Many of you may have the experience of balancing your notebook on the tip of a finger. Figure 1 illustrates a similar experiment that you can easily perform. Take an irregularshaped cardboard having mass M and a narrow tipped object like a pencil. You can locate by trial and error a point G on the cardboard where it can be balanced on the tip of the pencil. (The cardboard remains horizontal in this position.) This point of balance is the centre of gravity (CG) of the cardboard.
The tip of the pencil provides a vertically upward force due to which the cardboard is in mechanical equilibrium. As shown in the Fig. 1, the reaction of the tip is equal and opposite to Mg and hence the cardboard is in translational equilibrium. It is also in rotational equilibrium; if it were not so, due to the unbalanced torque it would tilt and fall. There are torques on the card board due to the forces of gravity like $m_1\mathbf{g}$, $m_2\mathbf{g}$ …. etc, acting on the individual particles that make up the cardboard.
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| Fig.1 |
Fig. 1: Balancing a cardboard on the tip of a pencil. The point of support, G, is the centre of gravity.
The CG of the cardboard is so located that the total torque on it due to the forces $m_1\mathbf{g}$, $m_2\mathbf{g}$ …. etc. is zero.
If $\mathbf{r_i}$ is the position vector of the ith particle of an extended body with respect to its CG, then the torque about the CG, due to the force of gravity on the particle is τ$_{i}$ = $\mathbf{r_i} \times m_i\mathbf{g}$. The total gravitational torque about the CG is zero, i.e.
τ$_{g}=\Sigma$ τ$_{i}=\Sigma \mathbf{r_i} \times m_i \mathbf{g}=0$
We may therefore, define the CG of a body as that point where the total gravitational torque on the body is zero.
We notice that in Eq. (7.33), g is the same for all particles, and hence it comes out of the summation. This gives, since g is non-zero, $\Sigma m_i \mathbf{r_i}=0$. Remember that the position vectors (ri ) are taken with respect to the CG. Now, in accordance with the reasoning given below Eq. (7.4a) in Sec. 7.2, if the sum is zero, the origin must be the centre of mass of the body. Thus, the centre of gravity of the body coincides with the centre of mass in uniform gravity or gravity free space. We note that this is true because the body being small, g does not vary from one point of the body to the other. If the body is so extended that g varies from part to part of the body, then the centre of gravity and centre of mass will not coincide. Basically, the two are different concepts. The centre of mass has nothing to do with gravity. It depends only on the distribution of mass of the body.
In Sec. 7.2 we found out the position of the centre of mass of several regular, homogeneous objects. Obviously the method used there gives us also the centre of gravity of these bodies, if they are small enough.
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| Fig.2 |
Figure 2 illustrates another way of determining the CG of an irregular shaped body like a cardboard. If you suspend the body from some point like A, the vertical line through A passes through the CG. We mark the vertical $AA_1$. We then suspend the body through other points like B and C. The intersection of the verticals gives the CG. Explain why the method works. Since the body is small enough, the method allows us to determine also its centre of mass.
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