Solving a Pendulum Problem with Exponential Solution

[archaic]

Homework Statement

We have two charged balls of masses $m_1=m_2$, and charges $+q_1=-q_2$, fixed to the ends of a very light rod of length $\ell$.
The center of the rod is mounted on a friction-free pivot, and the whole system is then immersed in a uniform electric field of magnitude $E$.
a) Determine the period of oscillation if the system is disturbed from its initial orientation by a small angle $\theta$.
b) What if $m_1<m_2$?

Relevant Equations

$$\vec F=q\vec E$$

a) This looks somewhat like a pendulum problem (length $\ell/2$).

I reasoned there will be a clockwise rotation, and that the acceleration is due to the force of magnitude$$F=-\left[(F_2\sin\theta-m_2g\cos\theta)+(F_1\sin\theta+m_1g\cos\theta)\right]=-(|q_2|+q_1)E\sin\theta=-2qE\sin\theta=ma$$
The way I thought of it is that the acceleration put to bring 2 up is added to the acceleration put to bring 1 down to make the rotation acceleration.
For small angles, I get this differential equation$$\ddot\theta=-\frac{4qE}{m\ell}\theta$$
and if I put $\theta=A\cos(\omega t+\phi)$, then I would get $\omega^2=\frac{4qE}{m\ell}$, and, from it, $T=\pi\sqrt\frac{m\ell}{qE}$.

b) If I use the same reasoning, then I get an exponential solution to my problem, since the force of gravity won't disappear, which makes me think that my solution for a) is wrong.
Any help would be great!

 

[etotheipi]

I don't think you are meant to account for an external gravitational field (for instance, the diagram could be a birds-eye view of the configuration).

 

[jbriggs444]

since the force of gravity won't disappear

I am quite sure that you are intended to ignore gravity. After all, you are not told whether the arrangement is horizontal with you looking down from above or vertical with you looking in from the side.

Edit: Lost to @etotheipi by a nose.

 

[etotheipi]

I reasoned there will be a clockwise rotation, and that the acceleration is due to the force of magnitude$$F=-\left[(F_2\sin\theta-m_2g\cos\theta)+(F_1\sin\theta+m_1g\cos\theta)\right]=-(|q_2|+q_1)E\sin\theta=-2qE\sin\theta=ma$$

But in any case, I don't think this is correct. You didn't include a contribution from the normal force from the hinge in the net component of force in the $-\hat{\theta}$ direction. With that included, $a$ would be the component of acceleration of the centre of mass of the system in the $-\hat{\theta}$ direction. But without knowing anything about $\vec{N}$, I don't think that's useful.

It's easier to just use $\sum \tau_z = I_z \ddot{\theta}$ about the hinge.

 

[kuruman]

But in any case, I don't think this is correct. You didn't include a contribution from the normal force from the hinge in the net component of force in the $-\hat{\theta}$ direction. With that included, $a$ would be the component of acceleration of the centre of mass of the system in the $-\hat{\theta}$ direction. But without knowing anything about $\vec{N}$, I don't think that's useful.

It's easier to just use $\sum \tau_z = I_z \ddot{\theta}$ about the hinge.

Why is the normal force needed? If gravity is not acting, as we all agree, the two electric forces form a couple so the net force on the arrangement is zero, hinge or no hinge, equal or unequal masses.

 

[jbriggs444]

Why is the normal force needed? If gravity is not acting, as we all agree, the two electric forces form a couple so the net force on the arrangement is zero, hinge or no hinge, equal or unequal masses.

Not so fast. The net electric force is zero. But if the pivot remains midway between the masses, rate of change of momentum is non-zero.

 

[kuruman]

Not so fast. The net electric force is zero. But if the pivot remains midway between the masses, rate of change of momentum is non-zero.

Rate of change of momentum is a vector. In what direction would that be in this case?

 

[etotheipi]

For instance, the centre of mass of the system is at the position $$\bar{x} = \frac{l(m_2 - m_1)}{2(m_1 + m_2)} \cos{\theta}$$where $l$ is the length of the rod. The net external force on the system in the $\hat{x}$ direction is $N_x$, since the electric forces cancel exactly. You end up with$$N_x = (m_1 + m_2)\ddot{\bar{x}}$$Now if $m_1 = m_2$, then $\ddot{\bar{x}}$ would be zero always. However in the general case where the masses are not equal, then it will not be.

 

[kuruman]

Yes, I spoke too fast. In the case of unequal masses the CM will accelerate as it rotates about the pivot.

 

[archaic]

Thank you @etotheipi and @jbriggs444! Though I think that I am to account for gravity as the problem is, originally, about a macroscopic model of a "dipolar molecule". In the first part of the problem, $E=450$ N/C, $m_1=m_2=0.015$ kg, and $q=0.8$ micro coulombs.
I will use the the relation mentioned by @etotheipi.
$$\tau_1=-\frac{\ell}{2}m_1g\sin\left(\theta+\frac{\pi}{2}\right)-\frac{\ell}{2}qE\sin\theta$$$$\tau_2=\frac{\ell}{2}m_2g\sin\left(\frac{\pi}{2}-\theta\right)-\frac{\ell}{2}qE\sin\theta$$$$\frac{\ell}{2}g\left(m_2\sin\left(\frac{\pi}{2}-\theta\right)-m_1\sin\left(\theta+\frac{\pi}{2}\right)\right)-\ell qE\sin\theta=(m_1+m_2)\frac{\ell^2}{4}\ddot\theta$$
The solution to this does give an oscillation period when the masses are equal, but elsewise it is an exponential function.

 

[archaic]

Oh, maybe in the second case I should disregard the electric field contribution? For b), I am given $m_2=2m_1$, which gives me $\frac{gm}{2qE}=204.375$.

No... I need a negative LHS.

 

[etotheipi]

$$\frac{\ell}{2}g\left(m_2\sin\left(\frac{\pi}{2}-\theta\right)-m_1\sin\left(\theta+\frac{\pi}{2}\right)\right)-\ell qE\sin\theta=(m_1+m_2)\frac{\ell^2}{4}\ddot\theta$$

N.B. that $\sin{\left(\frac{\pi}{2} - \theta\right)} = \sin{\left(\frac{\pi}{2} + \theta\right)} = \cos{\theta}$, which immediately suggests the simplification of the first term on the left to $\frac{lg(m_2 - m_1)}{2}\cos{\theta}$.

Oh, maybe in the second case I should disregard the electric field contribution?

I don't think so; if you want to solve it with a uniform gravitational field you still need both contributions.

 

[archaic]

For some reason, my brain didn't want to think that the macroscopic model can be setup so that we are looking from above. Thanks again, @jbriggs444.
You too, @etotheipi!

 

[archaic]

Okay, nope, that's not it. When $m_1=m_2$, the HW system accepted the answer, but rejected it for the other case.

 

[kuruman]

If you are convinced that you did the calculation correctly, consider this. Dipolar molecules don't have hinges. If this is supposed to be a model for a dipolar molecule, perhaps you should consider the "hinge" to be at the center of mass.

 

[etotheipi]

If you are convinced that you did the calculation correctly, consider this. Dipolar molecules don't have hinges. If this is supposed to be a model for a dipolar molecule, perhaps you should consider the "hinge" to be at the center of mass.

That is a nice idea, but the question does specifically say that the centre of the rod is mounted on a friction-free pivot.

 

[kuruman]

That is a nice idea, but the question does specifically say that the centre of the rod is mounted on a friction-free pivot.

As is often the case, unless we see OP's math work for part (b) we will be in the dark about why OP's solution was rejected.

 

[archaic]

@kuruman
for part a) (equal masses, so the gravity term vanishes), solving$$\frac{\ell}{2}g\left(m_2-m_1\right)-\ell qE\theta=(m_1+m_2)\frac{\ell^2}{4}\ddot\theta$$gives me $T=\pi\sqrt{\frac{(m_1+m_2)\ell}{qE}}$.
If I suppose that I am looking from above at the system, such that gravity plays no role in the rotation, then the solution is the same, and I only need to plug new values for the masses.
If, however, I suppose that gravity does intervene in this system, then I can put $f(t)=\frac{\ell}{2}g\left(m_2-m_1\right)-\ell qE\theta(t)$, which will eventually give me the same $T=\pi\sqrt{\frac{(m_1+m_2)\ell}{qE}}$.

 

[kuruman]

So in part (b) you used $T=\pi\sqrt{\dfrac{(m_1+m_2)\ell}{qE}}=\pi\sqrt{\dfrac{3m_1\ell}{qE}}$ and that answer was rejected?

 

[archaic]

So in part (b) you used $T=\pi\sqrt{\dfrac{(m_1+m_2)\ell}{qE}}=\pi\sqrt{\dfrac{3m_1\ell}{qE}}$ and that answer was rejected?

Yes.

 

[kuruman]

Yes.

Then I see three possibilities in order of likelihood.
1. You calculated the numerical answer incorrectly. Redo it very carefully, pay attention to powers of ten and don't forget to take the square root. If your answer matches the one that was rejected, proceed to the next step. If not, enter the new answer and hope for the best.
2. Read the statement of the problem carefully. Does it mention anywhere something like "A model of a dipolar molecule is two point charges separated by distance $\ell~\dots$" If that's the case, then the question in part (b) "What if $m_1<m_2$?" probably has the answer "then I need to consider the hinge to be at the center of mass and recalculate the period." Do the recalculation, enter the answer and hope for the best. If that doesn't work, proceed to the next step.
3. Somehow the correct answer did not get coded in the algorithm that is used to evaluate answers. This is not very likely, but I have seen it happen. The fix for that is to show you work to your professor and ask why both answers were rejected. If there is an error in the algorithm, you will be dong everyone a favor.

 

[archaic]

2. might be it, @kuruman.
$$m_1=15.0\,g\text{, }m_2=30.0\,g\text{, }\ell=\frac{15.0}{100}\,m$$

$$x_{cm}=\frac{15\times\frac{15}{100}}{15+30}=\frac{5}{100}=\frac{\ell}{3}$$$$\tau_1=-\frac{10}{100}m_1g\sin\left(\theta+\frac{\pi}{2}\right)-\frac{10}{100}qE\sin\theta$$$$\tau_2=\frac{5}{100}m_2g\sin\left(\frac{\pi}{2}-\theta\right)-\frac{5}{100}qE\sin\theta$$$$\frac{\ell}{3}g\left(m_2-2m_1\right)\cos\theta-\ell qE\sin\theta=\left[m_1\left(\frac{2\ell}{3}\right)^2+m_2\left(\frac{\ell}{3}\right)^2\right]\ddot\theta$$$$\implies-\ell qE\theta=\left[m_1\left(\frac{2\ell}{3}\right)^2+m_2\left(\frac{\ell}{3}\right)^2\right]\ddot\theta$$$$\implies-\frac{9qE}{(4m_1+m_2)\ell}\theta=\ddot\theta$$$$T=\frac23\pi\sqrt{\frac{(4m_1+m_2)\ell}{qE}}$$
What do you think? I only have one attempt left.. don't want to make silly mistakes.

 

[kuruman]

Looks fine, that's what I got. Try it and see if it works. If not, you will have to consider plan C. In any case, I am curious to know how this turns out.

By the way, two equal and opposite forces which are along two separate parallel lines form a couple. A couple has the property that the torque it exerts is independent of the origin; its magnitude is the magnitude of one of the forces (in this case $qE~$) multiplied by the perpendicular distance between the two parallel lines along which the two forces act (in this case $\ell \sin\theta~$). As always, the direction of the torque is given by the right hand rule. It is a useful shortcut that you can prove yourself if you wish.

 

[archaic]

Looks fine, that's what I got. Try it and see if it works.

It did! Thank you very much!

By the way, two equal and opposite forces which are along two separate parallel lines form a couple.

Right, I have seen this in a text.
$$\vec{OA}\times\vec{f_1}+\vec{OB}\times\vec{f_2}=\vec{OA}\times\vec{f_1}+\vec{OB}\times-\vec{f_1}=\left(\vec{OA}+\vec{BO}\right)\times\vec{f_1}=\vec{BA}\times\vec{f_1}$$