Find the Period of the rotation, the math gets tricky

  • Thread starter flyingpig
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  • #1
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Homework Statement



[PLAIN]http://img573.imageshack.us/img573/6932/72101948.png [Broken]

In the diagram above, a block is on a cone rotating above a height of h and is a distance r from the rotating axis. The cone is subjected to gravity, but is resisting due to the static friction on the cone (static friction because it is not going up or down). Find an expression of the period of this rotation.



The Attempt at a Solution



[PLAIN]http://img204.imageshack.us/img204/9218/28918584.png [Broken]

Breaking the components of the forces I get

(1) [tex]ncos(\alpha) - fsin(\alpha) = \frac{mv^2}{r}[/tex]

(2) [tex]nsin(\alpha) +fcos(\alpha)=mg[/tex]

(3) [tex]v^2 = \frac{4\pi^2 r^2}{T^2}[/tex]

Now, here is the problem(s)

Did I set it up right? Is there another equation missing?

Also, is it mathematically correctly to divide (1) by (2)

According to this thread https://www.physicsforums.com/showthread.php?t=465924 I can, but is only because the functions were linear? If I multiply (1) by [tex]sin(\alpha)[/tex] and (2) by cos(\alpha), can I add (1) to (2)?
 
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Answers and Replies

  • #2
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Can some mod edit my tex in cos\alpha...? I forgot to add [ tex] and [/tex]
 
Last edited:
  • #3
2,571
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Omg I solved it

[tex]f = \mu n[/tex]

And so

[tex](1)ncos\alpha - \mu nsin\alpha = \frac{mv^2}{r}[/tex]

[tex](2)nsin\alpha + \mu ncos\alpha = mg[/tex]

From Euclid's elements, I can divide (1) by (2) and I get

[tex]\frac{cos\alpha - \mu sin\alpha}{sin\alpha + \mu cos\alpha}[/tex]

Then substituting [tex]4\frac{\pi^2 r^2}{T^2} = v^2[/tex] and simplifying I get

T = [tex]\pm 2\pi\sqrt{\frac{r}{g}\frac{sin\alpha - \mu cos\alpha}{cos\alpha - \mu sin\alpha}}}[/tex]

And of course we reject negative period
 
  • #4
2,571
1
Omg I solved it

[tex]f = \mu n[/tex]

And so

[tex](1)ncos\alpha - \mu nsin\alpha = \frac{mv^2}{r}[/tex]

[tex](2)nsin\alpha + \mu ncos\alpha = mg[/tex]

From Euclid's elements, I can divide (1) by (2) and I get

[tex]\frac{cos\alpha - \mu sin\alpha}{sin\alpha + \mu cos\alpha} = \frac{v^2}{rg}[/tex]

Then substituting [tex]\frac{4\pi^2 r^2}{T^2} = v^2[/tex] and simplifying I get

[tex]T = \pm 2\pi\sqrt{\frac{r}{g}\left (\frac{sin\alpha - \mu cos\alpha}{cos\alpha - \mu sin\alpha}\right )}}[/tex]

And of course we reject negative period
 

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