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Chapter 12: Problem 111

Imagine that a narrow tunnel, just large enough for a baseball to fit, is drilled through the center of Earth. Suppose a standard baseball is dropped into the tunnel from the surface of Earth. Calculate the period of the subsequent motion of the baseball, assuming that Earth is a solid, uniform sphere that does not spin. Hint: Draw the free-body diagram for the ball and recall that the gravitational field inside a sphere is \(g(r)=-g_{0}(r / R)\), where \(g_{0}\) equals \(9.8 \mathrm{~m} / \mathrm{s}^{2}, R\) is the radius of Earth and equals \(6380 \mathrm{~km}\), and \(r\) is the distance from the center of Earth to the center of the ball. Newton's second law will lead you to the solution. SSM

Short Answer

Expert verified
The period of the subsequent motion of the baseball, assuming that the Earth is a solid, uniform sphere not spinning, is approximately 84.5 minutes.

Step by step solution

01

Apply the concept of gravitational field inside a sphere

The force acting on the ball at a given position inside the tunnel is its weight. As the ball moves inside the tunnel, its distance to the center of the Earth, \(r\), changes. We can thus, express the force \(F\) acting on the ball by \(F = m * g(r) = -m * g_{0} * (r / R)\), where \(m\) is the mass of the ball, \(r\) is its distance from the center of Earth, \(g_{0}\) is approximately 9.8 m/s虏 and \(R\) is the radius of the Earth, approximately 6380 kilometers.
02

Apply Newton's Second Law

According to Newton's second law, \(F = m * a = m * \frac{-d^2r}{dt^2}\), where \(a\) is the acceleration and \(t\) denotes time. We substitute the force from Step 1 to this equation: \(m * \frac{-d^2r}{dt^2} = -m * g_{0} * (r / R)\). One can simplify this to: \(\frac{d^2r}{dt^2} = g_{0} * (r / R)\). The negative sign is neglected because it does not affect the period of motion.
03

Solve the Differential Equation

This is a simple harmonic motion equation, and the solution is \(r(t) = Acos(\omega t + \phi)\), where \(A\) and \(\phi\) are constants, \(t\) is the time, and \(\omega\) is the angular frequency. From the differential equation, we can equate \(\omega^2 = g_{0} / R\).
04

Calculate the Period of Motion

The period \(T\) of the motion is calculated by \(T = \frac{2\pi}{\omega} = 2\pi * \sqrt{\frac{R}{g_{0}}}\), substituting the values for \(R\) and \(g_{0}\), where \(R = 6380 * 10^3 m\), \(g_{0} = 9.8 m/s虏\), gives a period \(T\) of approximately 84.5 minutes.

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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 of displacement. In the case of the baseball in the tunnel drilled through Earth, the motion of the baseball is akin to that of a simple harmonic oscillator. As the baseball moves closer or farther from Earth's center, the gravitational force changes, pulling it back towards the center.
To put it simply, SHM occurs because two conditions are met:
  • The force acting on the baseball is proportional to its displacement from its equilibrium position (the center of Earth in this case).
  • The force acts in the opposite direction of the displacement, leading to oscillatory motion.
This results in the baseball moving back and forth in the tunnel in a regular, predictable manner. This is described mathematically by the differential equation \[\frac{d^2r}{dt^2} = -\omega^2 r\]where \(\omega\) is the angular frequency. Overall, the understanding of SHM helps in predicting how the baseball will behave once it is dropped into the tunnel, moving rhythmically due to the gravitational pull.
Newton's Second Law
Newton's Second Law of Motion plays a critical role in understanding the dynamics of objects in motion, including our tunnel scenario. The law states that the force acting on an object is equal to the mass of the object times its acceleration, given by\[F = m * a\]For the baseball, we use this fundamental principle to express the gravitational force acting on it at any point inside the tunnel. The force is\[F = -m * g_{0} * \left(\frac{r}{R}\right)\]By equating this with \(F = m * \frac{d^2r}{dt^2}\), we can find a relationship between the force, displacement \(r\), and acceleration \(\frac{d^2r}{dt^2}\). Solving this relationship helps show that the motion of the baseball is simple harmonic in nature. What's particularly interesting here is how Newton's law enables us to derive the SHM formulae exactly, allowing us to predict the oscillatory motion of the baseball as it travels back and forth due to the altered gravitational forces within the Earth.
Period of Motion
The period of motion in SHM is the time it takes for one complete cycle of motion. For the baseball in our exercise, this means the time taken to travel from one end of the tunnel to the other and back. Understanding the period of motion is crucial as it provides insight into how long the basketball will take to complete its journey within the tunnel.
The formula for the period of motion \(T\) is given by:\[T = 2\pi * \sqrt{\frac{R}{g_{0}}}\]Where:
  • \(R\) is the radius of Earth, approximately 6380 km.
  • \(g_{0}\) is the surface gravitational acceleration, roughly 9.8 m/s虏.
Plugging in these values, the period \(T\) turns out to be approximately 84.5 minutes. This means that, ignoring Earth's rotation and assuming a uniform sphere, the baseball will take around 84.5 minutes to travel through the tunnel and return. Understanding this period helps students visualize the timing of simple harmonic motions and the effects of gravitational forces in a solid sphere such as Earth.

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Most popular questions from this chapter

(a) What is the period of a simple pendulum of length \(1 \mathrm{~m}\) at the top of Mt. Everest, \(8848 \mathrm{~m}\) above sea level. (b) Express your answer as a number times \(T_{0}\), the period at sea level where \(h\) equals 0 . The acceleration due to gravity in terms of elevation is $$ g=g_{0}\left(\frac{R_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}+h}\right)^{2} $$ where \(g_{0}\) is the average acceleration due to gravity at sea level, \(R_{\mathrm{E}}\) is Earth's radius, and \(h\) is elevation above sea level. Take \(g_{0}\) to be \(9.800 \mathrm{~m} / \mathrm{s}^{2}\) and Earth's radius to be \(R_{\mathrm{E}}\) is \(6.380 \times 10^{6} \mathrm{~m}\).

A block of wood floats in a basin of water. If it is pushed down slightly and released, the subsequent motion is oscillatory. Consider the buoyant force and gravitational force acting on the block to derive an algebraic expression for the period of this simple harmonic motion in terms of the surface area on which the block floats, the density of the water, the gravitational constant, the mass of the block, and constants.

A \(500-\mathrm{g}\) object is attached to a spring with a force constant of \(2.5 \mathrm{~N} / \mathrm{m}\). The object rests on a horizontal surface that has a viscous, oily substance spread evenly on it. The object is pulled \(15 \mathrm{~cm}\) to the right of the equilibrium position and set into harmonic motion. After \(3 \mathrm{~s}\), the amplitude has fallen to \(7 \mathrm{~cm}\) due to frictional losses in the oil. (a) Calculate the natural frequency of the system, (b) the damping constant for the oil, and (c) the frequency of oscillation that will be observed for the motion. (d) How much time will it take before the oscillations have died down to one-tenth of the original amplitude \((1.5 \mathrm{~cm})\) ?

Compare \(x(t)=A \cos \omega t\) to \(x(t)=A \cos (\omega t+\phi)\). What is the phase angle \(\phi\) and how does it change the solution to simple harmonic motion?

Geoff counts the number of oscillations of a simple pendulum at a location where the acceleration due to gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\), and finds that it takes \(25 \mathrm{~s}\) for 14 complete cycles. Calculate the length of the pendulum. SSM
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