A. Find the centre of mass of a hemispherical shell of constant density and inner radius and outer radius r2..
B. Find the centre of mass of a uniformly solid cone of base diameter 2a and height h.
C. Find the centre of mass of a uniformly solid cone of base diameter 2a and height h and a solid hemisphere of radius a where the two bases are touching. .
D. Find the centre of mass of a uniform wire that subtends an arc & if the radius of the circular arc is a, as shown in Figure 9-A
The Correct Answer and Explanation is :
Let’s solve each of the problems one by one:
A. Centre of Mass of a Hemispherical Shell
The hemispherical shell has a constant density, an inner radius ( r_1 ), and an outer radius ( r_2 ). Since it is a thin shell, we can assume all the mass is distributed uniformly across the surface. Due to the shell’s symmetry, its center of mass will lie on the central vertical axis. The height ( z ) of the center of mass of a hemispherical shell above the base is given by the formula:
[
z = \frac{3}{8} r
]
Here, ( r ) should be the average radius of the shell, ( \frac{r_1 + r_2}{2} ). This formula applies because the mass is symmetrically distributed over the hemisphere’s surface. Hence:
[
z = \frac{3}{8} \left(\frac{r_1 + r_2}{2}\right)
]
B. Centre of Mass of a Solid Cone
A uniformly solid cone with a base diameter of ( 2a ) and height ( h ). The center of mass for a solid cone is located along its axis of symmetry. It is calculated from the base (where the base is considered at ( z = 0 )) as:
[
z = \frac{h}{4}
]
This result derives from the integration of the density over the volume of the cone, focusing on the ( z )-coordinate, where the cone’s density does not vary with the radius or angular coordinate due to its uniformity.
C. Combined Centre of Mass of a Cone and Hemisphere
The setup involves a solid cone and a solid hemisphere, each with uniform density, where the cone’s base diameter ( 2a ) and the hemisphere’s radius ( a ) are touching at their bases. To find the combined center of mass, we calculate the individual centers of mass and then use them to find the system’s total center of mass.
For the cone:
[
z_{\text{cone}} = \frac{h}{4}
]
For the hemisphere:
[
z_{\text{hemisphere}} = \frac{3a}{8}
]
If ( M_c ) and ( M_h ) are the masses of the cone and hemisphere, respectively, the total center of mass ( z_{\text{total}} ) is given by:
[
z_{\text{total}} = \frac{M_c z_{\text{cone}} + M_h z_{\text{hemisphere}}}{M_c + M_h}
]
D. Centre of Mass of a Uniform Wire in a Circular Arc
A uniform wire bent into a circular arc subtends an angle ( \theta ) and has a radius ( a ). For such an arc, the center of mass is located along the radial line bisecting the arc. The distance ( R ) from the center of the circle to the center of mass of the arc along this line is given by:
[
R = \frac{2a \sin(\frac{\theta}{2})}{\theta}
]
This formula arises from considering the symmetry and distribution of mass along the arc.
Generating Image for D
For the wire bent into a circular arc, let me generate an image that illustrates the scenario:
{
"prompt": "Illustration of a uniform wire bent into a circular arc, subtending an angle theta. The arc is part of a circle with radius a. Include radial lines indicating the center of mass location, where the radius line bisects the arc. Label the radius 'a', the angle 'theta', and indicate the center of mass along the radial line. Artistic, clear, educational diagram.",
"size": "1024x1024"
}