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1800-102-2727Sphere is another continuous body whose geometry is symmetrical about its centre. If we think about a sphere, what are the types of spheres that we can imagine?
Complete solid sphere, solid hemisphere, complete hollow sphere, hollow hemisphere.
Since these all spheres are continuous, we can calculate its centre of mass by only considering its infinitesimal element and integrate it throughout.
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The position of the centre of mass depends on the way the mass is distributed through the sphere.
So, if the mass is distributed uniformly throughout the sphere then because of symmetry the centre of mass lies at the centre of the sphere.If we go in any direction from centre of the sphere we will come across the same mass and it can be considered that all the mass is concentrated at the geometrical centre of the sphere and whatever force we apply the acceleration of the centre will be equal to the whole sphere’s acceleration.
So, we can consider that sphere as a point mass concentrated at its centre and apply all the dynamics on that centre.
We are considering a solid hemisphere of mass M and has the radius R. The centre of mass will lie on the vertical line passing through the centre of the hemisphere, the vertical line is also normal to the base. In order to find the centre of mass, we have to consider an element.
We are taking an elemental disc at a height h from the base of the hemisphere. The mass of the elemental disc is dM and the width is dy.
The radius of the disc is-
—(1)
Mass of the disc-
Substitute equation (1) in equation (2),
Y-coordinate of Centre of mass-
Here is the y-coordinate which represents the height of the elemental disc from the base.
Putting the value of and calculating the centre of mass, we get
Centre of Mass of the solid hemisphere,
Here is the radius of the hemisphere.
We are having a hollow hemisphere of mass M and radius R. The centre of mass of the hollow hemisphere will lie on the y-axis, which is the line passing through the centre of the base of the hollow hemisphere.
We are considering an elemental strip of width and has a mass . The radius of the elemental ring is
The elemental mass
The y-coordinate of the centre of mass,
Putting the values of y and dM, we get
Therefore, the centre of mass of a hollow hemisphere will be at along the y-axis. Here is the radius of the hollow hemisphere.
Question.1 Two hollow hemispheres of mass and and the same radius are attached as shown in the figure. Find the location of the centre of mass of the whole system.
Answer. Centre of mass will lie along y- axis since along x-axis there is symmetry -
Applying the formulae of centre of mass along y- axis
So the centre of mass will lie distance above the x-axis.
Question.2 Find the centre of mass of a uniform plate having semicircular outer and inner boundaries of radius R1 and R2, respectively.
Answer
Given,
Outer radius = R1
Inner radius = R2
Consider an element plate of mass dm and thickness dr at a radius r as shown in the figure.
Let be the area density of the plate.
Mass of the smaller plate is as follows
Since the plate is symmetrical about the y-axis, the centre of mass of the plate lies on the y-axis.
Let the centre of mass be (0, ycom).
where centre of mass of the elemental plate
Therefore, the centre of mass of given plate is
Question.3 Three identical metal balls, each of radius r, are placed touching each other on a horizontal surface such that an equilateral triangle is formed when the centres of the three balls are joined. What is the location of the centre of mass?
Answer.
We know that the centre of mass of symmetric bodies lies at their geometric centre. Thus, the centre of mass of each of the balls will lie on their respective centres.
Thus, the system can be replaced by three point masses located at the centres of the balls as shown in the figure.
Let the centre of mass of the system be
Then,
Also,
Question.4 If a = 1 m, find the position of the centre of mass of the section with uniform mass distribution as shown in the figure.
Answer
In this figure, a square of side a has been removed from the larger square of side 2a.
Since both the squares of side a and 2a are uniform, their centre of masses (Co and C ) lie on their respective geometric centres as shown in the figure.
Thus,
Co = (a, a)
Let M be the mass of the original larger square and m be the mass of the smaller square.
Let be the area density of the section.
Mass of smaller section,
Let be the centre of mass of the given figure.
Since some mass is removed, we have to take the removed mass as negative in the formula.
Let x1 and x2 be the x-coordinates of the centre of masses of the original and removed part, respectively.
Then,
Therefore, the COM of the given figure is
If the length of each side is 2 m (a = 1), then the centre of mass is .
Question.1 The point of a system where an applied force causes linear acceleration without creating any angular acceleration is known as
Answer. B
Question.2 Which of the following statements is incorrect regarding the centre of mass.
a)The centre of gravity is the point through which the force of gravity acts on an object or a system.
b) Centre of mass of a body is a point at which the whole mass can be assumed as a point mass.
c) Centre of mass of the body will always be inside the body.
d)All of the above statements are correct .
Answer. c
Question.3 Where does the centre of mass of the solid sphere lie?
Answer. If the sphere's density is uniform or symmetrical about the centre, the centre of mass is at the centre.
Question.4 When does an object start rotating?
Answer. If we push on a rigid object at its centre of mass, then the object will always move as if it is a point mass. It will not rotate about any axis, regardless of its actual shape. If the object is subjected to an unbalanced force at some other point, then it will begin rotating about the centre of mass.