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Fundamentals Of Physics
Found in: Page 353
Fundamentals Of Physics

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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Short Answer

A cylindrical aluminum rod, with an initial length of 0.8000 m and radius 1000.0 μm , is clamped in place at one end and then stretched by a machine pulling parallel to its length at its other end. Assuming that the rod’s density (mass per unit volume) does not change, find the force magnitude that is required of the machine to decrease the radius to 999.9 μm . (The yield strength is not exceeded.)

The magnitude of force applied on the cylindrical rod by machine is 44N.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Understanding the given information

The initial length of the rod, L=0.8000 m

The initial radius of the rod, r=1000.0 μm10−6 m1 μm=1000.0×10−6 m

The final radius of the rod, =999.9 μm10−6 m1 μm=999.9×10−6 m

Step 2: Concept and formula used in the given question

You can use the concept of modulus of elasticity, and expression of density and volume of the rod. The formulas used are given below.

ρ=MVV=πr2hFA=EΔLL

Step 3: Calculation for the force magnitude that is required of the machine to decrease the radius to  999.9 mm

One end of the rod is clamped and the other end is stretched by the machine in such a way that the rod’s density remains the same after elongation of the rod.

The expression of the density of the rod is

ρ=MVV=Mρ=constantV=πr2L=constant

The expression of the volume of the rod before and after elongation is:

πr2L=πr'2L'r2L=r'2L'L'=r2Lr'2

Change in length of the rod due to the applied force by the machine is:

ΔL=L'−L=r2Lr'2−L=Lr2r'2−1

The expression of Young’s modulus of elasticity is:

FA=EΔLLF=AEΔLLF=πr2ELr2r'2−1LF=3.14×(1000×10−6 m)2×70×109 N/m2×(1000×10−6 m)2(999.9×10−6 m)2−1F=44 N

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