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Chapter 12: Problem 82

A uniform wire with mass \(M\) and length \(L\) is bent into a semicircle. Find the magnitude and direction of the gravitational force this wire exerts on a point with mass \(m\) placed at the center of curvature of the semicircle.

Short Answer

Expert verified
The gravitational force is \(\frac{2G m M \pi}{L^2}\) directed vertically towards the semicircle's plane.

Step by step solution

01

Understanding the Problem

We need to find the gravitational force exerted by a semicircular wire of mass \(M\) on a point mass \(m\) located at its center of curvature. The length of the wire is \(L\).
02

Identify the Elements for Gravitational Force

The wire is bent into a semicircle, so we'll consider a differential mass element \(dm\) of the wire exerting a force \(dF\) on the mass \(m\) at the center of curvature. The total gravitational force is the sum of these differential forces.
03

Define Differential Mass Element

Given the semicircle, we take an infinitesimal arc length \(ds = R d\theta \) where \(\theta\) is the angle from the center. The mass element is \(dm = \frac{M}{L} ds = \frac{M}{L} R d\theta\).
04

Elements of Gravitational Force

The differential gravitational force due to \(dm\) is \(dF = \frac{G m dm}{R^2} = \frac{G m}{R^2} \cdot \frac{M}{L} \cdot R d\theta\), where \(G\) is the gravitational constant, and \(R = \frac{L}{\pi}\), the radius of the semicircle.
05

Component Analysis of Force

Due to symmetry of the semicircle, the horizontal components of \(dF\) cancel out while the vertical components \(dF_y = dF \cdot \sin\theta\) add up. So, we need to integrate only \(dF_y\).
06

Integrate Force Components

Integrate \(dF_y = \sin\theta \cdot \frac{G m M}{L R} d\theta\) from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), giving total force \(F = \int_{-\pi/2}^{\pi/2} \frac{G m M}{L \cdot R} \sin\theta \cdot R d\theta\).
07

Calculate the Integral

The integral becomes \(F = \frac{G m M}{L} \int_{-\pi/2}^{\pi/2} \sin\theta d\theta\), which equals \(\frac{G m M}{L} \cdot 2\). This integral evaluates to \(F = \frac{2G m M}{\pi R}\).
08

Substitute the Radius

Substitute \(R = \frac{L}{\pi}\) into the force equation. We get \(F = \frac{2G m M}{L/\pi} = \frac{2G m M \pi}{L^2}\).
09

Direction of Force

The gravitational force acts along the vertical axis due to symmetry, pulling the mass \(m\) towards the plane of the semicircle.

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Key Concepts

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

Understanding Differential Mass Element
When dealing with a complex shape like a semicircular wire, breaking it down into simpler parts is key. A differential mass element, often denoted as \(dm\), is a small segment of mass that we consider when calculating effects like gravitational force. Think of it as a tiny slice of the entire wire. For our semicircular wire of total mass \(M\) and length \(L\), we describe \(dm\) in terms of an arc length \(ds\). Here's how it works:
  • First, identify the radius of the semicircle, \(R\), which is \(R = \frac{L}{\pi}\), since the total length of the semicircle is \(L\), covering half of a complete circle.
  • Next, express the arc length as \(ds = R \, d\theta\), where \(d\theta\) is a small angle measured in radians. This keeps things very small and precise.
  • The mass of \(dm\) becomes \(dm = \frac{M}{L} \cdot ds\), which simplifies to \(dm = \frac{M}{L} \cdot R \, d\theta\).
Using \(dm\), we can effectively sum up these small elements to find the total gravitational pull at the center of the semicircular wire.
It's like adding up many tiny influences to get the full picture.
Gravitational Force from a Semicircular Wire
The semicircular wire setup inspired an interesting challenge: finding out how this shape influences a point mass located at its center. Because the wire forms a semicircle, it has unique properties that affect the gravitational force we want to calculate. Here's the method:
  • We consider a small mass segment \(dm\), which exerts a tiny force \(dF\) on a central mass \(m\).
  • The force from each \(dm\) includes gravitational interactions: \(dF = \frac{G \, m \, dm}{R^2}\), where \(G\) is the gravitational constant and \(R\) is the semicircle's radius.
  • Substituting \(dm = \frac{M}{L} \cdot R \, d\theta\) into the force expression, we get \(dF = \frac{G \, m}{R^2} \cdot \frac{M}{L} \cdot R \, d\theta\).
This approach lets us focus on vertical components of force (since horizontal components cancel out, thanks to symmetry), which leads to a simpler integration.
By calculating these along the semicircle, we derive the overall gravitational force acting downward at the center.
Symmetry in Physics and Its Role
Symmetry is a powerful tool in physics, often simplifying what could otherwise be complex problems. In this context, the semicircular wire generates symmetrical forces that significantly influence our calculations. Why is symmetry important here?
  • The setup of a semicircle means that for each differential mass element \(dm\) on one side, there's a corresponding \(dm\) on the other side.
  • These pairs pull in opposite horizontal directions, which means they cancel out, leading to zero net horizontal force.
  • Only the vertical components of each force contribute to the total gravitational pull on the center mass \(m\).
Thus, symmetry helps us by reducing the complexity鈥攁llowing us to only need to integrate the vertical components across the angle covered by the semicircle from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
With symmetry, we focus on what truly matters, ensuring the solution is both elegant and efficient.

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Most popular questions from this chapter

A uniform, spherical, \(1000.0-\mathrm{kg}\) shell has a radius of 5.00 \(\mathrm{m}\) . (a) Find the gravitational force this shell exerts on a \(2.00-\mathrm{kg}\) point mass placed at the following distances from the center of the shell: (i) \(5.01 \mathrm{m},(\mathrm{ii}) 4.99 \mathrm{m},\) (iii) 2.72 \(\mathrm{m} .\) (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r=0\) to \(r \rightarrow \infty\) .

The mass of Venus is 81.5\(\%\) that of the earth, and its radius is 94.9\(\%\) that of the earth. (a) Compute the acceleration due to gravity on the surface of Venus from these data. (b) If a rock weighs 75.0 \(\mathrm{N}\) on earth, what would it weigh at the surface of Venus?

Titania, the largest moon of the planet Uranus, has \(\frac{1}{8}\) the radius of the earth and \(\frac{1}{100}\) the mass of the earth. (a) What is the acceleration due to gravity at the surface of Titania? (b) What is the average density of Titania? (This is less than the density of rock, which is one piece of evidence that Titania is made primarily of ice. \()\)

A thin, uniform rod has length \(L\) and mass \(M .\) A small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. 12.34\()\) . (a) Calculate the gravita- tional potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when \(x\) is much larger than \(L\) . (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix B. (b) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4 ). Show that your answer reduces to the expected result when \(x\) is much larger than \(L .\)

At a certain instant, the earth, the moon, and a stationary \(1250-\mathrm{kg}\) spacecraft lie at the vertices of an equilateral triangle whose sides are \(3.84 \times 10^{5} \mathrm{km}\) in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? You can ignore any gravitational effects due to the other planets or the sun.
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