What are (a) the x coordinate and (b) the y coordinate of the system’s center of mass

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In Fig. 9-39, three uniform thin rods, each of length L = 22 cm, form an inverted U. The vertical rods each have a mass of 14 g; the horizontal rod has a mass of 42 g. What are (a) the x coordinate and (b) the y coordinate of the system’s center of mass?

The Correct Answer and Explanation is:

Let’s analyze the problem carefully.

Problem Recap:

  • Three uniform thin rods, each length $L = 22 \text{ cm}$, form an inverted U shape.
  • Two vertical rods: each has mass $m_v = 14 \text{ g}$.
  • One horizontal rod (top connecting the vertical rods): mass $m_h = 42 \text{ g}$.

We need to find the center of mass (COM) coordinates of the system, i.e.,

  • $x_{\text{COM}}$
  • $y_{\text{COM}}$

Step 1: Visualizing the rods

  • The two vertical rods are placed vertically and separated by the length of the horizontal rod.
  • The horizontal rod connects the tops of the vertical rods.
  • Assume the rods are thin and uniform, so their center of mass is at their midpoint.
  • Define a coordinate system:
  • Place the origin $(0,0)$ at the bottom of the left vertical rod.
  • Let the rods be aligned such that the inverted U opens downward (vertical rods pointing upward).

Step 2: Coordinates of each rod’s center of mass

  • Left vertical rod:
  • Bottom at origin $(0,0)$.
  • Length $L=22 \text{ cm}$.
  • Its COM is at the midpoint: $$
    (x_1, y_1) = (0, \frac{L}{2}) = (0, 11 \text{ cm})
    $$
  • Right vertical rod:
  • It is placed horizontally at $x = L = 22 \text{ cm}$.
  • Bottom at $(22, 0)$.
  • Its COM is at: $$
    (x_2, y_2) = (22, \frac{L}{2}) = (22, 11 \text{ cm})
    $$
  • Horizontal rod:
  • It connects the tops of the vertical rods.
  • So it is placed horizontally at $y = L = 22 \text{ cm}$.
  • Length $L = 22 \text{ cm}$.
  • Its center is midway between the two vertical rods at: $$
    x_3 = \frac{0 + 22}{2} = 11 \text{ cm}
    $$
  • $y_3 = 22 \text{ cm}$

Step 3: Masses of the rods

  • $m_1 = 14 \text{ g}$ (left vertical)
  • $m_2 = 14 \text{ g}$ (right vertical)
  • $m_3 = 42 \text{ g}$ (horizontal)

Step 4: Calculate total mass

$$
M = m_1 + m_2 + m_3 = 14 + 14 + 42 = 70 \text{ g}
$$

Step 5: Calculate $x_{\text{COM}}$

$$
x_{\text{COM}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}
= \frac{14 \cdot 0 + 14 \cdot 22 + 42 \cdot 11}{70}
$$

Calculate numerator:

  • $14 \cdot 0 = 0$
  • $14 \cdot 22 = 308$
  • $42 \cdot 11 = 462$

Sum = $0 + 308 + 462 = 770$

Now divide by total mass:

$$
x_{\text{COM}} = \frac{770}{70} = 11 \text{ cm}
$$

Step 6: Calculate $y_{\text{COM}}$

$$
y_{\text{COM}} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M}
= \frac{14 \cdot 11 + 14 \cdot 11 + 42 \cdot 22}{70}
$$

Calculate numerator:

  • $14 \cdot 11 = 154$
  • Another $14 \cdot 11 = 154$
  • $42 \cdot 22 = 924$

Sum = $154 + 154 + 924 = 1232$

Divide by total mass:

$$
y_{\text{COM}} = \frac{1232}{70} \approx 17.6 \text{ cm}
$$

Final answers:

  • (a) $x_{\text{COM}} = 11 \text{ cm}$
  • (b) $y_{\text{COM}} \approx 17.6 \text{ cm}$

Explanation (300 words):

To find the center of mass of the inverted U formed by three rods, we treat each rod as a uniform mass distributed evenly along its length. The center of mass for each rod is at its midpoint. We set a coordinate system with the bottom of the left vertical rod as the origin $(0,0)$.

For the two vertical rods, the centers of mass lie halfway up their length, both 11 cm above the origin, but one is at $x=0$ (left rod), and the other is at $x=22 \text{ cm}$ (right rod). The horizontal rod spans the top of the two vertical rods, so its center of mass is positioned at the midpoint between the vertical rods horizontally, $x=11 \text{ cm}$, and at the full rod length vertically, $y=22 \text{ cm}$.

Next, we consider the masses of each rod: the vertical rods weigh 14 g each, and the horizontal rod weighs 42 g. These masses, combined with their positional coordinates, allow us to find the combined center of mass using weighted averages.

The $x$-coordinate of the center of mass is weighted toward the horizontal rod’s center and the two vertical rods. Since the horizontal rod is centered at 11 cm and the two vertical rods at 0 cm and 22 cm, the resulting $x_{\text{COM}}$ is exactly at 11 cm, indicating symmetry in the horizontal direction.

For the $y$-coordinate, the center of mass is higher than the midpoints of the vertical rods due to the heavier horizontal rod at the top. Calculations give a vertical center of mass approximately at 17.6 cm, which is between the midpoint of the vertical rods (11 cm) and the top of the horizontal rod (22 cm), but closer to the top because of the horizontal rod’s higher mass.

This approach neatly combines geometry, symmetry, and mass distribution principles to determine the system’s center of mass in two dimensions.

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