How Do You Calculate the Angle in a Spinning Mass Problem?

[Zhalfirin88]

Spinning mass

Homework Statement

1) A mass of 7.10 kg is suspended from a 1.21 m long string. It revolves in a horizontal circle as shown in the figure. The tangential speed of the mass is 2.90 m/s. Calculate the angle between the string and the vertical.

Picture 1) http://psblnx03.bd.psu.edu/res/msu/...Force_Motion_Adv/graphics/prob03_pendulum.gif

The Attempt at a Solution

For 1) I have the equation down to:

[tex] \frac{gL}{v^2} - \frac{gL}{v^2}cos^2 \theta - cos \theta = 0 [/tex] How would I put this into the quadratic formula? (L = length of string)

 

[Jebus_Chris]

You should show what you did. For both of them you just need to draw an fbd and find the net force.

 

[Zhalfirin88]

You should show what you did. For both of them you just need to draw an fbd and find the net force.

I have shown what I did? If I were to show everything that would be a hell of a lot of typing. I've already done FBD, all I need is what has been asked.

Problem 1:

[tex] T_y - mg = 0 [/tex] X-Direction: [tex] T sin \theta = \frac{mv^2}{r} [/tex]

[tex] T cos\theta = mg [/tex] X-Direction: [tex] T sin \theta = \frac{mv^2}{Lsin \theta} [/tex]

[tex] T = \frac{mg}{cos \theta} [/tex] X-Direction: [tex] T = \frac{mv^2}{Lsin^2 \theta} [/tex]

[tex] \frac{mg}{cos \theta} = \frac {mv^2}{Lsin^2 \theta} [/tex]

[tex] mgL sin^2 \theta = mv^2 cos \theta [/tex]

[tex] gL(1-cos^2 \theta) = v^2 cos \theta [/tex] Thus:

[tex]
\frac{gL}{v^2} - \frac{gL}{v^2}cos^2 \theta - cos\theta = 0
[/tex]

 

[Jebus_Chris]

I have no idea what you did for (1). What were you x and y equations for force?

 

[Zhalfirin88]

I have no idea what you did for (1). What were you x and y equations for force?

I edited above for #1. The x-direction is on the right side, y on the left.

 

[Jebus_Chris]

Alright, when I did it I had the radius as L >>
In your final equation just set cos[tex]\theta[/tex] = x.

 

[Zhalfirin88]

I did that but wouldn't you get something like

[tex] \frac{1 \pm \sqrt{1^2 - 4(1.41)(1.41)} }{2(1.41)} [/tex]

Which would simplify to:

[tex] \frac{1 \pm \sqrt {-6.9524}}{2.82} [/tex]

?

Edit: For reference the equation would be [tex] 1.41x^2 - x + 1.41 [/tex] Edited out 2nd question.

 

[Jebus_Chris]

[tex]

\frac{gL}{v^2} - \frac{gL}{v^2}cos^2 \theta - cos\theta = 0

[/tex]
So the quadratic equation would be
[tex]
-1.41x^2 - x + 1.41
[/tex]
You had a positive when it is actually negative.