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:
• 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
(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 mi 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 mi 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
1. Fig.7.102(a) below shows an irregular shaped cardboard
• 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
1. Consider the irregular flat object shown in fig.7.101(a) below:
 |
| Fig.7.101 |
• 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 mi 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 mi 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
Experimental method to find CG of irregular laminar bodies
 |
| 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
In the next section, we will see some solved examples
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