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Chapter 13: 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 downward gravitational force magnitude is \(\frac{2GmM}{\pi R^2}\).

Step by step solution

01

Understand the Problem

We have a wire bent into a semicircle with a uniform mass distribution. Our goal is to find the gravitational force it exerts on a mass placed at its center of curvature. The problem involves integrating the contributions of gravitational forces from different sections of the wire.
02

Describe the System

Consider the semicircular wire with radius \(R\). The wire's mass \(M\) is uniformly distributed along its length \(L\). The wire forms a semicircle with the central point at its center of curvature. The wire subtends an angle of \(\pi\) radians at this center.
03

Identify Differential Mass Element

We take a small element of the wire, \(d\theta\), at an angle \(\theta\) from the horizontal diameter. The arc length of this element is expressed as \(Rd\theta\). Since the mass is uniformly distributed, the mass of this element \(dm\) can be defined: \[ dm = \frac{M}{\pi R} \cdot Rd\theta \] This simplifies to: \[ dm = \frac{M}{\pi} d\theta \]
04

Calculate Gravitational Force by dm

The gravitational force \(dF\) due to the element \(dm\) on mass \(m\) is given by Newton's law: \[ dF = \frac{Gm \, dm}{R^2} \] Substitute for \(dm\): \[ dF = \frac{GmM}{\pi R^2} d\theta \]
05

Integrate to Find Total Force

The forces from different elements have different directions. Due to symmetry, horizontal components cancel out, and only vertical components sum up. The vertical component of \(dF\) is \(dF \sin \theta\). Integrate over half the circle, from \(-\pi/2\) to \(\pi/2\): \[ F = \int_{-\pi/2}^{\pi/2} \left( \frac{GmM}{\pi R^2} \sin \theta \right) d\theta \] This evaluates to: \[ F = \frac{2GmM}{\pi R^2} \]
06

Direction of the Force

Due to symmetry, the net force is vertical and directed towards the center of the curvature (downward). Thus, the gravitational force is directed downwards along the radius.

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Key Concepts

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

Uniform Mass Distribution
When discussing uniform mass distribution, we mean that the mass is spread out evenly along the length of an object. In this case, our wire has a mass, denoted as \( M \), and is bent into a semicircle.
This semicircle forms half a circle and subtends an angle of \( \pi \) radians at its center. Since the wire's mass is uniformly distributed, every small segment of the wire has the same mass per unit length as every other part.
  • Even distribution simplifies calculations, as each unit length has consistent properties.
  • In problems like these, you work with small sections known as differential elements to calculate forces separately before integrating them.
Understanding this distribution is crucial because it allows us to define and use differential mass elements efficiently in our calculations.
Semicircular Wire
The semicircular wire poses interesting challenges when calculating forces. With its shape defined by a radius \( R \), the wire extends over an arc. The semicircular shape is significant because its symmetry affects how forces balance.
  • Any point on the wire can be described using an angle \( \theta \) from the horizontal diameter.
  • This angle helps to identify the position of each differential mass element.
For the semicircle, the whole wire subtends \( \pi \) radians, allowing us to simplify our approach by focusing only on the vertical component of forces as horizontal forces cancel out. This symmetry helps in integrating gravitational forces efficiently as seen in the step-by-step calculation.
Integration of Forces
Integration of forces involves adding up the small gravitational forces exerted by each section of the wire. With each section exerting a tiny force on the point mass \( m \) at the center, we use integrals to calculate these forces.
  • Each force \( dF \) is a product of gravitational constant \( G \), mass \( m \), and differential mass \( dm \) over radius squared \( R^2 \).
  • The vertical component \( dF \sin \theta \) is directly relevant due to the wire's symmetry.
We sum these vertical components over \( \theta \) ranging from \(-\pi/2\) to \(\pi/2\) using integration. The resulting total force is given by \( F = \frac{2GmM}{\pi R^2} \), indicating how the entire wire collectively affects the central point.
Newton's Law of Gravitation
Newton's law of gravitation is foundational in calculating gravitational forces. It states that every point mass attracts every other point mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.
This is expressed by the formula: \[ F = \frac{G m_1 m_2}{r^2} \]
  • \( G \) represents the universal gravitational constant.
  • \( m_1 \) and \( m_2 \) are the masses involved.
  • \( r \) denotes the distance between the mass centers.
When applying Newton's law to our semicircular wire and center point, each differential mass element contributes a small gravitational force \( dF \) on \( m \). By integrating these contributions, we can determine the total gravitational effect of the wire on the point mass at its center of curvature.

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Most popular questions from this chapter

An astronaut is standing at the north pole of a newly discovered, spherically symmetric planet of radius \(R .\) In his hands he holds a container full of a liquid with mass \(m\) and volume \(V .\) At the surface of the liquid, the pressure is \(p_{0} ;\) at a depth \(d\) below the surface, the pressure has a greater value \(p .\) From this information, determine the mass of the planet.

An object in the shape of a thin ring has radius \(a\) and mass \(M .\) A uniform sphere with mass \(m\) and radius \(R\) is placed with its center at a distance \(x\) to the right of the center of the ring, along a line through the center of the ring, and perpendicular to its plane (see Fig. E13.33). What is the gravitational force that the sphere exerts on the ring-shaped object? Show that your result reduces to the expected result when \(x\) is much larger than \(a\).

An experiment is performed in deep space with two uniform spheres, one with mass 50.0 \(\mathrm{kg}\) and the other with mass 100.0 \(\mathrm{kg} .\) They have equal radii, \(r=0.20 \mathrm{m} .\) The spheres are released from rest with their centers 40.0 \(\mathrm{m}\) apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. (a) Explain why linear momentum is conserved. (b) When their centers are 20.0 m apart, find (i) the speed of each sphere and (ii) the magnitude of the relative velocity with which one sphere is approaching the other. (c) How far from the initial position of the center of the 50.0-kg sphere do the surfaces of the two spheres collide?

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at \(30,000 \mathrm{km} / \mathrm{s}\) . (a) How far are these clumps from the center of the black hole? (b) What is the mass of this black hole, assuming circular orbits? Express your answer in kilograms and as a multiple of our sun's mass. (c) What is the radius of its event horizon?

A landing craft with mass \(12,500 \mathrm{kg}\) is in a circular orbit \(5.75 \times 10^{3} \mathrm{m}\) above the surface of a planet. The period of the orbit is 5800 s. The astronauts in the lander measure the diameter of the planet to be \(9.60 \times 10^{6} \mathrm{m} .\) The lander sets down at the north pole of the planet. What is the weight of an 85.6 -kg astronaut as he steps out onto the planet's surface?
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