In Fig. 9-39, three uniform thin rods, each of length L = 15 cm, form an inverted U. The vertical rods each have a mass of 19 g; the horizontal rod has a mass of 35 g. What is the y coordinate of the system’s center of mass? (Give your answer in cm)
The Correct Answer and Explanation is:
Sure! Let’s analyze the problem step by step.
Problem Recap:
- Three uniform thin rods form an inverted U shape.
- Each rod has length $L = 15 \, \text{cm}$.
- Two vertical rods each have mass $m_v = 19 \, \text{g}$.
- The horizontal rod has mass $m_h = 35 \, \text{g}$.
- Find the y-coordinate of the system’s center of mass.
Step 1: Understand the shape and coordinate system
The rods form an inverted U, so:
- Two vertical rods stand upright, separated by the length of the horizontal rod.
- The horizontal rod connects the top ends of the vertical rods.
- We take the bottom of the vertical rods as $y = 0$.
- The rods are uniform, so their centers of mass are at their geometric centers.
Step 2: Locate the centers of mass of each rod along y-axis
- Vertical rods: Length = 15 cm
Mass = 19 g each
Since they stand vertically, their center of mass is halfway up, at $y = \frac{L}{2} = \frac{15}{2} = 7.5 \, \text{cm}$. - Horizontal rod: Length = 15 cm
Mass = 35 g
This rod is horizontal and at the top connecting the two vertical rods, so its vertical position is at the top of the vertical rods: $y = L = 15 \, \text{cm}$.
Step 3: Find total mass of the system
$$
M = 2 \times m_v + m_h = 2 \times 19 + 35 = 38 + 35 = 73 \, \text{g}
$$
Step 4: Calculate weighted average $y$ position (center of mass)
$$
y_{\text{cm}} = \frac{\sum m_i y_i}{\sum m_i} = \frac{m_v \times y_v + m_v \times y_v + m_h \times y_h}{M}
$$
Since both vertical rods have the same y-coordinate of center of mass:
$$
y_{\text{cm}} = \frac{19 \times 7.5 + 19 \times 7.5 + 35 \times 15}{73}
= \frac{(19 \times 7.5) \times 2 + 35 \times 15}{73}
$$
Calculate each term:
$$
19 \times 7.5 = 142.5
$$
$$
142.5 \times 2 = 285
$$
$$
35 \times 15 = 525
$$
Sum numerator:
$$
285 + 525 = 810
$$
Divide by total mass:
$$
y_{\text{cm}} = \frac{810}{73} \approx 11.1 \, \text{cm}
$$
Final answer:
$$
\boxed{y_{\text{cm}} \approx 11.1 \, \text{cm}}
$$
Detailed Explanation (300+ words):
This problem involves finding the vertical position of the center of mass of a composite system consisting of three rods arranged in an inverted U shape. Each rod is uniform, meaning its mass is distributed evenly along its length.
The center of mass of a uniform rod is at its midpoint. Therefore, for the two vertical rods, each 15 cm long, their centers of mass are located at half their length, or 7.5 cm above the bottom of the rods. The horizontal rod, also 15 cm long but heavier, connects the tops of the vertical rods. Its center of mass is located at its midpoint, which in this configuration lies 15 cm above the bottom of the system (since it is at the top of the vertical rods).
To find the overall center of mass, we calculate the weighted average of the individual rods’ centers of mass, using their masses as weights. The total mass is the sum of all three rods’ masses: 19 g for each vertical rod, and 35 g for the horizontal rod, totaling 73 g.
The combined vertical moment of the rods about the bottom is the sum of the mass times their respective center of mass positions. For the two vertical rods, each contributes $19 \times 7.5 = 142.5$ g·cm, summing to 285 g·cm. The horizontal rod contributes $35 \times 15 = 525$ g·cm.
Summing these moments gives 810 g·cm. Dividing by the total mass (73 g) gives the y-coordinate of the center of mass as approximately 11.1 cm.
This means that the system’s center of mass lies above the midpoints of the vertical rods but below the top of the horizontal rod, reflecting the heavier mass and position of the horizontal rod at the top.
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