# Friction and spinning mass

1. Sep 25, 2009

### Zhalfirin88

Spinning mass

1. The problem statement, all variables and given/known data
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.

3. The attempt at a solution

For 1) I have the equation down to:

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

Last edited by a moderator: Apr 24, 2017
2. Sep 25, 2009

### Jebus_Chris

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

3. Sep 25, 2009

### Zhalfirin88

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:

$$T_y - mg = 0$$ X-Direction: $$T sin \theta = \frac{mv^2}{r}$$

$$T cos\theta = mg$$ X-Direction: $$T sin \theta = \frac{mv^2}{Lsin \theta}$$

$$T = \frac{mg}{cos \theta}$$ X-Direction: $$T = \frac{mv^2}{Lsin^2 \theta}$$

$$\frac{mg}{cos \theta} = \frac {mv^2}{Lsin^2 \theta}$$

$$mgL sin^2 \theta = mv^2 cos \theta$$

$$gL(1-cos^2 \theta) = v^2 cos \theta$$ Thus:

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

Last edited: Sep 25, 2009
4. Sep 25, 2009

### Jebus_Chris

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

5. Sep 25, 2009

### Zhalfirin88

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

6. Sep 25, 2009

### Jebus_Chris

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

7. Sep 25, 2009

### Zhalfirin88

I did that but wouldn't you get something like

$$\frac{1 \pm \sqrt{1^2 - 4(1.41)(1.41)} }{2(1.41)}$$

Which would simplify to:

$$\frac{1 \pm \sqrt {-6.9524}}{2.82}$$

?

Edit: For reference the equation would be $$1.41x^2 - x + 1.41$$ Edited out 2nd question.

Last edited: Sep 25, 2009
8. Sep 25, 2009

### Jebus_Chris

$$\frac{gL}{v^2} - \frac{gL}{v^2}cos^2 \theta - cos\theta = 0$$
So the quadratic equation would be
$$-1.41x^2 - x + 1.41$$
You had a positive when it is actually negative.