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

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Question: Calculate the period of the motion of a floating cylindrical can that has a mass, including the ballast, of 800.0 g and is displaced vertically in water by 1.00 cm. The damping effect is ignored, assuming the motion is simple harmonic. Answer: The period of the motion is approximately 0.324 seconds.

Step by step solution

01

Calculate the buoyant force in the equilibrium position

To start, let's first calculate the buoyant force acting on the can when it is in its equilibrium position. The buoyant force (FB) is given by the formula: \(F_B = V_\text{submerged} \times \rho_\text{water} \times g \) where \(V_\text{submerged}\) is the volume of the submerged part of the can, \(\rho_\text{water}\) is the density of water, and \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^2}\) or \(980 \mathrm{cm/s^2}\)). In the equilibrium position, the buoyant force is equal to the weight of the can and ballast. Therefore, we can write: \(F_B = m_\text{can} \times g\) where \(m_\text{can}\) is the mass of the can and ballast (800 g). Solving for the buoyant force, we get: \(F_B = 800 \times 980 = 784 \times 10^3 \mathrm{dynes}\).
02

Calculate the submerged volume at equilibrium

Now that we have the buoyant force at the equilibrium position, we can find the submerged volume at equilibrium using the buoyant force equation: \(V_\text{submerged} = \frac{F_B}{\rho_\text{water} \times g}\) Plugging in the values, we get: \(V_\text{submerged} = \frac{784 \times 10^3 \mathrm{dynes}}{1\cdot 980 \ \mathrm{g\ cm^{-3} \times cm\ s^{-2}}}\) \(V_\text{submerged} = 800 \ \mathrm{cm^3}\)
03

Calculate the change in submerged volume due to displacement

The can is lifted 1.00 cm from its equilibrium position, causing a change in the submerged volume. The change in submerged volume (\(\Delta V\)) can be calculated as: \(\Delta V = A \times \Delta h\) where \(A\) is the cross-sectional area of the can, and \(\Delta h\) is the vertical displacement (1.00 cm). The cross-sectional area can be calculated using the diameter (10.0 cm): \(A = \frac{\pi d^2}{4} = \frac{\pi \times (10.0)^2}{4} = 78.5 \ \mathrm{cm^2}\) Now, we can calculate the change in submerged volume: \(\Delta V = 78.5 \ \mathrm{cm^2} \times 1.0 \ \mathrm{cm} = 78.5 \ \mathrm{cm^3}\)
04

Determine the spring constant

The can experiences a force due to the buoyancy change when displaced, which we can consider like a spring force. To relate this force with the displacement, we can use the formula: \(F_\text{spring} = -k \Delta h\) where \(k\) is the spring constant. The spring force can also be calculated as the change in the buoyant force, which is: \(F_\text{spring} = \Delta F_B = \Delta V \times \rho_\text{water} \times g\) Substituting the values, we get: \(F_\text{spring} = 78.5 \ \mathrm{cm^3} \times 1 \ \mathrm{g\ cm^{-3}} \times 980 \ \mathrm{cm\ s^{-2}} = 76.93 \times 10^3 \ \mathrm{dynes}\) Now, we can find the spring constant: \(k = - \frac{F_\text{spring}}{\Delta h} = - \frac{76.93 \times 10^3 \ \mathrm{dynes}}{1 \ \mathrm{cm}} = 76.93 \times 10^3 \ \mathrm{dyn \ cm^{-1}}\)
05

Calculate the time period of the motion

Since the motion of the can is simple harmonic, we can use the formula: \(T = 2 \pi \sqrt{\frac{m_\text{can}}{k}}\) where \(T\) is the time period of the motion. Substituting the mass of the can and ballast (800 g) and the spring constant, we get: \(T = 2 \pi \sqrt{\frac{800 \ \mathrm{g}}{76.93 \times 10^3 \ \mathrm{dyn \ cm^{-1}}}}\) \(T \approx 0.324 \ \mathrm{s}\) The period of the motion is approximately 0.324 seconds.
06

Determine the vertical displacement as a function of time

Since the motion is simple harmonic, the vertical displacement (\(y(t)\)) as a function of time can be represented as: \(y(t) = Y_0 \cos(\omega t)\) where \(Y_0\) is the initial displacement (1.00 cm), \(\omega\) is the angular frequency, and \(t\) is the time. The angular frequency can be calculated as: \(\omega = \frac{2 \pi}{T}\) Substituting the period calculated in step 5, we get: \(\omega \approx 19.38 \ \mathrm{rad/s}\) Therefore, the vertical displacement as a function of time is: \(y(t) = 1 \ \mathrm{cm} \cos(19.38 t)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simple Harmonic Motion
Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that displacement. It's like a swinging pendulum or a mass attached to a spring that moves back and forth. The classic hallmark of simple harmonic motion is this repetitive, oscillating behavior.
In physics problems, simple harmonic motion can be analyzed using formulas that describe the displacement, velocity, and acceleration of an object in SHM. One key formula involves calculating the period (T) of the motion, which is the amount of time it takes to complete one full cycle of oscillation. This can be determined by the equation:
\[ T = 2 \pi \sqrt{\frac{m}{k}} \]
Here, 'm' represents the mass of the object in motion, and 'k' is the spring constant, which is a measure of the stiffness of the spring in spring-mass systems. The angular frequency (ω), given by \( \omega = \frac{2 \pi}{T} \), describes how many radians the object passes through per unit of time. For a cylinder oscillating vertically in water, as described in the exercise, we treat the buoyant force as a 'restoring force' much like the force exerted by a spring. When the can is displaced from its equilibrium position and released, it exhibits SHM, oscillating about this central point.
Buoyancy in Physics
Buoyancy is the upward force exerted by a fluid, such as water, that opposes the weight of an immersed object. It's the reason why objects like ships, buoys, and even our cylindrical can in the exercise stay afloat. Physically, buoyancy is a consequence of fluid pressure differences: the pressure at the bottom of an object submerged in a fluid is greater than the pressure at the top, due to the weight of the fluid above.
To calculate the buoyant force (FB), we use Archimedes' principle, which states that the buoyant force on a submerged object is equal to the weight of the fluid displaced by the object:
\[ F_B = V_{\text{submerged}} \times \rho_{\text{water}} \times g \]
Where \(V_{\text{submerged}}\) is the volume of the fluid displaced by the object, \(\rho_{\text{water}}\) is the density of the fluid, and \(g\) is the acceleration due to gravity. In the exercise provided, we apply Archimedes' principle to find the equilibrium condition and use the concept of buoyancy to analyze the motion of the cylindrical can in the water.
Oscillations in Mechanics
Oscillations refer to back and forth motion around a central point, called the equilibrium position. In mechanics, oscillatory motion is seen in various systems, such as pendulums, springs, and even atoms in molecules. The oscillation is a fundamental concept that applies to many areas of physics including acoustics, optics, and quantum mechanics.
In the context of our problem with the cylindrical can, when the can is displaced and released, it exhibits oscillatory motion by moving up and down in the water. This motion fits into oscillations in mechanics because it follows a predictable pattern governed by Newton’s laws. To fully analyze this behavior and predict the can's position at any given time, we invoke the principles of harmonic motion. The oscillation can be described by the displacement function:
\[ y(t) = Y_0 \cos(\omega t) \]
Here, \(y(t)\) is the vertical displacement at time \(t\), \(Y_0\) is the initial displacement (amplitude), and \(\omega\) is the angular frequency. Using these concepts, we can predict how the can will move over time, which is crucial for fields like engineering where understanding the dynamics of structures in fluid environments is necessary.

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Most popular questions from this chapter

An \(80.0-\mathrm{kg}\) bungee jumper is enjoying an afternoon of jumps. The jumper's first oscillation has an amplitude of \(10.0 \mathrm{~m}\) and a period of \(5.00 \mathrm{~s}\). Treating the bungee cord as spring with no damping, calculate each of the following: a) the spring constant of the bungee cord. b) the bungee jumper's maximum speed during the ascillation, and c) the time for the amplitude to decrease to \(2.00 \mathrm{~m}\) (with air resistance providing the damping of the oscillations at \(7.50 \mathrm{~kg} / \mathrm{s})\)

A mass \(M=0.460 \mathrm{~kg}\) moves with an initial speed \(v=3.20 \mathrm{~m} / \mathrm{s}\) on a level frictionless air track. The mass is initially a distance \(D=0.250 \mathrm{~m}\) away from a spring with \(k=\) \(840 \mathrm{~N} / \mathrm{m}\), which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance \(d\), before reversing direction. After bouncing off the spring the mass travels with the same speed \(v\), but in the opposite dircction. a) Determine the maximum distance that the spring is compressed. b) Find the total elapsed time until the mass returns to its starting point. (Hint: The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.)

A 2.0 -kg mass attached to a spring is displaced \(8.0 \mathrm{~cm}\) from the equilibrium position. It is released and then oscillates with a frequency of \(4.0 \mathrm{~Hz}\) a) What is the energy of the motion when the mass passes through the equilibrium position? b) What is the speed of the mass when it is \(20 \mathrm{~cm}\) from the equilibrium position?

The figure shows a mass \(m_{2}=20.0\) g resting on top of a mass \(m_{1}=20.0 \mathrm{~g}\) which is attached to a spring with \(k=10.0 \mathrm{~N} / \mathrm{m}\) The coefficient of static friction between the two masses is 0.600 . The masses are oscillating with simple harmonic motion on a frictionless surface. What is the maximum amplitude the oscillation can have without \(m_{2}\) slipping off \(m_{1} ?\)

A mass \(m=1.00 \mathrm{~kg}\) in a spring-mass system with \(k=\) \(1.00 \mathrm{~N} / \mathrm{m}\) is observed to be moving to the right, past its equi. librium position with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\) at time \(t=0 .\) a) Ignoring all damping, determine the equation of motion. b) Suppose the initial conditions are such that at time \(t=0\), the mass is at \(x=0.500 \mathrm{~m}\) and moving to the right with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine the new equation of motion. Assume the same spring constant and mass.

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