Question

A cylindrical can of diameter \(10.0 \mathrm{~cm}\) contains some ballast so that it floats vertically in water. The mass of can and ballast is \(800.0 \mathrm{~g}\), and the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\). The can is lifted \(1.00 \mathrm{~cm}\) from its equilibrium position and released at \(t=0 .\) Find its vertical displacement from equilibrium as a function of time. Determine the period of the motion. Ignore the damping effect due to the viscosity of the water.

Short answer

The displacement function with respect to time is: \(y(t) = cos(\sqrt{\frac{k}{M_{can}}} \cdot t)\) and the period of motion is: \(T = 2\pi\sqrt{\frac{1}{25\pi g/\text{cm}}}\).

Step-by-step solution

Calculate the cross-sectional area of the cylinder

Since we are given the diameter of the cylindrical can, we can find the radius: $$r = \frac{10.0\,\text{cm}}{2} = 5.0\, \text{cm}$$ Now we can calculate the cross-sectional area (A) of the cylinder: $$A = \pi r^2 = \pi (5.0\,\text{cm})^2 = 25\pi\,\text{cm}^2$$.

Determine the buoyant force

Apply Archimedes' principle to find the buoyant force (F_B) acting on the cylinder. The principle states that the upward buoyant force is equal to the weight of the fluid that the body displaces: $$F_B = \rho_{water} \cdot V_{displaced} \cdot g$$

Equate the buoyant force to the weight to find the displaced volume

As the can is floating, its weight will be balanced by the upward buoyant force. Therefore, we can equate the two and find the displaced volume (V_displaced) of the water: $$F_B = M_{can}\cdot g\implies \rho_{water} \cdot V_{displaced} \cdot g = M_{can}\cdot g$$ Cancel out the gravitational constant (\(g\)) and solve for \(V_{displaced}\): $$V_{displaced} = \frac{M_{can}}{\rho_{water}} = \frac{800.0\,\text{g}}{1.00\,\frac{\text{g}}{\text{cm}^3}} = 800.0\,\text{cm}^3$$.

Obtain the equilibrium depth

To find the equilibrium depth, we need to relate the cross-sectional area of the cylinder to the displaced volume. We can do this by multiplying the area by the depth (h): $$V_{displaced} = A\cdot h$$ Now, we can find the depth at equilibrium by substituting the known values: $$h = \frac{V_{displaced}}{A} = \frac{800.0\,\text{cm}^3}{25\pi\,\text{cm}^2} = \frac{32}{\pi}\,\text{cm}$$.

Calculate the spring constant

Using Hooke's Law, we can calculate the spring constant (k) for the can: $$F_B = k \cdot h \implies k = \frac{\rho_{water} \cdot V_{displaced} \cdot g}{h}$$ Substitute the known values: $$k = \frac{800.0\,\text{g} \cdot g}{\frac{32}{\pi}\,\text{cm}} = \frac{25\pi\,\text{g} \cdot g}{\text{cm}}$$.

Setting up the differential equation for displacement

We can set up a differential equation for displacement (\(y\)) as follows: $$M_{can} \cdot \frac{d^2y}{dt^2} = F_{spring} = -k\cdot y$$ Rearranging and simplifying: $$\frac{d^2y}{dt^2} = -\frac{k}{M_{can}} \cdot y$$

Solving the differential equation and finding the displacement function

From the equation above, we can infer that the displacement function is a simple harmonic motion with angular frequency \(\omega = \sqrt{\frac{k}{M_{can}}}\). Given that the can is lifted 1 cm from its equilibrium position and released at \(t=0\), we can express the displacement function as: $$y(t) = Acos(\omega t)$$ Since it is released from 1 cm above equilibrium: $$y(0) = A=1\,\text{cm}$$ Thus, the displacement function is: $$y(t) = cos(\sqrt{\frac{k}{M_{can}}} \cdot t)$$

Calculate the period

We can find the period (T) of the simple harmonic motion using the formula: $$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{M_{can}}{k}}$$ Substitute the values for mass and spring constant: $$T = 2\pi\sqrt{\frac{800.0\,\text{g}}{25\pi\,\text{g} \cdot g/\text{cm}}} = 2\pi\sqrt{\frac{1}{25\pi g/\text{cm}}}$$ The period of motion of the cylindrical can in water, ignoring the damping effects, is \(T = 2\pi\sqrt{\frac{1}{25\pi g/\text{cm}}}\). The displacement function with respect to time is \(y(t) = cos(\sqrt{\frac{k}{M_{can}}} \cdot t)\).

Key concepts

Archimedes' principle

Archimedes' principle is a fundamental law of physics that describes how objects behave when submerged in a fluid. According to this principle, when an object is partially or fully submerged, it experiences an upward force known as the buoyant force. This force is equal to the weight of the fluid displaced by the object.

In practical terms, Archimedes' principle allows us to calculate whether an object will float or sink and how much of it will remain above the surface when placed in a fluid. For instance, if an object has a lesser density than the fluid, it will float because the buoyant force exceeds the object's weight. Conversely, if it's denser, it will sink because its weight is greater than the buoyant force.

When we look at the exercise where a cylindrical can is involved, the principle is used to determine the buoyant force that keeps it afloat. This force helps to establish a balance with the gravitational force acting on the mass of the can and ballast, leading to the floating and subsequent simple harmonic motion when disturbed.

Buoyant force

Buoyant force is an upward force exerted by a fluid that opposes the weight of an object immersed in it. This force plays a crucial role in determining the motion and stability of objects in fluids, such as boats, submarines, and even floating cans, as shown in our exercise.

The magnitude of the buoyant force can be calculated using Archimedes' principle, and it is essential in analyzing situations related to floating and submerged objects. It's also interesting to note that the buoyant force is responsible for the 'weightless' feeling we experience when swimming. The force is not affected by the depth of the object in the fluid; rather, it depends solely on the volume of the displaced fluid.

In the case of the cylindrical can in the given exercise, after calculating the buoyant force, we use it to understand the can's equilibrium position. This force is balanced against the weight of the can, leading to a definition of equilibrium from which simple harmonic motion occurs when the can is displaced.

Differential equation

Differential equations are mathematical tools used to describe relationships involving rates of change. They are central to much of contemporary science and engineering because they allow us to model physical systems evolving over time or space.

In the context of simple harmonic motion, the differential equation takes the form of Newton's second law of motion, which relates the net force acting on an object to its acceleration. This relationship can be used to describe the oscillatory motion of systems, such as springs, pendulums, and, as shown in our exercise, floating objects acting under a buoyant force. The general solution to the differential equation of simple harmonic motion reveals the position of the object as a function of time, helping us predict future states of the system.

By setting up a differential equation for the displacement of the floating can in the exercise, we identify the motion as simple harmonic with a specific angular frequency. Solving the equation provides an expression for the can’s displacement as a function of time, which shows how the object will oscillate about the equilibrium point over time.