Problems: Moment of Inertia | Answer Key - Edubirdie

University:
Massachusetts Institute of Technology
Course:
18.02SC | Multivariable Calculus
Academic year:

2021
Views:

132

Pages:

1
Author:

eu3p1n

Moment of inertia 1. Let R be the triangle with vertices (0, 0), (1, 0), (1, polar moment of inertia. √ 3) and density δ = 1. Find the Answer: The region R is a 30, 60 , 90 triangle. y √ 3 r = sec θ r x 1 The polar moment of inertia is the moment of inertia around the origin (that is, the z-axis). The ﬁgure shows the triangle and a small square piece within R. If the piece has area dA then its polar moment of inertia is dI = r2 δ dA. Summing the contributions of all such pieces and using δ = 1, dA = r dr dθ, we get the total moment of inertia is �� I= r2 δ dA = r2 r dr dθ = r3 dr dθ. R R R Next we ﬁnd the limits of integration in polar coordinates. The line x = 1 ⇔ r cos θ = 1 ⇔ r = sec θ. So, using radial stripes, the limits are: (inner) r from 0 to sec θ; (outer) θ from 0 to π/3. Thus, � π/3 � sec θ I= 0 Inner integral: sec4 θ r3 dr dθ. 0 . 4 Outer integral: Use sec4 θ = sec2 θ sec2 θ = (1 + tan2 θ) d(tan θ) ⇒ the outer integral is � √ � √ � ��π/3 1 tan3 θ �� 1 √ ( 3)3 3 tan θ + = 3+ = . � 4 3 4 3 2 0 √ The polar moment of inertia is 3 . 2

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