Problem 1 - Calculus

Question

Compute the mass \(M\) along the \(x\) axis, the moment \(M_{y}\) around \(x=0,\) and the center of \(\operatorname{mass} \bar{x}=M_{y} / M\). $$ m_{1}=2 \text { at } x_{1}=1, m_{2}=4 \text { at } x_{2}=2 $$

Step-by-Step Solution

Answer

Mass \( M \) is 6, moment \( M_y \) is 10, center of mass \( \bar{x} = \frac{5}{3} \).

Step 1: Calculate Total Mass (M)

The total mass \( M \) along the \( x \) axis is the sum of the individual masses. We have two masses given: \( m_1 = 2 \) at \( x_1 = 1 \) and \( m_2 = 4 \) at \( x_2 = 2 \). The formula to find the total mass is: \[ M = m_1 + m_2 = 2 + 4 = 6. \] Thus, the total mass \( M \) is 6.

Step 2: Calculate Moment (M_y) around x=0

The moment \( M_y \) is the sum of the products of each mass and its respective distance from the origin \( x=0 \). We calculate it as follows: \[ M_y = m_1 \cdot x_1 + m_2 \cdot x_2 = 2 \cdot 1 + 4 \cdot 2 = 2 + 8 = 10. \] Thus, the moment \( M_y \) around \( x=0 \) is 10.

Step 3: Calculate Center of Mass (\(\bar{x}\))

The center of mass \( \bar{x} \) is given by the formula \( \bar{x} = \frac{M_y}{M} \). With \( M_y = 10 \) and \( M = 6 \), we find: \[ \bar{x} = \frac{10}{6} = \frac{5}{3}. \] Therefore, the center of mass \( \bar{x} \) is \( \frac{5}{3} \).

Key Concepts

Total Mass

To understand how the total mass is calculated, think about it as simply gathering all the weights you have at different points. Imagine you have two objects with specific masses placed on a straight line (the x-axis). When you add these weights together, you get the total mass of the system. In our exercise, we have:

• Mass at point 1: \( m_1 = 2 \)
• Mass at point 2: \( m_2 = 4 \)

To find the total mass \( M \), you just add these values:\[ M = m_1 + m_2 = 2 + 4 = 6. \]So, the total mass here is 6. The concept is really just about adding up all the individual masses to get one big number that represents the whole system.

Moment Calculation

The moment around a point tells us about the distribution of mass relative to that point. It's like asking how much influence each mass exerts based on its distance from a point, typically the origin. To find the moment \( M_y \), you multiply each mass by its distance from your reference point (which is the origin, \( x=0 \) in this instance).

• For the first mass: \( m_1 \cdot x_1 = 2 \cdot 1 = 2 \)
• For the second mass: \( m_2 \cdot x_2 = 4 \cdot 2 = 8 \)

Add these products together to get the total moment:\[ M_y = 2 + 8 = 10. \]This tells us the combined effect of all the masses as if trying to rotate the system about the reference point. Moments give a sense of the "twist" or "turning" effect caused by the weights around their position.

Center of Mass

The center of mass allows us to determine where the system balances perfectly, assuming it could pivot without friction. It's similar to finding the 'average' position of all the mass. You compute it by dividing the total moment by the total mass.To illustrate using our earlier work:

• We have the total moment \( M_y = 10 \)
• We have the total mass \( M = 6 \)

The formula to find the center of mass \( \bar{x} \) is:\[ \bar{x} = \frac{M_y}{M} = \frac{10}{6} = \frac{5}{3}. \]Thus, the center of mass \( \bar{x} \) is located at \( \frac{5}{3} \). This spot is where the system would perfectly balance on a fulcrum, much like balancing a seesaw.