Cubic polynomial for the motion of a heavy symmetric top

  • I
  • Thread starter Happiness
  • Start date
  • #1
673
28
In the second paragraph after the expression of ##f(u)## below, it wrote "there are three roots to a cubic equation and three combinations of solutions". However, the combination of having three equal real roots was not mentioned. Why?

In the next paragraph, in the second sentence, it wrote "at points ##u=\pm1##, ##f(u)## is always negative, except for the unusual case where ##u=\pm1## is a root". But LHS of (5.62') ##=\dot{u}^2=\dot{\theta}^2\sin^2\theta=0## when ##u=\cos\theta=\pm1##,i.e., ##\theta=0## or ##\pi##. Thus RHS should also be 0. That means ##u=\pm1## is always a root to ##f(u)=0##. What's wrong?

Screen Shot 2016-04-29 at 11.24.41 pm.png

Screen Shot 2016-04-29 at 11.25.02 pm.png

Screen Shot 2016-04-29 at 11.25.19 pm.png

Screen Shot 2016-04-29 at 11.26.09 pm.png

Screen Shot 2016-04-29 at 11.26.27 pm.png
 

Answers and Replies

  • #2
18,408
8,222
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

Related Threads on Cubic polynomial for the motion of a heavy symmetric top

Replies
13
Views
3K
Replies
3
Views
948
  • Last Post
Replies
2
Views
3K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
661
Replies
2
Views
1K
  • Last Post
Replies
0
Views
1K
Replies
3
Views
2K
Replies
3
Views
2K
Top