Finding charges in a pendulum in equilibrium


by Zayin
Tags: charges, equilibrium, pendulum
Zayin
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#1
Nov12-12, 03:37 AM
P: 4
1. The problem statement, all variables and given/known data

Two small metallic spheres, each of mass 0.20 g, are suspended as pendulums by light strings from a common point. The spheres are given the same electric charge, and it is found that the two come to equilibrium when each string is at an angle of 5.0° with the vertical. If each string is 30.0 cm long, what is the magnitude of the charge on each sphere?


2. Relevant equations

[tex]F = k \ \frac{q_1 q_2}{r^2} \ \mbox{ or } \ \frac{1}{4 \pi \epsilon_0} \ \frac{q_1 q_2}{r^2}[/tex]
[tex]\begin{equation*} \begin{split}
\frac{1}{4 \pi \epsilon_0} = k & = & & 8.988 \times 10^9 \ N \cdot m^2 / C^2 \\ \\
\end{split} \end{equation*}[/tex]


3. The attempt at a solution

Finding the horizontal distance between the two spheres to give r:
1. An isosceles triangle is formed with the two strings, so the base angles are the same.
2. Therefore, the base angles are ((180-2×5)/2) = 85°.
3. Using the sine rule, r = 0.3×sin(2×5)/sin(85).

Finding the force of gravity that is acting horizontally on the two spheres to bring them to the vertical, which gives us F:
1. Treating the 30 cm strings as radii of a circle with centre from the common point, and the spheres as points on the circle, a straight line tangent to a sphere intersects the vertical.
2. The angle from one of the spheres to the other sphere and then to where the tangent intersects the vertical is thus 5° (radii are perpendicular to a tangent, adjacent angles).
3. Therefore, right angle trigonometry tells us that the force of gravity acting horizontally on a sphere is mg/tan(5) = 0.196/tan(5).

q_1 = q_2 (given)

So:[tex]F = k \ \frac{q^2}{r^2}[/tex]
Therefore,[tex]q = \sqrt{\frac{Fr^2}{k}}[/tex]

And when I sub my values in I get q = 8.3 x 10^-7 C, which is apparently incorrect, the correct answer being 7.2 x 10^-9 C. Help?
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Simon Bridge
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#2
Nov12-12, 04:02 AM
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Quote Quote by Zayin View Post
Finding the horizontal distance between the two spheres to give r:
1. An isosceles triangle is formed with the two strings, so the base angles are the same.
2. Therefore, the base angles are ((180-2×5)/2) = 85°.
3. Using the sine rule, r = 0.3×sin(2×5)/sin(85).
Good thinking.
Best to keep symbols as long as possible though. Helps with troubleshooting.
Finding the force of gravity that is acting horizontally on the two spheres to bring them to the vertical, which gives us F:
I'm sorry - what? Gravity is not horizontal - from the following, you mean tangential?
1. Treating the 30 cm strings as radii of a circle with centre from the common point, and the spheres as points on the circle, a straight line tangent to a sphere intersects the vertical.
2. The angle from one of the spheres to the other sphere and then to where the tangent intersects the vertical is thus 5° (radii are perpendicular to a tangent, adjacent angles).
3. Therefore, right angle trigonometry tells us that the force of gravity acting horizontally on a sphere is mg/tan(5) = 0.196/tan(5).
If I'm reading you right you are finding the component of a component here but you only have one trig factor.

I think you have over-complicated this - what is wrong with drawing a free-body diagram for each mass and summing the forces to zero?
(Put y-axis along the tension.)
ap123
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#3
Nov12-12, 04:19 AM
P: 265
It would be easier to find r by using a single right-angle triangle.
Then r = 2Lsinθ, (where L is the string length).

In all problems like this, it's always best to start with a free-body diagram, ( you only need to do this for 1 sphere because of symmetry ). I would put the y-axis along the direction of the weight.
This gives 2 equations allowing you to eliminate the string tension and leaving the charge as the only unknown.
As Simon Bridge says, you are making this more complicated than you have to.

Simon Bridge
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Nov12-12, 04:35 AM
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Finding charges in a pendulum in equilibrium


I would put the y-axis along the direction of the weight.
That would be traditional ("y=up/down" and all that)... however, if you put the y-axis along the string, you don't need to eliminate the tension :) But the cost is that you have two vectors to resolve instead of one ... play to your strengths.
Zayin
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#5
Nov12-12, 04:52 AM
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Quote Quote by ap123 View Post
It would be easier to find r by using a single right-angle triangle.
Then r = 2Lsinθ, (where L is the string length).
Heh heh. I know, but using the sine rule makes me feel fancy. :<

Quote Quote by Simon Bridge View Post
what is wrong with drawing a free-body diagram for each mass and summing the forces to zero?
(Put y-axis along the tension.)
So I did this, and solved along the x axis:

[tex]\begin{equation*} \begin{split}
k \ \frac{q^2}{r^2}cos5° - mgsin5° & = & & 0 \\

∴ kq^2cos5° & = & & r^2mgsin5° \\

∴ q & = & & \sqrt{\frac{r^2mgsin5°}{kcos5°}} \\

& = & & \sqrt{\frac{\frac{0.3sin10°}{sin85°}^2×0.196×sin5°}{8.988\times 10^9×cos5°}} \\

& = & & 7.2\times 10^{-8} \ C \\

\end{split} \end{equation*}[/tex]
But... the answer is supposedly 7.2 x 10^-9 C.

I can't seem to find the missing 10^-1. Am I overlooking it, or is the answer provided simply incorrect?

Thanks so much for your help!
ap123
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#6
Nov12-12, 05:13 AM
P: 265
Remember that the mass given is 0.2 grams, so mg=0.00196, not 0.196
Zayin
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#7
Nov12-12, 05:29 AM
P: 4
*headdesk*

Thanks.
Simon Bridge
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#8
Nov12-12, 07:27 PM
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Nice to see people using LaTeX though - kudos.
Just a tip: to get the trig to format correctly you should put a "\" in front of it ... because ##kq^2\cos(5^\circ)## reads better than ##kq^2cos(5^\circ)## ... technically units should be in normal, rather than italic, text ... which you can do, in math-mode, with the \text{} so 15m/s will be ##15\text{ms}^{-1}## rather than ##15ms^{-1}## ... but it is still much nicer than trying to format in plain text innit?

On your math - the other thing to try was to realize that the sum of the force vectors being zero means that the tension (T), gravity (mg), and coulomb force (F) vectors, form a right-angle triangle with the tension on the hypotenuse, and the half-apex angle (##\alpha =5^\circ##) is between mg and T. In which case, ##F=mg\tan(\alpha)## right off ... from there you can see where you originally made your mistake.

The power in the fbd method is that it does not rely on you drawing the right triangle right at the start.


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