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Chapter 6: Problem 27

A small spherical ball of radius \(r\) is released from its completely submerged position (as shown in the figure) in a liquid whose density varies with height \(h\) (measured from the bottom) as \(\rho_{L}=\rho_{0}\left[4-\left(3 h / h_{0}\right)\right]\). The density of the ball is \((5 / 2) \rho_{0} .\) The height of the vessel is \(h_{0}=12 / \pi^{2}\) Consider \(r \ll h_{0}\) and \(g=10 \mathrm{~m} / \mathrm{s}^{2}\). The motion of the ball in the vessel is (1) oscillatory but not simple harmonic motion (2) simple harmonic motion with time period \(2 \mathrm{~s}\) (3) simple harmonic motion with time period \(1 \mathrm{~s}\) (4) the motion is not oscillatory

Short Answer

Expert verified
The motion is simple harmonic with a time period of 2 seconds, option (2).

Step by step solution

01

Identify the Forces Involved

When the ball is released, the primary forces acting on it are the weight of the ball and the buoyant force due to the displaced liquid. The weight of the ball is given by the product of its volume, density, and gravitational acceleration: \( F_{\text{gravity}} = \frac{4}{3} \pi r^3 \left(\frac{5}{2} \rho_0 \right) g \).
02

Calculate the Buoyant Force

The buoyant force, \( F_{b} \), depends on the density of the liquid at the height where the ball is submerged. It is calculated as the product of the volume of the displaced liquid, its variable density \( \rho_L \), and gravitational acceleration. At a height \( h \), the density of the liquid is \( \rho_{L} = \rho_{0} \left[ 4 - \frac{3h}{h_0} \right] \). Therefore, \( F_{b} = \frac{4}{3} \pi r^3 \rho_L g = \frac{4}{3} \pi r^3 \rho_0 \left[ 4 - \frac{3h}{h_0} \right] g \).
03

Set Up the Net Force Equation

The net force on the ball is the difference between the buoyant force and the gravitational force: \( F_{\text{net}} = F_b - F_{\text{gravity}} = \frac{4}{3} \pi r^3 \rho_0 \left[ 4 - \frac{3h}{h_0} \right] g - \frac{4}{3} \pi r^3 \left(\frac{5}{2} \rho_0 \right) g \).
04

Simplify the Net Force Equation

Simplifying the expression for net force, we get: \( F_{\text{net}} = \frac{4}{3} \pi r^3 \rho_0 g \left[ 4 - \frac{3h}{h_0} - \frac{5}{2} \right] \). Further simplification yields: \( F_{\text{net}} = \frac{4}{3} \pi r^3 \rho_0 g \left[ -\frac{3}{2} - \frac{3h}{h_0} \right] \).
05

Analyze Acceleration and Motion Type

Apply Newton's second law: \( F_{\text{net}} = m_{\text{ball}} \cdot a \), where the mass of the ball \( m_{\text{ball}} = \frac{4}{3} \pi r^3 \left(\frac{5}{2} \rho_0 \right) \). The acceleration \( a = \frac{F_{\text{net}}}{m_{\text{ball}}} = \frac{g}{5} \left( -\frac{3}{2} - \frac{3h}{h_0} \right) \), leading to \( a = -\frac{3g}{5} \left( \frac{h + h_0 / 2}{h_0} \right) \).
06

Determine the Nature of Motion

The acceleration equation resembles that of simple harmonic motion (SHM), which is of the form \( a = -\omega^2 x \), where \( x \) is displacement. Here, \( \omega^2 = \frac{3g}{5h_0} \), which indicates that the motion is indeed simple harmonic. The time period \( T \) for SHM is given by \( T = 2 \pi \sqrt{\frac{1}{\omega^2}} \).
07

Calculate the Time Period

\( T = 2 \pi \sqrt{\frac{5h_0}{3g}} \). Substituting \( h_0 = \frac{12}{\pi^2} \) and \( g = 10 \), we find \( T = 2 \pi \sqrt{\frac{5 \times \frac{12}{\pi^2}}{3 \times 10}} = 2 \pi \sqrt{\frac{2}{\pi^2}} = 2 \).
08

Conclusion

Since the calculated time period is \(2 \text{ seconds}\), the motion of the ball is simple harmonic motion with a time period of 2 seconds.

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Key Concepts

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

Buoyancy
Buoyancy is the upward force that a fluid exerts on an object submerged in it. This force opposes the weight of the object and can cause it to float.
  • For a completely submerged object, the buoyant force is equal to the weight of the liquid displaced by the object.
  • This is governed by Archimedes' principle, which states that the buoyant force equals the weight of the fluid that the object displaces.
In the exercise, the buoyant force depends on the height at which the ball is submerged because the liquid's density changes with height. The ball experiences an upward buoyant force that varies with the height due to the variable density of the liquid. As the ball rises or falls, the changing density directly impacts the buoyance experienced.
Density Variation
Density variation in a fluid refers to the changes in density throughout the fluid column. In the given exercise, the liquid's density decreases linearly with height.
  • The density formula provided is \[ \rho_{L} = \rho_{0} \left[4-\left(\frac{3h}{h_{0}}\right)\right] \]where \( \rho_{0} \) is the reference density, \( h \) is the height, and \( h_{0} \) is the height of the vessel.
  • This means that as the ball moves upwards, it experiences less buoyancy because the density of the liquid decreases.
Understanding density variation is crucial because it affects the magnitude of the buoyant force acting on objects submerged in such fluids. This plays a critical role in determining the nature of motion of the ball.
Newton's Second Law
Newton's Second Law of motion states that the acceleration of an object depends on the net force acting on it and its mass. This can be expressed as\[ F_{ ext{net}} = m imes a \]In the context of this exercise:
  • The net force acting on the ball is the difference between the buoyant force and its weights.
  • By applying Newton's Second Law, the acceleration \( a \) is obtained using the equation \( a = \frac{F_{ ext{net}}}{m} \), where\( m \) is the mass of the ball.
  • This relationship helps in determining if the motion of the ball is simple harmonic or not.
Understanding this law allows us to analyze the motion dynamics of the ball as it moves through the liquid.
Oscillatory Motion
Oscillatory motion occurs when an object moves back and forth around an equilibrium position. In the exercise, the ball undertakes simple harmonic motion, a specific form of oscillatory motion.
  • For simple harmonic motion, the acceleration \( a \) is directly proportional to the displacement \( x \) and is always directed towards the equilibrium position, represented as \( a = -\omega^2 x \).
  • The given problem shows that the acceleration equation matches that of simple harmonic motion due to the form \( a = -\frac{3g}{5}\left(\frac{h + h_0 / 2}{h_0}\right) \).
  • The angular frequency \( \omega \) is extracted to find the time period of the motion using \( T = 2\pi \sqrt{\frac{1}{\omega^2}} \).
The conclusion from the exercise reveals that the ball's motion is simple harmonic with a period of 2 seconds, reflecting the inherent properties of oscillatory systems.

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Most popular questions from this chapter

A spring balance reads \(W_{1}\) when a ball is suspended from it. A weighing machine reads \(W_{2}\) when a tank of liquid is kept on it. When the ball is immersed in the liquid, the spring balance reads \(W_{3}\) and the weighing machine reads \(W_{4}\). Then, which of the following are not correct? (1) \(W_{1}W_{3}\) (4) \(W_{2}>W_{4}\)

A tube of uniform cross-section has two vertical portions connected with a horizontal thin tube \(8 \mathrm{~cm}\) long at their lower ends. Enough water to occupy \(22 \mathrm{~cm}\) of the tube is poured into one branch and enough oil of specific gravity \(0.8\) to occupy \(22 \mathrm{~cm}\) is poured into the other. Find the distance(in \(\mathrm{cm}\) ) of the common surface \(E\) of the two liquids from point \(B\).

A block of silver of mass \(4 \mathrm{~kg}\) hanging from a string is immersed in a liquid of relative density \(0.72\). If relative density of silver is 10 , then tension in the string will be (1) \(37.12 \mathrm{~N}\) (2) \(42 \mathrm{~N}\) (3) \(73 \mathrm{~N}\) (4) \(21 \mathrm{~N}\)

A body is floating in a liquid in a beaker. If the whole system falls freely under gravity, the upthrust on the body due to the liquid is (1) zero (2) equal to the weight of the liquid displaced (3) equal to the weight of the body in air (4) equal to the weight of the immersed portion of the body

A small body of density \(\rho^{\prime}\) is dropped from rest at a height \(h\) into a lake of density \(\rho\), where \(\rho>\rho^{\prime} .\) Which of the following statements is/are correct if all dissipative effects are neglected (neglect viscosity)? (1) The speed of the body just entering the lake is \(\sqrt{2 g h}\). (2) The body in the lake experiences upward acceleration equal to \(\left\\{\left(\rho / \rho^{\prime}\right)-1\right\\} g\). (3) The maximum depth to which the body sinks in the lake is \(h \rho^{\prime} /\left(\rho-\rho^{\prime}\right)\) (4) The body does not come back to the surface of the lake.
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