6.8.2. Centre of gravity

Many of you may have the experience of balancing your notebook on the tip of a finger. Figure 6.24 illustrates a similar experiment that you can easily perform. Take an irregular-shaped 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.

This chunk explains how to find the center of gravity (CG) by balancing an irregular object. When you balance the cardboard on a pencil, you are finding the point G where the weight of the cardboard acts straight down. At this point, the gravitational pull is evenly distributed, resulting in stability. This means that the forces acting on the cardboard, like its weight, and the force from the pencil together create no torque, keeping the cardboard level.

Think of a seesaw at a playground. If you sit in the exact middle (CG), the seesaw stays balanced. If you sit off-center, one side will go down while the other side rises. The center of gravity ensures stability in various objects, and is similar to how balance works on a seesaw.

If ri 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 τττττi = ri × mi g. The total gravitational torque about the CG is zero, i.e., τ τ τ τgii im = = × =∑ ∑ r g 0 (6.33). We may therefore, define the CG of a body as that point where the total gravitational torque on the body is zero.

Here, we learn that the center of gravity (CG) can be mathematically defined through torque. When all the particle weights (m_i g) create torques (τ_i), for the CG, the total effect of these torques must balance out to zero. This means that if you were to sum all the torques created by the weights at their respective distances (r_i) from the CG, it would cancel out, indicating stability and balance.

Imagine trying to balance a long stick on your finger. If the CG of the stick is directly over your finger (like the point where all weights balance out), it stands up straight. If it's not, it will tip over. The concept of balancing torques relates directly to how we find this CG in real-life situations.

We notice that in Eq. (6.33), g is the same for all particles, and hence it comes out of the summation. This gives, since g is non-zero, ∑mi ir = 0. Remember that the position vectors (ri) are taken with respect to the CG. Now, in accordance with the reasoning given below Eq. (6.4a) in Sec. 6.2, 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.

In this segment, we explore the relationship between the center of gravity and the center of mass. Since the gravitational field (g) does not change across small objects, the CG aligns with the center of mass (CM). We summarize that when an object's mass is uniformly distributed (like a solid ball), its CG is also the point where mass is evenly distributed about it.

Picture a perfectly balanced seesaw where the mass on both sides is equal. The point in the middle where it balances is both the center of gravity and the center of mass, demonstrating how evenly distributed mass translates to balance.

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