Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A thin rod with mass $M = 5.00 kg$ M= is bent in a semicircle of radius $R = 0.650 m$ . (Fig. 13-56). (a) What is its gravitational force (both magnitude and direction on a particle with mass $m = 3.0 \times 10^{- 3} kg$ at P, the center of curvature? (b) What would be the force on the particle the rod were a complete circle?

The magnitude of the gravitational force on the particle $= 1.5 \times 10^{- 12} N$ and it is directed upwards (along the +Y axis).
The force on the particle if the rod is a complete circledata-custom-editor="chemistry" $= 0 N$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Listing the given quantities

The mass of the rod $M = 5.00 kg$ .

The mass of the particle $m = 3.00 \times 10^{- 3} kg$ and is placed at P, the centre of the semicircle.

The radius of the semicircle $R = 0.650 m$ .

Step 2: Understanding the concept of Newton’s laws

We can use Newton's law of gravitation and the concept of integration to find the gravitational force.

Formula:

$F = \frac{GMm}{r^{2}}$

Step 3: (a) Calculations for magnitude and direction of the gravitational force

To determine the force exerted by the rod, we will have to calculate its mass per unit length since the rod is in the form of a semicircle.

Mass per unit length $μ = \frac{M}{L}$ , where L = length of wire = $πR$

Now, consider a small section dl of the rod at an angle θ.

The mass of this section is $μdl$ .

The force exerted by the section dl on the particle at P is given by .

The total force by all such elements will be directed radially. Hence the net force will be sum of the vertical (sin) components of these forces. All the horizontal (cos) components get cancelled in pairs as they will be directed opposite to each other.

Hence, the net force on particle at P due to semicircular wire is

$\begin{array}{rcl} F & = & \int F^{'} sinθ \\ = & \int \frac{GMμdlsinθ}{R^{2}} \end{array}$

From the figure, we have $dl = Rdθ$ .

Hence we write

$\begin{array}{rcl} F & = & \int \frac{GMμRdθsinθ}{R^{2}} \\ = & \frac{GmM}{{πR}^{2}} \int_{0}^{π} sinθ d θ \end{array}$

On integrating, we get

$\begin{array}{rcl} F & = & \frac{2 GmM}{{πR}^{2}} \\ = & \frac{2 \times 5 \times 3 \times 10^{- 3} \times 6.67 \times 10^{- 11}}{3.14 \times {(0.650)}^{2}} \\ = & 1.51 \times 10^{- 12} N \end{array}$

This force will be directed upwards along the + Y axis

Step 4: (b) Calculations of the force on the particle if the rod is complete circle

If the rod is now made in the form of a complete circle, the net force acting on the particle at P will be zero. As we have seen in part (a), all the horizontal components get cancelled in pairs. So, all the vertical components will also get cancelled.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Listing the given quantities

Step 2: Understanding the concept of Newton’s laws

Step 3: (a) Calculations for magnitude and direction of the gravitational force

Step 4: (b) Calculations of the force on the particle if the rod is complete circle

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Fundamentals Of Physics

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Step by Step Solution

Step 1: Listing the given quantities

Step 2: Understanding the concept of Newton’s laws

Step 3: (a) Calculations for magnitude and direction of the gravitational force

Step 4: (b) Calculations of the force on the particle if the rod is complete circle

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