Question

Given particles of masses \(m, 2 m\), and \(3 m\) at the points \((0.1),(1,0)\), and \((2,3)\), tind the point \(P\) about which their total moment of inertia will be least. (Recill that to tind the moment of inertia of \(m\) about \(P\), you multiply \(m\) by the square of its distance front \(P\).)

Short answer

\( \left(\frac{4}{3}, \frac{5}{3} \right) \)

Step-by-step solution

- Identify the coordinates and masses

The particles have masses and coordinates as follows: - Mass: \(m\), Coordinates: \((0,1)\) - Mass: \(2m\), Coordinates: \((1,0)\) - Mass: \(3m\), Coordinates: \((2,3)\).

- Use the formula for the center of mass

The coordinates of the center of mass (COM) for a system of particles are given by \(x_{COM} = \frac{\sum m_{i} x_{i}}{\sum m_{i}}\) and \(y_{COM} = \frac{\sum m_{i} y_{i}}{\sum m_{i}}\).

- Calculate the sum of masses

Sum of the masses: \(m + 2m + 3m = 6m\).

- Calculate the sum of moments for x-coordinate

Sum of the moments for the x-coordinate: \(0 \cdot m + 1 \cdot 2m + 2 \cdot 3m = 0 + 2m + 6m = 8m\).

- Calculate the x-coordinate of the center of mass

Using the formula for the x-coordinate of the COM: \(x_{COM} = \frac{8m}{6m} = \frac{4}{3}\).

- Calculate the sum of moments for y-coordinate

Sum of the moments for the y-coordinate: \(1 \cdot m + 0 \cdot 2m + 3 \cdot 3m = m + 0 + 9m = 10m\).

- Calculate the y-coordinate of the center of mass

Using the formula for the y-coordinate of the COM: \(y_{COM} = \frac{10m}{6m} = \frac{5}{3}\).

- Identify the point P

The point \(P\) about which the total moment of inertia is least is the center of mass: \( \left(\frac{4}{3}, \frac{5}{3} \right) \).

Key concepts

moment of inertia

The moment of inertia is a fundamental concept in physics, especially in rotational dynamics. It represents the rotational equivalent of mass for linear motion. In simple terms, it quantifies how much torque (rotational force) is needed for a given angular acceleration around a specific axis.

For a point mass, the moment of inertia can be calculated using the formula: \(I = m \times r^2\), where: \(m\) is the mass of the particle.

\(r\) is the distance from the axis or point of rotation.

Understanding the moment of inertia helps determine the rotational motion of objects. For example, a figure skater can spin faster by pulling their arms close to their body, reducing their moment of inertia.

In the given exercise, the goal is to find the point about which the total moment of inertia of the system of particles is the least. This is achieved by locating the center of mass (COM) of the particles, as the moment of inertia about the COM is minimal.

particle masses and coordinates

Knowing the masses and coordinates of each particle is crucial in solving problems related to the center of mass and moment of inertia. Given particles of varying masses at different coordinates affect the overall calculations of these properties.

In the provided exercise, we have three particles with:

• Particle 1: Mass \(m\), Coordinates \((0,1)\)
• Particle 2: Mass \(2m\), Coordinates \((1,0)\)
• Particle 3: Mass \(3m\), Coordinates \((2,3)\)

These details help in calculating the sum of masses and moments, which are essential for determining the coordinates of the center of mass.

It's important to carefully identify these properties to avoid errors in subsequent calculations.

coordinate calculations

Coordinate calculations are necessary for locating the center of mass of a system of particles. The center of mass (COM) is the point where the system's mass is equally distributed in all directions.

The formula for calculating the x-coordinate and y-coordinate of the COM is given by: \[ x_{COM} = \frac{\text{sum of moments for x}}{\text{sum of masses}}, \ y_{COM} = \frac{\text{sum of moments for y}}{\text{sum of masses}} \]

In the exercise, we calculated the sum of masses: \((m + 2m + 3m = 6m)\), the sum of moments for the x-coordinate: \((0 \times m + 1 \times 2m + 2 \times 3m = 8m)\), and the sum of moments for the y-coordinate: \((1 \times m + 0 \times 2m + 3 \times 3m = 10m)\).

The x-coordinate and y-coordinate of the COM were then found using these sums: \[ x_{COM} = \frac{8m}{6m} = \frac{4}{3}, \ y_{COM} = \frac{10m}{6m} = \frac{5}{3} \]

Thus, the point about which the total moment of inertia of the particles is the least is \(\frac{4}{3}, \frac{5}{3}\). This point is the balance point or the center of mass for the system.