Problem 17 - Applied Physics

Question

Find the volume of a brake cylinder whose diameter is \(4.00 \mathrm{~cm}\) and whose length is \(4.20 \mathrm{~cm}\).

Step-by-Step Solution

Answer

The volume of the brake cylinder is approximately \(52.75\,\text{cm}^3\).

Step 1: Identify the Cylinder Dimensions

The problem states that the brake cylinder has a diameter of \(4.00\,\text{cm}\) and a length of \(4.20\,\text{cm}\). Since the diameter is given, the radius \(r\) can be found using the formula \(r = \frac{d}{2}\), where \(d\) is the diameter. Thus, \(r = \frac{4.00}{2} = 2.00\,\text{cm}\).

Step 2: Write Down the Volume Formula for a Cylinder

The volume \(V\) of a cylinder is calculated using the formula \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height (or length) of the cylinder.

Step 3: Substitute the Radius and Length into the Formula

Substitute \(r = 2.00\,\text{cm}\) and \(h = 4.20\,\text{cm}\) into the formula, giving: \[ V = \pi (2.00)^2 (4.20) \].

Step 4: Calculate the Volume

First, calculate \((2.00)^2 = 4.00\). Then, calculate \(4.00 \times 4.20 = 16.80\). Finally, the volume is \( V = \pi \times 16.80 \). Using \(\pi \approx 3.14\), we find \( V \approx 3.14 \times 16.80 = 52.75\,\text{cm}^3 \).

Key Concepts

geometry

Geometry is a fascinating branch of mathematics that deals with shapes, sizes, and the properties of space. Cylinders, like the brake cylinder in our problem, are three-dimensional geometric shapes. They have a circular base and a specific height. Classically, geometry helps us understand and calculate measurements, such as area and volume.
For a cylinder, understanding its geometry is crucial to finding how much space it can hold or the material needed to create it. The knowledge of diameter, radius, and height are essential components here.

• Diameter: The full distance across the circle.
• Radius: Half of the diameter, which is a key measure for calculations.
• Height (Length): The distance between the two circular bases.

The study of geometry not only helps in mathematical applications but also in practical life situations, making it an invaluable tool in various fields.

volume formula

The volume of any three-dimensional shape is a measure of the space it occupies. For a cylinder, the volume formula is \[ V = \pi r^2 h \] where \( V \) stands for volume, \( r \) is the radius of the circular base, and \( h \) is the height (or length) of the cylinder.
This formula was derived by the principle of stacking infinitely thin circular disks from the bottom to the top. When these disks are aggregated over the entire height, we get the cylinder's volume.
Understanding this formula means learning how the cross-sectional area (\( \pi r^2 \)) multiplies by the length of the cylinder to create volume. It simplifies real-world tasks like determining the capacity of cylindrical containers, from a small pen to a massive oil tank.

cylinder dimensions

Cylinder dimensions are fundamental when it comes to calculating volume and surface area. They dictate how much space a cylinder can possess. Here are key aspects of cylinder dimensions:

• Radius: Calculated by dividing the diameter by two. In our example, the radius is \( 2.00 \mathrm{~cm} \).
• Diameter: The given full width of the circular base, which is \( 4.00 \mathrm{~cm} \) for the brake cylinder.
• Height (or Length): Given as \( 4.20 \mathrm{~cm} \), which is the distance between the two bases.

Knowing these dimensions helps not only in finding volumes but also in manufacturing, logistics, and crafting specific objects to fit a desired spatial constraint. The dimensions are foundational and guide all the subsequent calculations in geometric problems.

mathematical problem solving

Mathematical problem solving is a skill that involves reasoning, critical thinking, and applying mathematical principles to reach a solution. In our exercise, solving for the cylinder's volume required several straightforward steps:

• Identify given values: Recognize the diameter and height from the problem statement.
• Calculate the radius: Use the formula \( r = \frac{d}{2} \) to find the radius.
• Apply the formula: Input these values into the volume formula \( V = \pi r^2 h \).
• Perform calculations: Work through arithmetic operations and substitute approximate values for constants like \( \pi \).

Such a methodical approach doesn't just solve problems efficiently but also builds confidence in handling different types of mathematical challenges. Practice in mathematical problem solving allows students to adapt this method to varied scenarios and enhances their logical reasoning abilities.