More Solved examples involving Torque and Center of gravity

This page shows two more solved examples in continuation of the examples that we saw here: Solved examples involving torque.

Example 1
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body
Solution:
1. The uniform circular disc can be supported at it's center as shown in fig.1(a) below:

Fig.1

• But when a hole is cut on the right side, the disc will tilt towards the left and fall off. This is shown in fig.1(b)
2. If we shift the support towards the left (along the diameter), the disc will balance again
• This is shown in fig.2(a) below:

Fig.2

• In fig.2(a), the disc is supported at the new CG
• We have to find the position of the new CG
3. In fig.2(b), the small circle is painted red and glued back to get the original disc
• The CG of this 'composite disc' will be same as the center of the original disc 
4. In such a situation, we apply the conditions of equilibrium
• Let the brown portion in fig.2(b) be indicated as 'Brown'
• Let the red portion in fig.2(b) be indicated as 'Red'
• Let 'mass per unit area' of the disc be M
• Since the disc is uniform, this 'M' will be same at all points on the disc
5. Fig.3(a) below shows the dimensions

Fig.3

• Area of Red = $\mathbf\small{\pi\,\left(\frac{R}{2}\right)^2=\frac{\pi R^2}{4}}$
    ♦ So $\mathbf\small{W_{Red}=\frac{\pi R^2Mg}{4}}$
• Area of Brown = $\mathbf\small{\pi\,R^2-\frac{\pi R^2}{4}=\frac{3\pi R^2}{4}}$
    ♦ So $\mathbf\small{W_{Brown}=\frac{3\pi R^2Mg}{4}}$ 
6. Fig.3(b) shows the forces
• The distance between the required CG and the support is denoted as 'x'
7. Since the composite disc is in translational equilibrium, we have:
R - W_Brown - W_Red = 0
8. Since the composite disc is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support 
(i) Torque created by W_Brown about the support = W_Brown × x (anti clockwise)
(ii) Torque created by R about the support = 0
(iii) Torque created by W_Red about the support = W_Red × ^R⁄₂ (clockwise)
9. So applying the condition, we get:
-(W_Brown × x) + (W_Red × ^R⁄₂) = 0
$\mathbf\small{\Rightarrow x=\frac{R\,W_{Red}}{2\,W_{Brown} }}$
• Substituting the known values, we get: $\mathbf\small{x=\frac{\pi R^3Mg}{4}\div \frac{6\pi R^2Mg}{4}}$
$\mathbf\small{\Rightarrow x=\frac{\pi R^3Mg}{4}\times \frac{4}{6\pi R^2Mg}=\frac{R}{6}}$

Example 2
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Solution:
1. Given that, the metre stick is balanced on a knife edge at its centre 
• That means, if the support is given at the exact center, the metre stick will balance
• That means, it is a uniform scale
• This is shown in fig.2(a) below:

Fig.2

2. The condition when the two coins are placed at the 12 cm mark is shown in fig.b
• The forces and dimensions are also shown
• We see that, for equilibrium, the support has to be shifted by 5 cm towards the left
3. Since the system in fig.b is in translational equilibrium, we have:
R - W_Coin - W_Stick = 0
8. Since the system is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support 
(i) Torque created by W_Coin about the support = W_Coin × 0.33 (anti clockwise)
(ii) Torque created by R about the support = 0
(iii) Torque created by W_Stick about the support = W_Stick × 0.05 (clockwise)
9. So applying the condition, we get:
-(W_Coin × 0.33) + (W_Stick × 0.05) = 0
W_Stick = (W_Coin × ^0.33⁄_0.05 ) = (10 ×  ¹⁄₁₀₀₀ ×  ^0.33⁄_0.05 ) = 0.066 kg = 66 g

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Chapter 7.22 - Solved examples involving Torque

In the previous section, we saw center of gravity. In this section we will see some solved examples

Solved example 7.21
The metal bar in fig.7.103(a) is 0.70 m long and has a mass of 4 kg. It is supported on two knife edges placed 0.10 m from each end. A 6 kg mass is suspended from a point P, which is 0.20 m from the left support

Fig.7.103

Find the reactions at the supports. Assume the bar is of uniform cross section and homogeneous. Take g = 9.8 ms^-2
Solution:
1. Let us name the supports as A and B
■ The weight of the rod acts at it’s CG
• Given that, the metal bar is of uniform cross section and homogeneous.
■ So the CG of the rod will be at it’s geometric center
2. The detailed measurements are shown in fig.b
The CG is marked as G
3. Since the bar is in translational equilibrium, we have:
R_A + R_B - 6g - 4g = 0
⇒ R_A + R_B = 10g = 98.0 N.
4. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support A
(i) Torque created by R_A about A = zero
(ii) Torque created by 6g about A = 6g × 0.2 = 1.2g Nm (clockwise)
(iii) Torque created by 4g about A = 4g × 0.25 = 1.0g Nm (clockwise)
(iv) Torque created by R_B about A = R_B × 0.5 Nm (anti clockwise)
5. So applying the condition, we get:
1.2g + g - 0.5R_B = 0
⇒ R_B = 43.12 N
• Substituting this value of R_B in (3), we get: R_A = (98-43.12) = 54.88 N

Solved example 7.22
A 3 m long ladder having a mass of 20 kg, leans on a frictionless wall. It’s feet rest on the ground 1 m from the wall as shown in fig.7.104(a) below:

Fig.7.104

Find the reaction forces on the wall and the floor. Take g = 9.8 ms^-2
Solution:
1. Let the end points of the ladder be A and B
• A is 1 m from the wall. This is shown in fig.b
• Let C be the foot of the wall
2. The reaction from the floor at A will be normal to the floor
• This reaction is denoted as R_A
3. The frictional force prevents the point A from moving away from C
• This frictional force is denoted as F. It pulls the ladder towards C. Other wise the ladder will slip
• Also recall that, the frictional force is always parallel to the surface
4. The reaction from the wall at B will be normal to the wall
• This reaction is denoted as R_B
5. Given that, the wall is frictionless
• If there was friction, a force F would have acted parallel to the wall in the upward direction
• This force would have made some contribution towards: 'preventing the movement of B towards C'   • In other words, this force would have made some contribution towards: 'preventing the ladder from slipping'
• In addition to that, this force would have made some contribution towards resisting the vertical force (20 g) of the ladder
    ♦ Where 'g' is the acceleration due to gravity
• But in this problem there is no such force
    ♦ The slipping is prevented entirely by the horizontal force F at A
    ♦ The vertical load 20 g is resisted entirely by the vertical force R_A at A
6. The weight of the ladder is 20 g
• It acts downwards at the CG of the ladder
• The CG is marked as G in fig.b
7. The above steps gives us all the 4 forces acting on the ladder
• Now we can apply the conditions of equilibrium
• Since the ladder is in translational equilibrium, we have:
(i) In the x direction: F - R_B = 0
(ii) In the y direction: R_A - 20 g = 0
⇒ R_A = 20 g = 196 N
8. Since the bar is in rotational equilibrium, we have: net torque = 0
Let us take the torques about A
(i) Torque created by R_A about A = zero
(ii) Torque created by F about A = zero
(iii) Torque created by R_B about A = R_B × BC
So we want the length of BC
• Applying Pythagoras theorem to the right triangle ABC, we get:
$\mathbf\small{BC=\sqrt{AB^2-AC^2}=\sqrt{3^2-1^2}=2\sqrt{2}\;\text{m}}$
Thus the required torque = $\mathbf\small{2\sqrt{2}R_B\;\text{N m}}$ (clockwise) 
(iv) Torque created by 20 g about A = 20 g × AD
• So we want the length of AD
• D is the foot of the perpendicular drawn from G
• Consider the similar triangles ABC and AGD
• We have: $\mathbf\small{\frac{AD}{AC}=\frac{AG}{AB}}$
$\mathbf\small{\Rightarrow \frac{AD}{1}=\frac{1.5}{3}}$
⇒ AD = 0.5 m
• Thus the torque = 20 g × 0.5 × 9.8 = 98 N m (anti clockwise)
9. Applying the condition, we get:
$\mathbf\small{2\sqrt{2}R_B-98=0}$
⇒ R_B = 34.65 N
10. Substituting this value of R_B in 7(i), we get: F = 34.65 N.
11. At the point A, two forces are acting on the ladder:
• R_A vertically and F horizontally. They are shown in fig.c
• The resultant of the two forces = $\mathbf\small{\sqrt{(R_A)^2+F^2}=\sqrt{196^2+34.65^2}=}$ 199.04 N
• Let this resultant make an angle α with the horizontal
• Then $\mathbf\small{\alpha=\tan^{-1}\frac{R_A}{F}=\tan^{-1}\frac{196}{34.65}=}$ 79.97^o

Solved example 7.23 
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig.7.105(a) below:

Fig.7.105

The angles made by the strings with the vertical are 36.9^o and 53.1^o respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from it’s left end
Solution:
1. The free body diagram is shown in fig.b
• Let T₁ and T₂ be the tensions in the strings
• In fig.a, we see that, the left string makes an angle of 36.9^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles 
• In fig.a, we see that, the right string makes an angle of 53.1^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles
• Now we can resolve the tensions into horizontal and vertical components. This is shown in fig.c
2. Since the bar is in translational equilibrium, we have:
(i) In the x direction: - T₁ sin θ₁ + T₂ sin θ₂  = 0
⇒ T₁ sin θ₁ = T₂ sin θ₂.
⇒ T₁ sin 36.9 = T₂ sin 53.1 = T₂ cos (90 - 36.9) = T₂ cos 36.9
⇒ $\mathbf\small{\frac{T_2}{T_1}=\tan36.9 =0.75}$
⇒ T₂ = 0.75 T₁
(ii) In the y direction: T₁ cos θ₁ + T₂ cos θ₂ - W = 0
⇒ T₁ cos 36.9 + T₂ cos 53.1 = W
⇒0.8 T₁ + 0.6 T₂ = W
3. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about G
(i) Torque created by T₁ sin θ₁ about G = zero
(ii) Torque created by T₁ cos θ₁ about G = T₁ cos θ₁ × d (clockwise)
(iii) Torque created by W about G = zero
(iv) Torque created by T₂ cos θ₂ about G = T₂ cos θ₂ × (2-d) (anti clockwise)
(i) Torque created by T₂ sin θ₂ about G = zero
4. Applying the condition, we get:
[T₁ cos θ₁ × d] - [T₂ cos θ₂ × (2-d)] = 0
⇒ [T₁ × cos 36.9 × d] - [0.75 T₁ × cos 53.1 × (2-d)] = 0
(∵ from 2(i), we have: T₂ = 0.75 T₁)
⇒ [0.8 T₁ d] - [0.45 T₁ (2-d)] = 0
⇒ 0.8 T₁ d - 0.9 T₁ + 0.45 T₁ d = 0
⇒ 1.25 T₁ d = 0.9 T₁
⇒ 1.25 d = 0.9
⇒ d = 0.72 m

Solved example 7.24
A car weighs 1800 kg. The distance between it’s front and back axles is 1.8 m. It’s center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel
Solution:
1. Fig.7.106 below shows the schematic diagram

Fig.106

• The small grey circles at the center of the wheels denote the axles
• The front and rear axles are named as A and B respectively
2. When the car is in translational equilibrium, we have:
R_A + R_B -1800g = 0
3. When the car is in rotational equilibrium, we have: net torque = 0
Let us take the torques about G
(i) Torque created by R_A about G = 1.05 R_A (clockwise) 
(ii) Torque created by 1800g about G = zero
(iii) Torque created by R_B about G = 0.75 R_B (anticlockwise)
4. Applying the condition, we get:
1.05 R_A - 0.75 R_B = 0
1.05 R_A = 0.75 R_B
R_B = 1.4 R_A
5. Substituting this in (2), we get:
R_A + 1.4R_A = 1800 × 9.8
⇒ R_A = 7350 N
• So R_B = 1.4 × 7350 = 10290 N
6. The reaction on the front axle A is 7350 N
• But the load from this axle is resisted by two front wheels
• So reaction on each of the front wheels = ⁷³⁵⁰⁄₂ = 3675 N
7. The reaction on the back axle B is 10290 N
• But the load from this axle is resisted by two back wheels
• So reaction on each of the back wheels = ¹⁰²⁹⁰⁄₂ = 5145 N

Solved example 7.25
As shown in fig.7.107(a), the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE 0.5 m long is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. Take g = 9.8 ms^-2 [Hint: Consider the equilibrium of each side of the ladder separately]

Fig.7.107

Solution:
1. The detailed measurements are shown in fig.b
• Given:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
• Let ∠ABC = θ
    ♦ Since DE is parallel to the ground, we get: ∠ADE = θ   
2. Let us calculate the other required dimensions:
• A perpendicular is dropped from A onto BC
    ♦ This is shown as red dashed line 
    ♦ The foot of the perpendicular on BC is O
    ♦ The foot of the perpendicular on DE is G
• Applying Pythagoras theorem to the right triangle ADG, we get:
$\mathbf\small{AG=\sqrt{AD^2-DG^2}=\sqrt{0.8^2-0.25^2}=0.76\;\text{m}}$
3. In the similar triangles ABO and ADG, we get:
• $\mathbf\small{\frac{AD}{AB}=\frac{AG}{AO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.76}{AO}}$
⇒ AO = 1.52 m
4. Again in the same similar triangles ABO and ADG, we get:
$\mathbf\small{\frac{AD}{AB}=\frac{DG}{BO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.25}{BO}}$
⇒ BO = 0.5 m
5. A perpendicular is dropped from F onto AG
• The foot of this perpendicular is H
• In the similar triangles AFH and ADG, we get:
$\mathbf\small{\frac{AF}{AD}=\frac{FH}{DG}}$
$\mathbf\small{\Rightarrow \frac{0.4}{0.8}=\frac{FH}{0.25}}$
⇒ FH = 0.125 m
6. Let us write the above distances together:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
    ♦ AG =0.76 m
    ♦ AO = 1.52 m
    ♦ BO = 0.5 m
    ♦ FH = 0.125 m
• Thus we obtained all the distances that we will soon require for torque calculations
7. The forces are shown in fig.c below:

Fig.7.107 (c) & (d)

• R_B is the normal reaction at B
• R_C is the normal reaction at C
• Note that the tensions T in the string are internal forces and so will cancel each other
8. For translational equilibrium, we have:
• R_B + R_C - (40 × 9.8) = 0
⇒ R_B + R_C - 392 = 0
⇒ R_B + R_C = 392
9. For rotational equilibrium, we need to calculate the torques first:
• Let us find the torques about O
(i) Torque created by R_B about O = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by the 40 kg load about O = (40 × 9.8) × FH = 392 0.125 = 49 Nm (anti clockwise)
(iii) Torque created by R_C about O = R_C × CO = R_C × 0.5 = 0.5R_C Nm (anti clockwise)
10. Applying the condition, we get:
⇒ 0.5R_B - 49 - 0.5R_C = 0
⇒ 0.5R_B - 0.5R_C = 49
⇒ R_B - R_C = 98
11. Solving the equations in (8) and (10), we get:
R_B = 245 N, R_C = 147 N
12. Next we want to find T
• In the fig.d above, the FBD of side AB is shown separately
• The forces acting are: R_B, T and the load of 40 kg
• Let us calculate the torques about A:
(i) Torque created by R_B about A = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by T about A = T × AG = T × 0.76 = 0.76T Nm (anti clockwise)
(ii) Torque created by the 40 kg load about A = (40 × 9.8) × FH = 392 × 0.125 = 49 Nm (anti clockwise)
13. Applying the condition, we get:
0.5R_B - 0.76T - 49 = 0
0.5 × 245 - 0.76T - 49 = 0
T = 96.7 N

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Two more solved examples can be seen here

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In the next section, we will see moment of inertia

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Chapter 7.21 - Center of Gravity

In the previous section, we saw principle of moments. In this section we will see center of gravity

1. Consider the irregular flat object shown in fig.7.101(a) below:

Fig.7.101

• It has small but uniform thickness (for example, a cardboard)
• It's surface is parallel to the xy-plane
• It is somewhat transparent so that, we can see the support on which it is resting
• We see that, the support is a ‘pointed end’, like the tip of a pencil
2. Let the point (on the object) at which it is supported be 'G' 
• We see that there is no rotation or translation for the object
• It may seem amazing that an object (and that too, an irregular one) is able to rest at equilibrium on a pointed end
• But once we analyze the situation scientifically, it will no longer be a source of amazement
3. The analysis can be done as follows:
• The object contains infinite number of particles. Let us assume that there are n particles
• Three of them are shown as small red spheres in fig.7.101(b)
4. Let the mass of the particles be m₁, m₂, m₃, . . . so on . . .
• Then their weights will be $\mathbf\small{m_1\vec{g},\;m_2\vec{g},\;m_3\vec{g}\,\;.\;.\;.}$so on . . .
• These weights are, forces acting in the -ve z-direction
5. So total force in the -ve z–direction = $\mathbf\small{m_1\vec{g}+m_2\vec{g}+m_3\vec{g}\,+\;.\;.\;.\;+\;m_i\vec{g}\,+\;.\;.\;.\;+\;m_n\vec{g}}$
= $\mathbf\small{(m_1+m_2+m_3\,+\;.\;.\;.\;+\;m_i\,+\;.\;.\;.\;+\;m_n)\vec{g}=M\,\vec{g}}$
• Where M is the mass of the whole object
6. The direction of $\mathbf\small{M\,\vec{g}}$ will be towards the -ve z direction because $\mathbf\small{\vec{g}}$ acts in the -ve z-direction
• So we can write $\mathbf\small{M\,\vec{g}}$ as $\mathbf\small{-\left(M\,|\vec{g}|\right)\hat{k}}$  
7. For translational equilibrium:
• $\mathbf\small{\sum{\vec{F}_{z,ext}}=\vec{0}}$
• Let the reaction at the support point be $\mathbf\small{\vec{R}_G}$
• If this condition is to be satisfied, the vector sum ($\mathbf\small{M\,\vec{g}+\vec{R}_G}$) must be a zero vector

• This condition tells us that, if there is to be translational equilibrium, $\mathbf\small{\vec{R}_G}$ is indeed vertical

• The reason can be written in 5 steps:
(i) If $\mathbf\small{\vec{R}_G}$ is inclined, it will have a vertical as well as a horizontal component
(ii) But we see no other horizontal components to counter it
(iii) So, if there is to be translational equilibrium, $\mathbf\small{\vec{R}_G}$ must be truly vertical
(iv) Also, it must be directed towards the +ve side of the y-axis
(v) Only then we will get ($\mathbf\small{M\,\vec{g}+\vec{R}_G}$) equal to a zero vector
8. Since we know the directions, we can write the above equation in terms of unit vectors also:
$\mathbf\small{-\left(M\,|\vec{g}|\right)\hat{k}+|\vec{R}_G|\hat{k}=\vec{0}}$
$\mathbf\small{\Rightarrow \left(-M\,|\vec{g}|+|\vec{R}_G|\right)\hat{k}=\vec{0}}$
• In the above equation, we have vectors on both sides. So their magnitudes must be equal
• Thus the condition becomes:
$\mathbf\small{\left(-M\,|\vec{g}|+|\vec{R}_G|\right)=0}$
That means the two magnitudes are equal:
$\mathbf\small{M\,|\vec{g}|=|\vec{R}_G|}$
9. Now we can consider the rotational equilibrium:
• We want the net torque to be zero
• In this problem, the forces do not lie in a single plane. So we want all the 3 conditions (related to rotational equilibrium) to be satisfied
• That is, we want:
$\mathbf\small{\sum{\vec{\tau}_{x,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{y,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{z,ext}}=\vec{0}}$
• In general, we want: $\mathbf\small{\sum{\vec{\tau}_{ext}}=\vec{0} }$
10. Let 'G' be the origin 'O' from where position vectors are drawn
    ♦ Let $\mathbf\small{\vec{r}_1}$ be the position vector of the first particle
    ♦ Let $\mathbf\small{\vec{r}_2}$ be the position vector of the second particle
    ♦ Let $\mathbf\small{\vec{r}_3}$ be the position vector of the third particle . . . so on . . .
• Then we get: 
    ♦ Torque created by the first particle = $\mathbf\small{\vec{r}_1 \times m_1\vec{g}}$ 
    ♦ Torque created by the second particle = $\mathbf\small{\vec{r}_2 \times m_2\vec{g}}$
    ♦ Torque created by the third particle = $\mathbf\small{\vec{r}_3 \times m_3\vec{g}}$  . . . so on . . .
[These torques are called gravitational torques because, the 'forces causing the torques' are gravitational forces]
11. So total torque
= $\mathbf\small{(\vec{r}_1\times m_1\vec{g})+(\vec{r}_2\times m_2\vec{g})+(\vec{r}_3\times m_3\vec{g})\,+\;.\;.\;.\;+\;(\vec{r}_i\times m_i\vec{g})\,+\;.\;.\;.\;+\;(\vec{r}_n\times m_n\vec{g})}$
= $\mathbf\small{\sum{(\vec{r}_i\times m_i\vec{g}})}$
• The $\mathbf\small{\vec{g}}$ is the same for all particles. So it can be taken out of the brackets
• Thus we get:
Total torque = $\mathbf\small{\vec{g}\sum{\vec{r}_i\times m_i}}$
• We want this total torque to be equal to zero vector. So we can write:
$\mathbf\small{\vec{g}\sum{(\vec{r}_i\times m_i)}=\vec{0}}$
• $\mathbf\small{\vec{g}}$ cannot be zero. So we get:
• $\mathbf\small{\sum{(\vec{r}_i\times m_i)}=\vec{0}}$
12. This is the condition which is to be satisfied, if we want rotational equilibrium for the body
• We will write it in words:
(i) There is a unique point 'G' in the body
(ii) We consider that 'G' as the origin 'O' 
• From that 'O', we draw position vectors to every particle of the body
(iii) For each particle we calculate $\mathbf\small{(\vec{r}_i\times m_i)}$ 
(the above product is not a cross product. Because m_i is a scalar) 
(iv) Then we take their vector sum: $\mathbf\small{\sum{(\vec{r}_i\times m_i)}}$
(v) That vector sum should be a zero vector. Then only we get rotational equilibrium for the body
13. So our next aim is to find the location of 'G'
• In an earlier section of this chapter, we saw Eq.7.4: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$ 
• This equation is used to find the center of mass (C) of any object (Details here).
• For using this method, we first fix a convenient point as the origin 'O', and then take the position vectors from that 'O'
• The $\mathbf\small{\vec{r}_C}$ in Eq.7.4, is the position vector of 'C' from that 'O'
■ Now, what if we take the 'C' itself as the origin 'O'
• Then we will get: $\mathbf\small{\vec{r}_C=\vec{0}}$
• The Eq.7.4 will then become: $\mathbf\small{\vec{0}=\frac{\sum{\,m_i\,\vec{r}_i}}{M}}$
$\mathbf\small{\Rightarrow \vec{0} \times M={\sum{ \,m_i\,\vec{r}_i}}}$
$\mathbf\small{\Rightarrow \vec{0} ={\sum{m_i\,\vec{r}_i}}}$
14. We will write the above equation in words:
(i) We take the 'C' of the body
(ii) We consider that 'C' as the origin 'O' 
• From that 'O', we draw position vectors to every particle of the body
(iii) For each particle we calculate $\mathbf\small{(\vec{r}_i\times m_i)}$ 
(the above product is not a cross product. Because m_i is a scalar) 
(iv) Then we take their vector sum: $\mathbf\small{\sum{(\vec{r}_i\times m_i)}}$
(v) That vector sum will be a $\mathbf\small{\vec{0}}$
■ That means, if we take 'C' as 'O', then $\mathbf\small{\sum{(\vec{r}_i\times m_i)}}$ will definitely be $\mathbf\small{\vec{0}}$
15. Comparing (12) and (14), we find that:
• The 'G' in fig.7.101 is nothing else but 'C'
■ We can write: 'G' and 'C' are the same
16. We have seen the methods to find 'C' of various systems and bodies, regular or irregular
• So we can indeed find 'G' of any such system or body, regular or irregular
■ This 'G' is called the center of gravity of the body
• The 'center of gravity' is abbreviated as: CG 
17. We saw that, CG can be determined by the same methods that we saw for 'C'
• But there is a constraint
• The $\mathbf\small{\vec{g}}$ is not constant every where. For example, $\mathbf\small{\vec{g}}$ is slightly greater at the poles than at the equator
• So for very large bodies, we cannot take $\mathbf\small{\vec{g}}$ out of the brackets in step (11) above
■ If the body is so large that, the $\mathbf\small{\vec{g}}$ varies from part to part of that body, then the 'C' and 'CG' will not coincide
■ The 'C' has nothing to do with 'CG'. The 'C' depends only on the distribution of mass of the body

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Experimental method to find CG of irregular laminar bodies       

1. Fig.7.102(a) below shows an irregular shaped cardboard

Fig.7.102

• A hole is made at any convenient point near the edge. That hole is marked as 'A'
2. A thin pin is passed through the hole and the cardboard is suspended from the point S
• Now the cardboard is perpendicular to the floor of the room
• The hole A should be just large enough so that, the cardboard is able to rotate freely about the pin
• If the hole is too small, friction will arise between the pin and the cardboard, thereby obstructing free rotation
• Also, when suspended from S, the cardboard should not touch side walls or other objects 
3. When the suspended cardboard becomes stable, draw a vertical dashed line through A. This is shown in red color in fig.a
■ The ‘CG’ of the cardboard lies somewhere on the red dashed line
Let us see the reason. It can be written in 5 steps:
(i) The particles on the left side of the red dashed line create anti clockwise 'gravitational torques' about G
(ii) The particles on the right side of the red dashed line create clockwise 'gravitational torques' about G
(iii) The particles on the red dashed line create no torques at all about G
(iv) The anti clockwise torques must balance out the clockwise torques. Other wise, the cardboard will rotate
(v) So we can be sure that, when the cardboard is stable, the G lies somewhere on the red vertical dashed line
4. Next, make a similar hole at another point B
• Repeat the experiment
• We get the green vertical dashed line shown in fig.b
• The CG lies somewhere on the green dashed line
5. So we have two information:
(i) The CG lies on the red dashed line
(ii) The CG lies on the green dashed line
■ So obviously, the CG lies at the point of intersection of the red and green lines
6. To make a check:
• Make a similar hole at a third point C
• Repeat the experiment
• We get the yellow vertical dashed line shown in fig.c
■ This yellow line will pass through the CG obtained in fig.b
7. Note that, since the cardboard is small, all particles of the cardboard will be experiencing the same $\mathbf\small{\vec{g}}$ 
■ So the CG that we determined above, is also the center of mass ('C') of the cardboard

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In the next section, we will see some solved examples

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