis.Determine the X-coordinate of the center.
The Correct Answer and Explanation is :
Let’s break down and address each of your questions:
1. Polar Moment of Inertia of a Hollow Circular Section
For a hollow circular section with external diameter ( D ) and internal diameter ( d ), the polar moment of inertia ( J ) can be calculated using the formula:
[
J = \frac{\pi}{2} \left( D^4 – d^4 \right)
]
This formula is derived from subtracting the moment of inertia of the inner circle (radius ( d/2 )) from the moment of inertia of the outer circle (radius ( D/2 )) around the center of the section. The polar moment of inertia represents the resistance to torsion of the object.
2. Principal Axes and Principal Moment of Inertia
Principal Axes: These are the axes about which the product of inertia is zero. They represent the axes where the distribution of mass is most symmetrical and provide the principal moments of inertia.
Principal Moment of Inertia: The moments of inertia about the principal axes are the principal moments of inertia. These are the minimum and maximum values of the moment of inertia for a given object, and they represent the axes around which the object’s mass distribution is most optimal for rotational motion.
3. Moment of Inertia of a Semicircular Lamina
For a semicircular lamina with radius ( r = 2 ) m, the moment of inertia about its centroidal axes can be calculated using the formula for a semicircular lamina:
[
I_{\text{centroid}} = \frac{1}{8} m r^2
]
Where:
- ( m ) is the mass of the semicircular lamina.
- ( r = 2 ) m is the radius of the semicircle.
To calculate the moment of inertia of the lamina about the centroidal axes, you would need to know the mass of the lamina. Without mass, we can only express it in terms of area:
[
I_{\text{centroid}} = \frac{1}{8} A r^2
]
Where ( A ) is the area of the semicircle, which is ( \frac{1}{2} \pi r^2 ).
For ( r = 2 \, \text{m} ), the area ( A ) is:
[
A = \frac{1}{2} \pi (2^2) = 2\pi \, \text{m}^2
]
So:
[
I_{\text{centroid}} = \frac{1}{8} \times 2\pi \times 2^2 = \frac{1}{8} \times 8\pi = \pi \, \text{kg m}^2
]
4. X-Coordinate of the Center of a Semicircular Area
For a semicircular area with a radius of 100 mm, whose diameter lies along the Y-axis, the centroid’s X-coordinate can be determined using the formula for the centroid of a semicircle:
[
x = \frac{4r}{3\pi}
]
Where:
- ( r = 100 ) mm or 0.1 m.
Thus:
[
x = \frac{4 \times 0.1}{3\pi} = \frac{0.4}{3\pi} \approx 0.0424 \, \text{m} \, (42.4 \, \text{mm})
]
So, the X-coordinate of the center of the semicircular area is approximately 42.4 mm.
Explanation Summary:
- The polar moment of inertia of a hollow circular section is determined by the difference between the moments of inertia of the external and internal circles.
- Principal axes are the axes where the mass distribution is most symmetrical, and the principal moments of inertia are the corresponding moments of inertia about these axes.
- For a semicircular lamina, the centroidal moment of inertia is calculated based on the lamina’s radius, and it depends on the area and mass.
- The X-coordinate of the centroid of a semicircular area with its diameter along the Y-axis is derived using standard formulas for centroid calculation, specifically for semicircles.
These concepts are fundamental in structural and mechanical engineering for determining the behavior of objects under rotational forces.