Moment of intertia

  • Thread starter kasse
  • Start date
  • #1
kasse
382
1

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
Homework Helper
1,648
4
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
Homework Helper
726
3


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...
 

Suggested for: Moment of intertia

Replies
1
Views
480
  • Last Post
Replies
1
Views
1K
Replies
11
Views
1K
Replies
2
Views
901
  • Last Post
Replies
4
Views
929
Replies
1
Views
2K
Replies
7
Views
4K
  • Last Post
Replies
1
Views
794
Replies
3
Views
2K
Replies
11
Views
276
Top