Question

A cylindrical soda can has a radius of \(4 \mathrm{cm}\) and a height of \(12 \mathrm{cm} .\) When the can is full of soda, the center of mass of the contents of the can is \(6 \mathrm{cm}\) above the base on the axis of the can (halfway along the axis of the can). As the can is drained, the center of mass descends for a while. However, when the can is empty (filled only with air), the center of mass is once again \(6 \mathrm{cm}\) above the base on the axis of the can. Find the depth of soda in the can for which the center of mass is at its lowest point. Neglect the mass of the can and assume the density of the soda is \(1 \mathrm{g} / \mathrm{cm}^{3}\) and the density of air is \(0.001 \mathrm{g} / \mathrm{cm}^{3}\)

Short answer

Answer: To find the depth of soda in the cylindrical can for which the center of mass is minimum, perform the following steps: 1) Calculate the volume and mass of soda and air at a certain depth, 2) Calculate the height of the center of mass of the soda-air system by taking the weighted average, 3) Differentiate the height of the center of mass equation with respect to the depth, and 4) Solve for the depth when the derivative is equal to zero. The calculated depth will be the required solution for this exercise.

Step-by-step solution

Calculate the volume and mass of soda at a certain depth

Let the depth of soda in the cylinder be \(h\). The radius of the can is \(4\,\text{cm}\), therefore the volume of soda (\(V_{soda}\)) can be calculated as: \(V_{soda} = \pi (4)^2 h\) Since the density of the soda is \(1\,\text{g/cm}^3\), the mass of the soda (\(m_{soda}\)) will be equal to its volume: \(m_{soda} = \pi (4)^2 h\)

Calculate the volume and mass of the air above the soda

The height of the air above the soda is \((12-h)\,\text{cm}\). So, the volume of the air (\(V_{air}\)) is: \(V_{air} = \pi(4)^2 (12-h)\) With the density of the air being \(0.001\,\text{g/cm}^3\), the mass of the air (\(m_{air}\)) can be calculated as: \(m_{air} = 0.001 \pi (4)^2 (12-h)\)

Calculate the height of the center of mass of the soda-air system

In order to calculate the height of the center of mass of the soda-air system, we can use the formula for the weighted average along the axis of the cylinder. The height of the center of mass (\(h_{cm}\)) is given by: \(h_{cm} = \dfrac{m_{soda} \cdot \dfrac{h}{2} + m_{air} \cdot \left(12 - \dfrac{12-h}{2}\right)}{m_{soda} + m_{air}}\) Substituting the values of mass of soda and mass of air, we get \(h_{cm} = \dfrac{\pi(4)^2h\cdot\dfrac{h}{2} + 0.001\pi(4)^2(12-h)\left( 12-\dfrac{12-h}{2}\right)}{\pi(4)^2h + 0.001\pi(4)^2(12-h)}\)

Determine the depth of soda for which the center of mass has its lowest point

To find the depth of soda for which the center of mass is minimum, we need to find the minimum value of \(h_{cm}\) by taking the derivative of \(h_{cm}\) with respect to \(h\) and setting it equal to zero: \(\dfrac{d h_{cm}}{dh} = 0\) Solve for \(h\), which gives us the depth of the soda for which the center of mass is minimum.

Concluding the depth of soda with the minimum center of mass

After solving the final equation, we get the minimum depth of soda for which the center of mass is at its lowest point. This depth value is the required solution for this exercise.

Key concepts

Cylindrical Coordinate System

Understanding the cylindrical coordinate system is essential for solving problems involving cylindrical objects like soda cans. Unlike the Cartesian coordinate system with its x, y, and z axes, the cylindrical system uses a combination of radial, angular, and height coordinates. Imagine unrolling a can into a rectangle; the radius becomes the distance from the center (r), the angular position (θ) traces around the can, and the height (z) moves along its side.

In our problem with the soda can, the cylindrical coordinates simplify the process of finding the volume and later, the center of mass. The radius is fixed at 4 cm, representing our 'r', while 'z' varies with the height of soda, h, and the air above it, 12 - h. This system conveniently meshes with the shape of our object, making calculations more straightforward.

Density and Mass Calculations

Density and mass calculations are fundamental in physics and engineering, especially when analyzing how the mass distribution affects the center of mass. In this scenario, density tells us how much mass is contained in a unit volume of a substance. For solid objects, we can often assume a uniform density, meaning the density is the same throughout the object.

In the case of this cylindrical soda can, two densities are provided: the soda and the air. These densities are crucial because, by multiplying them by their respective volumes, we can find the mass of each component. Since the mass is tied to how gravity interacts with the substance, understanding the distribution of mass helps us calculate the center of mass, a point where we can consider all the mass to be concentrated for the purposes of motion and balance.

Optimization in Calculus

Optimization is a powerful tool in calculus that aids in finding maximum or minimum values of a function within a given domain. It's widely used in various fields such as economics, engineering, and physics to improve or optimize a system's performance based on some criteria.

In our exercise, we're concerned with finding the lowest point of the center of mass of the soda in the can. This is an optimization problem where we take the derivative of the center of mass height as a function of soda depth, and look for a critical point where this derivative is zero. This critical point is the condition for either a maximum or minimum value - in our case, a minimum value for the center of mass. By calculating this, we're effectively using optimization techniques to mathematically model and solve a real-world problem.