Determine whether each approximation is an overestimate or underestimate and give a qualitative reason why. For each approximation, compute the percentage difference from the exact value. How would the percentage difference change if the bar was more massive or longer?

College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Overview: Here you will make successive approximations of the moment of inertia of a
constant-density 1D bar when the pivot point is at one extreme end of the bar. You will model
the bar first as a single point mass, followed by two-, three- and four-point mass approximations.
In each case, you will compare the approximation with the exact value.
The exact moment of inertia of a constant-density 1D rod of length L and mass M with pivot point
at one extreme end is calculated in Example 12.5 on page 302 and has a value of
lexact =
The moment of inertia is then:
1
Procedure: For simplicity, let's model a bar of mass M=1kg and length L = Imeter. To
approximate the continuous bar with N point masses, each point mass has a mass of mi = M/N.
We then place the masses symmetrically with respect to the actual bar at positions xi.
Iapprox
ML²
N
• Σmax7.
i=1
The positions x; can be worked out by inspecting the diagrams on the following page.
Transcribed Image Text:Overview: Here you will make successive approximations of the moment of inertia of a constant-density 1D bar when the pivot point is at one extreme end of the bar. You will model the bar first as a single point mass, followed by two-, three- and four-point mass approximations. In each case, you will compare the approximation with the exact value. The exact moment of inertia of a constant-density 1D rod of length L and mass M with pivot point at one extreme end is calculated in Example 12.5 on page 302 and has a value of lexact = The moment of inertia is then: 1 Procedure: For simplicity, let's model a bar of mass M=1kg and length L = Imeter. To approximate the continuous bar with N point masses, each point mass has a mass of mi = M/N. We then place the masses symmetrically with respect to the actual bar at positions xi. Iapprox ML² N • Σmax7. i=1 The positions x; can be worked out by inspecting the diagrams on the following page.
Continuous Bar
0
1 point-mass approximation
0
2 point-mass approximation
0
3 point-mass approximation
0
0
m=M/2
m₁=M/3
XI
4 point-mass approximation
m₁=M/4
x₁
X1
m₂ M/4
X2
m₁=M
X1
m₂=M/3
79000
X2
m₂=M/4
X3
m₂-M/2
X2
m3 M/3
X3
m4 M/4
L
For each approximation, compute the percentage difference from the exact value.
How would the percentage difference change if the bar was more massive or longer?
Extra gradit (um 10
L
L
L
Determine whether each approximation is an overestimate or underestimate and give a
qualitative reason why.
L
x
X
x
Transcribed Image Text:Continuous Bar 0 1 point-mass approximation 0 2 point-mass approximation 0 3 point-mass approximation 0 0 m=M/2 m₁=M/3 XI 4 point-mass approximation m₁=M/4 x₁ X1 m₂ M/4 X2 m₁=M X1 m₂=M/3 79000 X2 m₂=M/4 X3 m₂-M/2 X2 m3 M/3 X3 m4 M/4 L For each approximation, compute the percentage difference from the exact value. How would the percentage difference change if the bar was more massive or longer? Extra gradit (um 10 L L L Determine whether each approximation is an overestimate or underestimate and give a qualitative reason why. L x X x
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