Problem 5 - University Physics with Modern Physics

Question

A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?

Step-by-Step Solution

Answer

The ratio of the radius of the aluminum sphere to the lead sphere is approximately 1.61.

Step 1: Understand the relationship between mass, density, and volume

The mass of a sphere is given by \( m = \rho V \) where \( \rho \) is the density and \( V \) is the volume. The volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. Since both spheres have the same mass, their masses can be expressed as:\[ m = \rho_{\text{lead}} \cdot V_{\text{lead}} = \rho_{\text{aluminum}} \cdot V_{\text{aluminum}} \]

Step 2: Express each volume in terms of radius

Since the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), substitute for each material:- For lead: \( V_{\text{lead}} = \frac{4}{3} \pi r_{\text{lead}}^3 \)- For aluminum: \( V_{\text{aluminum}} = \frac{4}{3} \pi r_{\text{aluminum}}^3 \)

Step 3: Set up the equation for equal mass and simplify

Since both spheres have the same mass, \[ \rho_{\text{lead}} \cdot \frac{4}{3} \pi r_{\text{lead}}^3 = \rho_{\text{aluminum}} \cdot \frac{4}{3} \pi r_{\text{aluminum}}^3 \] Simplify the equation by cancelling \( \frac{4}{3} \pi \) from both sides: \[ \rho_{\text{lead}} \cdot r_{\text{lead}}^3 = \rho_{\text{aluminum}} \cdot r_{\text{aluminum}}^3 \]

Step 4: Solve for the radius ratio

Re-arrange the equation to solve for the ratio \( \frac{r_{\text{aluminum}}}{r_{\text{lead}}} \): \[ \left(\frac{r_{\text{aluminum}}}{r_{\text{lead}}}\right)^3 = \frac{\rho_{\text{lead}}}{\rho_{\text{aluminum}}} \] To find the ratio of the radii, take the cube root: \[ \frac{r_{\text{aluminum}}}{r_{\text{lead}}} = \sqrt[3]{\frac{\rho_{\text{lead}}}{\rho_{\text{aluminum}}}} \]

Step 5: Substitute densities and calculate the cube root

The density of lead is approximately \( 11.34 \, \text{g/cm}^3 \), and the density of aluminum is approximately \( 2.70 \, \text{g/cm}^3 \). Substitute these values into the equation: \[ \frac{r_{\text{aluminum}}}{r_{\text{lead}}} = \sqrt[3]{\frac{11.34}{2.70}} \] Calculate the ratio: \[ \frac{r_{\text{aluminum}}}{r_{\text{lead}}} \approx \sqrt[3]{4.2} \approx 1.61 \]

Step 6: Conclusion

Therefore, the ratio of the radius of the aluminum sphere to the radius of the lead sphere is approximately \( 1.61 \).

Key Concepts

Understanding Mass

Mass is a fundamental concept in physics that describes how much matter is present in an object. It is commonly measured in grams (g) or kilograms (kg), and it reflects the amount of substance in an object.

• Mass is a scalar quantity, meaning it has magnitude but no direction.
• It remains constant regardless of location, unlike weight, which can change with gravitational pull.
• In our problem, the mass of the two spheres, one made of aluminum and the other of lead, is the same.

This means that both spheres contain the same total amount of matter, irrespective of their size or density differences.

Exploring Volume

Volume is a measure of the space an object occupies. For three-dimensional objects like spheres, it is typically measured in cubic centimeters (cm³) or liters. The volume of a sphere can be calculated using the formula:\[ V = \frac{4}{3} \pi r^3 \]where \( r \) is the radius of the sphere. The formula shows that volume depends strongly on the radius.

• The larger the radius, the larger the volume.
• A small increase in the radius can lead to a significant increase in volume because the radius is cubed in the formula.

In the exercise, understanding volume is crucial for calculating how different densities affect the size of the spheres needed to have the same mass.

Significance of Sphere Radius

The radius is a key dimension of a sphere. It is the distance from the center of the sphere to any point on its surface. The radius has a profound impact on other properties of a sphere, such as its volume.In our exercise, both spheres have the same mass but are made of different materials with different densities:

• Higher density materials like lead result in smaller spheres for the same mass.
• Lower density materials like aluminum require larger radii to achieve the same mass.

The exercise tasked us with finding the ratio between these radii. Using the density values provided, and knowing the mass remains constant, we derived the ratio\[ \frac{r_{\text{aluminum}}}{r_{\text{lead}}} = \sqrt[3]{\frac{\rho_{\text{lead}}}{\rho_{\text{aluminum}}}} \]This calculation underscores how the density differences influence the sphere's form to maintain equivalent mass.