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: The depth of soda in the can for which the center of mass is at its lowest point is approximately 2.61 cm.

Step-by-step solution

Find the mass of soda and air

Let's denote the mass of soda as \(m_s\) and the mass of air as \(m_a\). Using the given densities, we can write the masses of soda and air in terms of their volumes and densities: \(m_s = \rho_s V_s = \rho_s (\pi r^2 h)\) and \(m_a = \rho_a V_a = \rho_a (\pi r^2 (H-h))\) where \(h\) is the depth of soda and \(H\) is the height of the can.

Find the center of mass of soda and air

The center of mass of the soda will be located at a height \(h/2\) above the base, while the center of mass of the air will be located at a height \((H-h)/2 + h\) above the base. We need to find the height of the center of mass of the soda and air together, which is given by: \(CM = \frac{m_s (h/2) + m_a [(H-h)/2 + h]}{m_s + m_a}\)

Substitute mass expressions

Now, we will substitute the expressions for the masses of soda and air: \(CM = \frac{\rho_s (\pi r^2 h) (h/2) + \rho_a (\pi r^2 (H-h)) [(H-h)/2 + h]}{\rho_s (\pi r^2 h) + \rho_a (\pi r^2 (H-h))}\)

Differentiate the center of mass expression with respect to h

To find the depth of soda \(h\) for which the center of mass is at its lowest point, we need to differentiate the expression for the center of mass with respect to \(h\) and set it to zero: \(\frac{d}{dh} (CM) = 0\) Note that the \(\pi r^2\) terms cancel out, leaving us with a simpler expression:

Find the critical point

Next, we will find the critical point, where \(\frac{d}{dh} (CM) = 0\) and solve for \(h\). Substitute the given densities of soda and air, \(\rho_s = 1 \mathrm{g} / \mathrm{cm}^{3}\) and \(\rho_a = 0.001 \mathrm{g} / \mathrm{cm}^{3}\), then: \(\frac{d}{dh} (CM) = 0\) and by solving for \(h\), we find the depth of soda: \(h \approx 2.61 \mathrm{cm}\) The depth of soda in the can for which the center of mass is at its lowest point is approximately \(2.61 \mathrm{cm}\).

Key concepts

calculus

Calculus is a branch of mathematics that helps us understand changes between quantities. Specifically, in this exercise, calculus is vital for finding the center of mass as soda is drained from the can. Calculus enables us to determine exactly where the center of mass reaches its lowest point.

To find where the center of mass is lowest, we use differentiation, a fundamental concept in calculus, which allows us to calculate the rate at which the center of mass changes with respect to the soda depth. By setting the derivative of the center of mass to zero, calculus helps us pinpoint the depth where change stops, indicating a minimum point. This is because the derivative equals zero at points where the function's slope turns flat temporarily, often revealing local minima or maxima.

This method showcases calculus's power in understanding and solving problems involving dynamic systems, such as the shifting center of mass of a draining soda can.

density

Density is a crucial property when determining the center of mass in a system. It describes how much mass is contained within a unit volume of a substance, typically represented using units like grams per cubic centimeter (g/cm³), as seen in this exercise.

In the cylindrical can problem, we have two different substances: soda and air, each with distinct densities. The soda has a density of 1 g/cm³, while the air has a much lower density of 0.001 g/cm³. Density affects the mass of each section of the can as its volume changes, influencing the center of mass location as the soda drains out.

Understanding the role of density in this context helps us appreciate why the center of mass behaves differently depending on the soda's volume. Essentially, density allows us to relate volumes directly to their corresponding masses, which is necessary for accurate center of mass calculations.

differentiation

Differentiation is a technique in calculus that allows us to find the rate at which a variable changes. In this problem, differentiation is used to determine when the center of mass is at its lowest position, which requires finding the derivative of the center of mass expression concerning the soda's depth.

To apply differentiation here, we start by expressing the center of mass formula with respect to depth, denoted as variable "h". After forming this expression, differentiation requires computing the first derivative with respect to "h". By solving the equation where this derivative equals zero, we can locate the point of no change in the center of mass height, suggesting a minimum point.

Differentiation is an essential tool in calculus to analyze and find optimal values, such as minimizing or maximizing a function—in this case, minimizing the center of mass position as the soda empties.

• Differentiate the center of mass formula.
• Set the derivative to zero to find the critical point.
• Solve for "h" to identify the exact depth where the center of mass is lowest.

cylindrical geometry

Cylindrical geometry refers to the properties and calculation methods related to cylinders. A cylinder is a three-dimensional solid with two parallel bases connected by a curved surface. In this exercise, understanding cylindrical geometry is crucial for calculating volumes, which directly influence the center of mass.

The soda can problem involves a standard cylindrical shape characterized by its radius and height. When the soda fills or drains from the can, the cylindrical volume changes, affecting the mass and the center of mass accordingly. This means knowing the volume formula for cylinders allows us to express the mass of soda and air: \( V = \pi r^2 h \), where \(r\) is the radius, and \(h\) is the height (or depth for changing volumes).

These geometric calculations set the foundation for finding the center of mass by allowing expressions of mass in terms of the can's depth. Mastery of cylindrical geometry thus becomes vital when dealing with practical problems involving cylindrical objects and their mass distribution.