Contents
Physics Topics are often described using mathematical equations, making them precise and quantifiable.
Is the Centre of Gravity Below the Centre of Mass?
It is known that an extended object is an aggregation of a large number of particles. If the masses of these constituents are m1, m2, m3,……, the earth pulls these particles with respective forces m1g, m2g, m3g, …… towards its centre [Fig.]. Resultant of these like parallel forces acting on the particles is the weight W of the object. This weight W, acts vertically downward through a definite point. This point is called the centre of gravity (G) of the body. Like the centre of mass, the centre of gravity may be located inside or out side the body.
The respective distances between the particles remain fixed. Hence, the resultant of the weights also remains the same both in magnitude and in direction and acts through the same point. Thus, it can be stated that the position of the centre of gravity of a body does not depend on the orientation of the body.
Definition: The force of gravity that is exerted on an extended object always acts through a unique point. This point is the centre of gravity of the object.
To find the position of the centre of gravity: The position of the centre of gravity (G) can be determined using the rules for finding the resultant of like parallel forces [see Section 1.4]. The magnitude of the resultant is equal to the weight of the body:
W = m1g + m2g + m3g + ….. = Mg
[Here, M = m1 + m2 + m3 + ….. = total mass of the object]
Let the coordinates of m1, m2, … be (x1, y1, z1), (x2, y2, z2), respectively and the point of action of the resultant W, i.e., the coordinates of the centre of gravity be (x, y, z).
Some Important information about the centre of gravity: To establish equation (1), the magnitude of g is assumed to be the same for all particles constituting the body. This assumption is acceptable for small objects only, where the variation of g is negligible. But if the body is very large, the magnitude of g varies from point to point and equation (1) is not applicable.
Equation (1) of this section and equation (2) of Section 1.2.1 are identical. Hence, if the magnitudes of acceleration due to gravity at all points within a body are same, the centre of mass and the centre of gravity become identical.
Uniqueness of centre of gravity: A body cannot have more than one centre of gravity. This is called the uniqueness of centre of gravity. Let us assume a body has two centres of gravity G and G’ [Fig.(a)]. From the definition of centre of gravity, irrespective of the orientation of the body, its weight should act through G and G’. But the weight of a body always acts vertically downwards. There fore, GG’ must always be a vertical line. But for all orientations of the body, GG’ cannot be vertical [Fig.(b)].
Hence, a body having two centres of gravity is not a possibility
Difference Between Centre of Mass And Centre of Gravity
i) Centre of mass of a body is such a point that a force applied at this point produces linear motion only and does not produce any rotational motion. Centre of gravity is the point through which the weight of a body acts vertically downwards.
ii) As the mass of a body remains constant anywhere in the universe, a body always has a definite centre of mass. But at places where the weight of the body becomes zero, there is no centre of gravity, e.g., a body inside an artificial satellite rotating around earth or a free falling body is weightless. So in these cases the body has no centre of gravity.
iii) If the magnitude of acceleration due to gravity remains the same at all points of a body, the centre of mass and the centre of gravity will be identical. But if the magnitude of gravitational acceleration is different at different points of a body, then the centre of mass will not coincide with the centre of gravity, e.g, for very large mountains like Himalaya acceleration due to gravity is greater at the bottom of the mountain, thats why even for an object of uniform density earths attraction is greater at the bottom. Again position of the centre of mass depends only on the density and not on the weight of the body. Thats why for massive and tall objects like these centre of gravity remains below the centre of mass.
Numerical Examples
Example 1.
A rod of weight W is kept horizontally on two knife edges with a distance d between them. Centre of gravity of the rod is at a distance x from A. Find the normal reactions at points A and B [Fig.].
Solution:
Let the centre of gravity of the rod be C. Normal reactions at A and B are R1 and R2. Weight W through C acting downwards, and R1 and R2 are the three coplanar forces that keep the rod in equilibrium.
Hence, W – R1 – R2 = 0 or, W = R1 + R2
Taking the moment of the forces about C,
-R1 ᐧ x + R2(d – x ) = 0 or R1x = R2(d – x)
or, x(R1 + R2) = R2d or, Wx = R2d
∴ R2 = W
and R1 = W – R2 = W – \(\frac{W x}{d}\) = W(1 – \(\frac{x}{d}\)).
Example 2.
Show that the centre of gravity of three equal weights suspended from three vertices of a triangle coincide with the centre of mass of the triangle.
Solution:
Let a median of the triangle ABC be AD [Fig.]. Identical weights W are hung from each of the vertices A, B and C. Resultant of the weights suspended from B and C = 2W and it acts from the point D. Resultant of the forces 2W at D and W at A is 3W. Suppose this resultant acts at G.
Here, W × AG = 2W × DG
or, AG = 2DG
Thus, G is the point of intersection of the medians of the triangle, which is the centre of mass.
Example 3.
A person of mass 80kg is standing on the top of a 18 kg ladder of length 6 m. Upper end of the ladder rests on a smooth vertical wall and the lower end is on the ground 3 m away from the vertical wall. What should be the minimum coefficient of friction between the floor and the ladder so that the system remains in equilibrium?
Solution:
Given that, for the ladder weight W acts through the centre of gravity, C (mid-point of AB) of the ladder [Fig.].
Weight W’ of the man acts at B downwards.
Normal reaction R’ by the wall acts at B.
Normal reaction R by the ground acts at A.
Limiting friction that acts along the floor at A, f = µR
where, µ is the coefficient of friction, required for equilibrium.
At equilibrium, the sum of all horizontal forces and the sum of all vertical forces will be zero separately.
µR – R’ = 0 or, µR = R’ ………. (1)
R – W – W’ = 0 or, R = W + W’ ……… (2)
Again the sum of the moments of all forces taken about A will be zero.
Example 4.
A uniform cylinder of diameter 8 cm is kept on a rough inclined plane, whose angle of inclination with the ground is 30°. What should be the maximum height of the cylinder so that it does not topple?
Solution:
The cylinder ABCD is kept on the plane of inclination 30°. Suppose the cylinder is on the verge of toppling over when its height is 2h. At this stage, the line of action of gravity must pass vertically through the end point A at the base of the cylinder.
From Fig.,
tan 30° = \(\frac{A E}{E G}\)
or, \(\frac{1}{\sqrt{3}}\) = \(\frac{r}{h}\), or, h = \(\sqrt{3}\)r = 4\(\sqrt{3}\) cm
Hence, maximum permissible height = 2h = 2 × 4\(\sqrt{3}\) = 8\(\sqrt{3}\) cm.
Example 5.
A hollow cylinder of height 100 cm and diameter 8 cm has its top end open. It contains water up to a height of 50 cm. Mass per square centimetre of the hollow cylinder is 9 g. Find the height of the centre of gravity of the water-cylinder system from the base of the cylinder.
Solution:
Centre of gravity of the curved surface of the empty cylinder is 50 cm above its base at P [Fig.].
Weight of the base of the empty cylinder
= π(4)2 × 9 × g dyn [g = acceleration due to gravity]
Weight of curved surface of the cylinder
= 2π × 4 × 100 × 9 × g dyn. Its centre of gravity is at G1, where PG1 = 50 cm.
Weight of water in the cylinder π(4)2 × 50 × 1 × g dyn and centre of gravity of this water is at G2 which is 25 cm above the base of the cylinder.
Suppose the centre of mass of the cylinder filled with water is at G, which is h cm above its base. The centre of gravity coincides with this centre of mass.
Taking moments of all the weights about P,
2π × 4 × 100 × 9 × G1P + π × 16 × 50 × 1 × G2P
= (π × 16 × 9 + 2π × 4 × 100 × 9 + π × 16 × 50 × 1) × GP
or, 2π × 4 × 100 × 9 × 50 + π × 16 × 50 × 1 × 25 = (π × 16 × 9 + 2π × 4 × 100 × 9 + π × 16 × 50 × 1) × h
or h = 46.7 cm.
Example 6.
A uniform narrow rod of mass M and length 2L is kept vertically, along the y-axis (as shown in Fig.), on a smooth horizontal plane. The lower end of the rod coincides with the origin (0, 0). Due to a slight disturbance at time t = 0, the rod slides along the positive x-axis and begins to fall. Determine
(i) the shift In the centre of gravity during the fall,
(ii) the equation of the locus of a point at a distance r from the lower end of the rod and also mention the shape of the locus.
Solution:
Initially (0, L) was the coordinates of the centre of gravity of the rod. Let B be the point at a distance r, from point O (0, 0), on the rod. So, (0, r) denotes the point B.
At t = 0, the rod is disturbed and it falls. During this fall there is no other force acting on it except the downward gravitational force.
i) We know that the centre of gravity is influenced only by an external force which is gravity in this case. So the centre of gravity will be shifted vertically downwards from (0, L) to (0, 0).
ii) Let (x, y) be the position of B’ at any moment during the fall of the rod, where B’ Is the point at a distance r, from the lower end O’ of the rod.
This is the equation of the locus of a point (x, y) and it is elliptical with its centre at the origin.
Example 7.
A circular metal plate of uniform thickness has a radius of 10 cm. A hole of radius 4 cm is punched on the plate a little away from its centre. Centres of the plate and of the hole are 5 cm apart Find the centre of gravity of the plate with the hole.
Solution:
Centre of gravity of the plate before the hole was punched was at its centre G [Fig.]. The centre of gravity of the circular part before the punching was at its centre G1.
Suppose the centre of gravity of the plate with the hole is at G2 on the line GG1.
Let W1 = weight of the plate with hole, W2 = weight of the disc that is cut off.
∴ W1 × GG2 = W2 × GG1 or, GG2 = \(\frac{W_2}{W_1}\) × GG1
But GG1 = 5 cm, W1 = π(102 – 42) × ρ (ρ = weight per unit area) and W2 = π × 42 × ρ
∴ GG2 = \(\frac{4^2}{10^2-4^2}\) × 5 = \(\frac{20}{21}\)cm
Hence, the centre of gravity of the system will be \(\frac{20}{21}\)cm to the left of the centre of the plate.
Example 8.
A mass of 10 kg is suspended using two strings. One string makes an angle of 60° with the vertical. What should be the angle made by the other string so that the tension in the string will be minimum? Find the tension in each wire. [HS ‘04]
Solution:
As the mass is in equilibrium, from the force diagram [Fig.] we get,
T1cosθ1 + T2cosθ2 = mg
or, T1cosθ1 = mg – T2cosθ2 ….. (1)
Also, T1sinθ1 = T2sinθ2 ……. (2)
From (2) and (1) we get,
\(\frac{\sin \theta_1}{\cos \theta_1}\) = \(\frac{T_2 \sin \theta_2}{m g-T_2 \cos \theta_2}\)
or, mg sinθ1 = T2(sinθ1cosθ2 + cosθ1sinθ2)
∴ T2 = \(\frac{m g \sin \theta_1}{\sin \left(\theta_1+\theta_2\right)}\) ….. (3)
As m, g and θ1 are constants, the minimum value of T2 corresponds to the maximum value of sin(θ1 + θ2) i.e., sin(θ1 + θ2) = 1 = sin90°
Example 9.
A chain of mass m and length l is kept on a horizontal frictionless table, such that \(\frac{1}{4}\)th of the length of the chain is hanging out of the table. Find the work done to pull the hanging part of the chain onto the table.
Solution:
Mass per unit length of the chain = \(\frac{m}{l}\)
∴ Mass of hanging part of the chain = \(\frac{m}{l}\) × \(\frac{l}{4}\) = \(\frac{m}{4}\)
∴ Centre of gravity of this portion is at \(\frac{l}{4 \times 2}\) = \(\frac{l}{8}\) below the top of the table.
∴ Work to be done to lift the hanging part of the chain
= \(\frac{m}{4}\) × g × \(\frac{l}{4}\) = \(\frac{m g l}{32}\)