Question

If a thin \(1-\mathrm{m}\) cylindrical rod has a density of \(\rho=1 \mathrm{g} / \mathrm{cm}\) for its left half and a density of \(\rho=2 \mathrm{g} / \mathrm{cm}\) for its right half, what is its mass and where is its center of mass?

Short answer

Answer: The mass of the cylindrical rod is 0.15 kg, and its center of mass is 0.5833 m from the left end.

Step-by-step solution

Convert the length to cm

As 1 meter is equal to 100 centimeters, the total length of the rod in cm is: \(1m = 100cm\) Now the rod is divided into two equal halves, each with different densities. So, each half has a length of \(50cm\)

Determine the volume of each half

Both halves of the rod have the same volume, since they have the same length (50 cm) and cross-sectional area. Also, we have a thin cylindrical rod with a constant cross-sectional area, so we can assume that its volume is proportional to its length. Hence, the volume of the left half (V1) and the right half (V2) is proportional to their lengths: \(V1 \propto 50\,cm\) \(V2 \propto 50\,cm\)

Determine the mass of each half

We can now find the mass of each half by using the respective densities and the proportional volumes: \( \text{Mass of the left half (M1)} = \rho_1 \cdot V1 = 1\ g/cm³ \cdot 50\ cm \propto 50\ g\) \( \text{Mass of the right half (M2)} = \rho_2 \cdot V2 = 2\ g/cm³ \cdot 50\ cm \propto 100\ g\)

Convert the masses back to kg

Since 1 kg is equal to 1000 g, we can convert the masses of the left and right halves to kilograms: \(M1 = 50\ g \cdot \frac{1\ kg}{1000\ g} = 0.05\ kg\) \(M2 = 100\ g \cdot \frac{1\ kg}{1000\ g} = 0.1\ kg\)

Calculate the total mass

Now, we can find the total mass of the rod (M) by adding the masses of both halves: \(M = M1 + M2 = 0.05\ kg + 0.1\ kg = 0.15\ kg\)

Calculate the center of mass

The center of mass of the rod can be found using the following equation: \(X_{\text{center of mass}} = \frac{M1 \cdot X1 + M2 \cdot X2}{M}\) Where \(X1\) and \(X2\) are the distances from the left end of the rod to the center of mass of the left and right halves respectively. Since the rod is divided into equal halves with lengths of 50 cm, \(X1 = \frac{50\ cm}{2} = 25\ cm\) \(X2 = 50\ cm + \frac{50\ cm}{2} = 75\ cm\) Now, substitute the values of \(M1\), \(X1\), \(M2\), \(X2\), and \(M\) into the formula: \(X_{\text{center of mass}} = \frac{0.05\ kg\cdot 25\ cm + 0.1\ kg\cdot 75\ cm}{0.15\ kg}\) \(X_{\text{center of mass}} = \frac{1.25\ kg\cdot cm + 7.5\ kg\cdot cm}{0.15\ kg}\) \(X_{\text{center of mass}} = \frac{8.75\ kg\cdot cm}{0.15\ kg} = 58.33\ cm\) To report the center of mass in meters, we can convert 58.33 cm back to meters: \(X_{\text{center of mass}} = 58.33\,cm \cdot \frac{1\,m}{100\,cm} = 0.5833\,m\)

Final Answer

The mass of the cylindrical rod is \(0.15\ kg\) and its center of mass is at a distance of \(0.5833\ m\) from the left end.

Key concepts

Center of Mass

The center of mass is the point at which the distribution of mass is balanced for an object. For a symmetric object with uniform density, the center of mass is located at its geometric center.
However, for objects with varying densities like a cylindrical rod with different density parts, the center of mass shifts towards the heavier portion.
In our exercise, we calculated the center of mass for a rod whose left half is less dense and right half is more dense. The formula used is:

• \( X_{\text{center of mass}} = \frac{M1 \cdot X1 + M2 \cdot X2}{M} \)

Where \( M1 \) and \( M2 \) are masses, and \( X1 \) and \( X2 \) are their distances from a reference point. This weighted average leads us to find the balancing point of the rod, which indicates the center of mass, situated at \( 0.5833 \, m \) from the left end in our specific example.

Density

Density is a measure of how much mass is contained in a given volume. It is defined as mass per unit volume and is expressed in g/cm³ or kg/m³.
For the cylindrical rod in the problem, density is not uniform. The left half has a density of \( 1 \, g/cm^3 \), while the right half is denser with \( 2 \, g/cm^3 \).
Understanding density differences helps us calculate mass correctly over sections of a rod. Here, we visually split the rod into two halves, taking into account the given densities to determine how the mass is distributed along its length, important for locating the center of mass.

Mass Calculation

Calculating mass involves understanding both density and volume. For uniform cross-sections like a cylindrical rod, volume relates directly to length for each segment.
The mass of a section is calculated by multiplying its density by its volume. When density varies across the rod, each section must be calculated separately:

• Left half's mass: \( \rho_1 \cdot V1 = 1 \, g/cm^3 \cdot 50 \, cm = 50 \, g \)
• Right half's mass: \( \rho_2 \cdot V2 = 2 \, g/cm^3 \cdot 50 \, cm = 100 \, g \)

Convert these masses to kilograms for standard use, making sure to sum them correctly for the complete mass of the object.

Cylindrical Rod

A cylindrical rod is a three-dimensional object with a circular cross-section and a specific length. They are commonly used in physics problems to simplify objects while focusing on linear properties like length and density.
In this exercise, our cylindrical rod is 1 meter in total length, separated into two equal halves with different densities. The shape and division allow us to assume constant cross-section, making density-related calculations straightforward.
By focusing on each half individually, we can explore how different densities affect the overall properties of the rod, demonstrating essential calculus applications for both academics and real-world scenarios.