ENGINEERING MECH DYNAMICS W/MASTREV
ENGINEERING MECH DYNAMICS W/MASTREV
14th Edition
ISBN: 9780135881187
Author: HIBBELER
Publisher: PEARSON
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Textbook Question
Chapter 17.1, Problem 1P

The rod's density ρ end cross-sectional area. A are constant. Express the result in terms of the rod’s total mass m.

Chapter 17.1, Problem 1P, The rod's density  end cross-sectional area. A are constant. Express the result in terms of the rods

Expert Solution & Answer
Check Mark
To determine

The moment of inertia Iy for the slender rod in terms of the rod’s total mass m .

Answer to Problem 1P

The moment of inertia Iy for the slender rod in terms of the rod’s total mass m is ml23 .

Explanation of Solution

Given:

The density of the rod is ρ .

The cross-sectional area of the rod is A .

Show the intersection of slender rod at the arbitrary point (x,y,z) as in Figure (1).

ENGINEERING MECH DYNAMICS W/MASTREV, Chapter 17.1, Problem 1P

Conclusion:

From the Figure 1,

Calculate the differential mass of the slender rod.

dm=ρdV (I)

Here, density of material is ρ , and the volume of the slender rod is V .

Substitute Adx for dV in Equation (I).

dm=ρAdx

Calculate the entire mass of the slender rod.

m=Mdm (II)

Substitute ρAdx for dm in Equation (II).

m=MρAdx=0lρAdx=ρA0ldx

=ρA(x)|0l=ρAl

Express the moment of inertia of the slender rod about the y axis.

Iy=Mx2dm (III)

Substitute ρAdx for dm in Equation (III).

Iy=Mx2(ρAdx)=0lx2(ρAdx)=ρA0lx2dx=ρA[x33]|0l

=ρAl33=13ρAl3=13l2(ρAl) (IV)

Substitute m for ρAl in Equation (IV).

Iy=13l2(m)=ml23

Hence, the moment of inertia Iy for the slender rod in terms of the rod’s total mass m is ml23 .

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Chapter 17 Solutions

ENGINEERING MECH DYNAMICS W/MASTREV

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