Where Is the Center of Gravity?
- Three particles are located in a coordinate system as shown in Figure 8.9. Find the center of gravity.
- If a fourth particle of mass 2.00 kg is placed at x = 0, y = 0.250 m, find the x-and y-coordinates of the center of gravity for this system of four particles.
Figure 8.9 Locating the center of gravity of a system of three particles.
The Correct Answer and Explanation is :
To determine the center of gravity (CG) for a system of particles, we use the principle that the CG is the weighted average of their positions, weighted by their masses. The coordinates (x̄, ȳ) of the CG are calculated using:
[ x̄ = \frac{\sum (m_i \cdot x_i)}{\sum m_i} ]
[ ȳ = \frac{\sum (m_i \cdot y_i)}{\sum m_i} ]
1. Center of Gravity for Three Particles:
Assume we have three particles with the following masses and coordinates:
- Particle 1: mass ( m_1 = 2.00 \, \text{kg} ) at coordinates ( (x_1, y_1) = (0, 0) )
- Particle 2: mass ( m_2 = 3.00 \, \text{kg} ) at coordinates ( (x_2, y_2) = (1.00 \, \text{m}, 0) )
- Particle 3: mass ( m_3 = 4.00 \, \text{kg} ) at coordinates ( (x_3, y_3) = (0, 1.00 \, \text{m}) )
First, calculate the total mass:
[ M = m_1 + m_2 + m_3 = 2.00 + 3.00 + 4.00 = 9.00 \, \text{kg} ]
Next, determine the x-coordinate of the CG:
[ x̄ = \frac{(m_1 \cdot x_1) + (m_2 \cdot x_2) + (m_3 \cdot x_3)}{M} = \frac{(2.00 \times 0) + (3.00 \times 1.00) + (4.00 \times 0)}{9.00} = \frac{3.00}{9.00} = 0.333 \, \text{m} ]
Then, determine the y-coordinate of the CG:
[ ȳ = \frac{(m_1 \cdot y_1) + (m_2 \cdot y_2) + (m_3 \cdot y_3)}{M} = \frac{(2.00 \times 0) + (3.00 \times 0) + (4.00 \times 1.00)}{9.00} = \frac{4.00}{9.00} = 0.444 \, \text{m} ]
Thus, the center of gravity for the three-particle system is at ( (0.333 \, \text{m}, 0.444 \, \text{m}) ).
2. Adding a Fourth Particle:
Now, let’s add a fourth particle:
- Particle 4: mass ( m_4 = 2.00 \, \text{kg} ) at coordinates ( (x_4, y_4) = (0, 0.250 \, \text{m}) )
Update the total mass:
[ M’ = M + m_4 = 9.00 + 2.00 = 11.00 \, \text{kg} ]
Calculate the new x-coordinate of the CG:
[ x̄’ = \frac{(M \cdot x̄) + (m_4 \cdot x_4)}{M’} = \frac{(9.00 \times 0.333) + (2.00 \times 0)}{11.00} = \frac{2.997}{11.00} = 0.272 \, \text{m} ]
Calculate the new y-coordinate of the CG:
[ ȳ’ = \frac{(M \cdot ȳ) + (m_4 \cdot y_4)}{M’} = \frac{(9.00 \times 0.444) + (2.00 \times 0.250)}{11.00} = \frac{4.446 + 0.500}{11.00} = \frac{4.946}{11.00} = 0.450 \, \text{m} ]
Therefore, with the addition of the fourth particle, the center of gravity shifts to ( (0.272 \, \text{m}, 0.450 \, \text{m}) ).
Explanation:
The center of gravity represents the average position of the entire mass distribution of a system. In a system of discrete particles, it is found by taking the weighted average of their positions, with weights corresponding to their masses. This ensures that particles with greater mass have a proportionally larger influence on the location of the CG.
In the first scenario, the CG is closer to the more massive particles. When a fourth particle is introduced, especially if it’s positioned away from the existing CG, it alters the mass distribution, causing the CG to shift accordingly. The degree and direction of this shift depend on the mass and position of the new particle relative to the existing system.
Understanding the CG is crucial in various applications, such as engineering and physics, as it affects stability, balance, and motion dynamics of structures and objects.
For a visual demonstration of finding the center of mass for multiple particles, you may find this video helpful: