Moment of intertia

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Homework Statement



A wire is shaped like the astroid x=cos3(t), y=sin3(t), t[0, 2*pi] and has constant density = k. Find its moment of intertia I0 around the origin.

2. The attempt at a solution

To find the MOE we must integrate k* (x2 + y2)ds along the curve. We differentiate and find that ds can be written 3*cos(t)*sin(t).The final expression is:

3k * Int (cos7(t)*sin(t) + sin7(t)*cos(t))dt for t [0, 2*pi]. But this is zero! What/where is my mistake?
 

Answers and Replies

  • #2
dynamicsolo
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Be careful when integrating periodically symmetric functions through a full period: your dt cycles through both positive and negative signs. You may be better off integrating only one quarter of the astroid and quadrupling that result. (Were you to express the moment of inertia as C · M · (R^2) , the moment for each quadrant would have the same constant C as the entire figure.)
 
  • #3
mjsd
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3k * Int (cos7(t)*sin(t) + sin7(t)*cos(t))dt for t [0, 2*pi]. But this is zero! What/where is my mistake?


Your mistake is: changing the function for ds when doing your simplification!

[tex]
ds = \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} \;dt
= 3\sqrt{\cos^2(t)\sin^2(t)}\; dt \;\neq \;3 \cos(t)\sin(t) \;dt
[/tex]

[tex]
3\sqrt{\cos^2(t)\sin^2(t)}\; dt = 3|\cos(t)\sin(t)|\;dt
[/tex]

that's why you have the two contributions cancelling each other out... you have forgotten the absolute value sign...
 

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