Problem 42 - College Physics

Question

A spherical balloon has volume \(V\) and radius \(R .\) By what factor is its radius reduced if you let enough air out of the balloon to reduce its volume by a factor of \(8 ?\) (Consult Chapter 0 if necessary.)

Step-by-Step Solution

Answer

The radius is reduced by a factor of 2.

Step 1: Understand the Relationship Between Volume and Radius

The volume of a sphere is given by the formula \( V = \frac{4}{3} \pi R^3 \), where \( V \) is the volume and \( R \) is the radius. This relationship will help us understand how changes in volume affect the radius.

Step 2: Set Up the Volume Reduction

We know that the volume of the balloon is reduced by a factor of 8. If the initial volume is \( V \), then the new volume \( V' = \frac{V}{8} \). Substitute this into the volume formula to find the new radius: \( \frac{4}{3} \pi R'^3 = \frac{V}{8} \).

Step 3: Substitute the Initial Volume Expression

Substitute \( V = \frac{4}{3} \pi R^3 \) into the equation for the reduced volume: \( \frac{4}{3} \pi R'^3 = \frac{1}{8} \times \frac{4}{3} \pi R^3 \). Simplify this to \( R'^3 = \frac{R^3}{8} \).

Step 4: Solve for the New Radius

To find \( R' \), take the cube root of both sides of the equation: \( R' = \sqrt[3]{\frac{R^3}{8}} = \frac{R}{\sqrt[3]{8}} = \frac{R}{2} \). This shows that the new radius is half of the original radius.

Key Concepts

Sphere volume formula

The volume of a sphere can be calculated using the formula \( V = \frac{4}{3} \pi R^3 \). This equation tells us that the volume \( V \) is directly proportional to the cube of the sphere's radius \( R \). Because of this cubed relationship, any change in the radius will significantly impact the volume of the sphere.
With this formula, you can see that if the radius increases, the volume grows rapidly since it involves the cube of the radius. Conversely, decreasing the radius will drastically drop the volume.
This core principle is crucial for solving problems regarding changes in sphere dimensions, like the balloon problem at hand.

Radius reduction factor

The radius reduction factor is how much the sphere's radius is decreased when its volume changes. Here, the volume is reduced by a factor of 8, which implies a significant shrinkage.
To find out the factor by which the radius is reduced, we leverage the formula \( R'^3 = \frac{R^3}{8} \) from the original solution. Taking the cube root, we find that \( R' = \frac{R}{2} \).
This means the radius is reduced to half its original size when the volume is cut by a factor of 8. It's important to understand this factor as it simplifies how we calculate size changes in practical questions involving spheres.

Cube root calculation

Cube root calculation is a mathematical method used to determine what number, when cubed, gives the original number. Here, it's pivotal to determine the new radius of the balloon, \( R' \), after the volume reduction.
In our context, we are tasked to find \( R' \) from \( R'^3 = \frac{R^3}{8} \). By taking the cube root of \( \frac{R^3}{8} \), we get \( R' = \frac{R}{\sqrt[3]{8}} \). Since \( \sqrt[3]{8} \) equals 2, \( R' = \frac{R}{2} \).
Understanding cube root calculations helps in finding the scale or proportion changes in spherical dimensions.

Relationship between volume and radius

The relationship between volume and radius is squared in complexity. For spheres, the volume is not simply scaled to the radius but exponentially linked, as given by \( V = \frac{4}{3} \pi R^3 \). This exponential relation means even small changes in the radius will disproportionately affect the volume.
For our balloon, reducing the volume by a factor of 8 required us to carefully calculate the corresponding radius reduction. Because the volume formula involves radius cubed, halving the radius resulted in an eight-fold decrease in volume.
This understanding is particularly useful in geometric context or physics, allowing predictions and calculations of dimension and volume changes due to unknown alterations.