Find the Period of the rotation, the math gets tricky

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The discussion centers on determining the expression for the period of rotation of a block on a cone subjected to gravity and static friction. The user sets up equations based on force components, including normal force and friction, and derives a relationship involving the period. They confirm the validity of dividing one equation by another and simplify the resulting expression for the period. Ultimately, they arrive at the formula T = ±2π√(r/g * (sinα - μcosα)/(cosα - μsinα)), rejecting the negative solution. The thread highlights the importance of correctly applying physics principles to solve the problem.
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Homework Statement



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

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

Breaking the components of the forces I get

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

(2) nsin(\alpha) +fcos(\alpha)=mg

(3) v^2 = \frac{4\pi^2 r^2}{T^2}

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 sin(\alpha) and (2) by cos(\alpha), can I add (1) to (2)?
 
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Can some mod edit my tex in cos\alpha...? I forgot to add [ tex] and [/tex]
 
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Omg I solved it

f = \mu n

And so

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

(2)nsin\alpha + \mu ncos\alpha = mg

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

\frac{cos\alpha - \mu sin\alpha}{sin\alpha + \mu cos\alpha}

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

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

And of course we reject negative period
 
Omg I solved it

f = \mu n

And so

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

(2)nsin\alpha + \mu ncos\alpha = mg

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

\frac{cos\alpha - \mu sin\alpha}{sin\alpha + \mu cos\alpha} = \frac{v^2}{rg}

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

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

And of course we reject negative period
 
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