GREPhysics.NET: GR9277 #83

GR9277 #83

Problem

GREPhysics.NET Official Solution

\prob{83}

Two pith balls of equal mass M and equal charge q are suspended from the same point on long massless threads of length L as shown in the figure above. If k is the Coulomb's law constant, then for small values of $\theta$ , the distance d between the charged pith balls at equilibrium is

$\left(\frac{2kq^2L}{Mg}\right)^{1/3}$
$\left(\frac{kq^2L}{Mg}\right)^{1/3}$
$\left(\frac{2kq^2L}{Mg}\right)^{1/2}$
$\left(\frac{kq^2L}{Mg}\right)^{1/3}$
$L/4$

Electromagnetism $\Rightarrow$ }Forces

Sum of the forces for one of the mass in the x (horizontal) and y (vertical) directions gives,

$\begin{eqnarray} \sum F_x = 0 = T\sin\theta - kq^2/d^2 \\ \sum F_y = 0 = T\cos\theta - mg <br /> \end{eqnarray}$

For small angles, $\cos\theta \approx 1 \Rightarrow T \approx mg$ . From the geometry, one can deduce that $\sin\theta = (d/2)/L$ .

Thus, the x equation yields $T(d/2)/L=kq^2/d^2 \Rightarrow d^3 = 2kq^2L/(mg)$ (since $T\approx mg$ from the y equation for small angles). This is choice (A).

Alternate Solutions

cjohnson415 2013-08-19 01:40:11	Here's an alternative solution that skips the trig altogether: Due to small $\theta$ , we can consider each as a simple pendulum, a special case of the simple harmonic oscillator: $\omega = \sqrt{\frac{k}{m}} = \frac{2 \pi} {T} = \sqrt{\frac{g}{L}}$ Thus $k = \frac{mg}{L}$ , which is the restoring force constant ( $F = -kx$ ), so: $\frac{mgd}{2L} = \frac{k_{c}q^2}{d^2}$ The rest is the same algebra, just a different/potentially quicker/potentially prettier way to arrive at the force balance. Reply to this comment
neo55378008 2012-08-18 12:27:15	Dimensional analysis helps! Since $\\F=\frac{kq^2}{r^2}$ is in N, $\\k q^2$ will have units of $\\N m^2$ Mg has units of N so $\frac{kq^2L}{Mg}$ will have units of $\\m^3$ This leaves only A and B. Knowing that solving for $\theta$ will give half of d, take the solution with an extra 2 in it! (A) Reply to this comment
redmomatt 2011-11-06 08:25:21	Heres another take without the trig functions... Since $\theta$ is small, we can remove all of the trig functions. Thus, we just need to equate the outward Coulomb force $F_c$ to the force of gravity along the horizontal $F_x$ . $\theta = \frac{F_x}{mg} \Rightarrow F_x = mg\theta = \frac{mgd}{2L}$ , since $L\theta = d/2$ . Hence, $F_c = F_x$ $\Rightarrow$ $\frac{ k q^2 }{d^2} = \frac{mgd}{2L}$ $\Rightarrow$ $d^3 = \frac{2kq^2L}{mg} \Rightarrow d = \left(\frac{2kq^2L}{mg}\right)^{1/3}$ , which is (B) . Reply to this comment
Andresito 2006-03-29 14:56:46	It is a bit faster to equate F(electric) = TSin(theta), mg = T*Cos(theta) and divide them. Tan(theta) ~ theta Reply to this comment

Comments

astropolo
2014-07-29 11:55:16

I have a quick question about the small angle approximations. Why can you approximate that Cos~1 while you use the geometry for Sin rather than saying sin~theta?

JoshWasHere
2014-08-20 08:20:22

When you expand cosine and sine, you see that the first two terms in the cosine and sine expansion are cos $\theta$ = 1 - $\frac{\theta ^2}{2!}$ and sin $\theta$ = $\theta$ - $\frac{\theta ^3}{3!}$ . Because of the small angle, we select only the first order terms, and drop all higher order terms (with the condition that $\theta$ < 5 degrees).

Reply to this comment

cjohnson415
2013-08-19 01:40:11

Here's an alternative solution that skips the trig altogether:

Due to small $\theta$ , we can consider each as a simple pendulum, a special case of the simple harmonic oscillator:

$\omega = \sqrt{\frac{k}{m}} = \frac{2 \pi} {T} = \sqrt{\frac{g}{L}}$

Thus $k = \frac{mg}{L}$ , which is the restoring force constant ( $F = -kx$ ), so:

$\frac{mgd}{2L} = \frac{k_{c}q^2}{d^2}$

The rest is the same algebra, just a different/potentially quicker/potentially prettier way to arrive at the force balance.

Reply to this comment

greatspirits
2012-11-02 15:49:28

Energy conservation anyone? The change in gravitational potential energy should equate to potential between the charges. So... $2mg\delta h$ should equate to... hmmm well I usually prefer the energy method but the force way seems to be more effecient here.

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neo55378008
2012-08-18 12:27:15

Dimensional analysis helps!
Since $\\F=\frac{kq^2}{r^2}$ is in N, $\\k q^2$ will have units of $\\N m^2$
Mg has units of N
so $\frac{kq^2L}{Mg}$ will have units of $\\m^3$
This leaves only A and B. Knowing that solving for $\theta$ will give half of d, take the solution with an extra 2 in it! (A)

henryb
2016-09-12 18:14:40

thx a lot. this is the simplest solution

Reply to this comment

redmomatt
2011-11-06 08:25:21

Heres another take without the trig functions...

Since $\theta$ is small, we can remove all of the trig functions. Thus, we just need to equate the outward Coulomb force $F_c$ to the force of gravity along the horizontal $F_x$ .

$\theta = \frac{F_x}{mg} \Rightarrow F_x = mg\theta = \frac{mgd}{2L}$ , since $L\theta = d/2$ .

Hence,

$F_c = F_x$ $\Rightarrow$ $\frac{ k q^2 }{d^2} = \frac{mgd}{2L}$ $\Rightarrow$ $d^3 = \frac{2kq^2L}{mg} \Rightarrow d = \left(\frac{2kq^2L}{mg}\right)^{1/3}$ , which is (B) .

Reply to this comment

Dodobird
2010-11-12 11:57:31

tan(theta) = d/2L
tan(theta) = kq*q/d*d*MG

d/2L = kq*q/d*d*MG

gives answer A.

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FortranMan
2008-10-03 14:02:49

Also note because theta is so small here, we can approximate the vector component of the gravity that's NOT inducing tension to be d/2.

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JAS
2007-11-01 09:20:45

It seems easier to equate torques around the pivot point, where the torque on the right mass due to the repulsive electric force isrnrn $L\frac{kq^2}{d^2}\cos\theta$ counterclockwisernrnand the torque on the right mass due to its weight is:rnrn $MgL\sin\theta$ clockwisernrn $\cos \theta \approx 1$ and $\sin \theta \approx \tan \theta = \frac{d}{2L}$ rnrnFrom this, choice A follows.

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relain
2007-10-23 19:21:51

typo in the question, option D should be a square not a cube root. It's currently identical to option B.

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Andresito
2006-03-29 14:56:46

It is a bit faster to equate F(electric) = T*Sin(theta), m*g = T*Cos(theta) and divide them. Tan(theta) ~ theta

Reply to this comment

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