0
$\begingroup$

I've got a problem from my physics course which I am a little stuck on.

A dumbell consisting of two identical masses m=5.8 kg attached to the ends of a thin (massless) rod of length a=0.4 m that is pivoted at its centre. The masses have charge +q and -q and the system is placed in a uniform electric field E=1.6 N/C in the positive x direction. Initially the dumbell is at rest and makes a 60o angle with the x axis. The dumbell is the released.

a) Calculate the horizontal displacement (in meters) of the mass with positive charge since the dumbell was released until the point where the dipole is temporarily aligned with the electric field (Use 3 significant figures in your answer).

b) Calculate the charge q knowing that when the dipole is momentarily aligned with the electric field, its kinetic energy is K=4.9x10^-3 J.

c)Calculate the angular velocity of the dumbell when its position is temporarily aligned with the electic field.

So I've approached the problem by torques. I've tried to find the perpendicular force due to the electric field, and used that to calculate the torque. I then use angular torque*angular displacement = change in angular kinetic energy. This approach doesn't seem to give me the right answer, and what's even more confusing is part c). If I try to answer part c) using K=4.9*10^-3 J as my angular K to get my angular velocity my answer isn't correct, which suggests to me the K given in b) is just linear kinetic energy.

I suspect this question has something to do with electric potential energy, however we haven't covered this content yet, making me think that I've missed something critical.

For reference here's a picture

enter image description here

$\endgroup$
0
$\begingroup$

Work done by the electric field is force times displacement along the direction of force. Center of mass doesn't move, so the only motion is rotation; horizontal spacing of charges goes from $a\cos\theta$ to $a$, so work done is $E q a(1-\cos\theta)$. Half of that work is done on the positively charged mass (which moves half that distance, i.e. $\frac{a}{2}(1-\cos\theta)$), and the other half on the negative one. And the total work done is equal to the total kinetic energy.

Now the moment of inertia is $2 m (\frac{a}{2})^2$; and the relationship between angular velocity and (rotational) kinetic energy is

$$E = \frac12 I \omega^2$$

Note that when the masses are aligned with the electric field, they have no horizontal component of velocity; but at any angle, you can think of their motion of being purely rotational.

That should be enough to get you going. It sounds like you went about this the right way, and that you lost a factor 2 somewhere? Without seeing the details of your work it is hard to guess.

update

Implementing the above, I get the following answers.

a) displacement of positive charge is 0.1 m (distance from center of rotation 0.2 m, rotating through 60 degrees).

b) Kinetic energy of the entire dipole is 4.9 $\cdot 10^{-3}$ J; the work done on one of the charges is half of that. So we find $$qEd = K.E. = 0.5\cdot 4.9\cdot 10^{-3}\;\mathrm{ J}\\ q = \frac{0.5\cdot 4.9\cdot 10^{-3}\; \mathrm{J}}{1.6\; \mathrm{N/C} \cdot 0.1\; \mathrm{m}} = 0.015 \;\mathrm{C}$$

c) $$\frac12 I \omega^2 = K.E.\\ \omega = \sqrt{\frac{2 K.E.}{2 m r^2}} = \sqrt{\frac{2\cdot 4.9\cdot 10^{-3}}{2\cdot 5.8 \cdot 0.2^2}}=0.145\;\mathrm{s^{-1}}$$

Alternatively,

$$\omega = \frac{v}{r}\\ v = \sqrt{\frac{2\cdot (0.5 * K.E.)}{m}} = 0.029\; \mathrm{m/s}\\ \omega = \frac{0.041\;\mathrm{m/s}}{0.2\;\mathrm{m}} = 0.145\; \mathrm{rad/s}$$

Either way, we get the same result.

$\endgroup$
  • 1
    $\begingroup$ Thanks Floris, when I did my original working I went down a similar track, but I didn't consider the work done on the negative end. I'll try again and see how it goes :) $\endgroup$ – istigatrice Apr 30 '15 at 13:06
  • $\begingroup$ So apparently this isn't the right way to do it. I managed to answer part b) using the regular (linear) work principles and using my displacement as the radius r, so q=kE/(r*E) So the displacement I'm using is r=0.2, which is the difference between the final and initial positions (the geometry makes an equilateral triangle. This is weird as the direction of this displacement vector isn't straight down, implying v isn't straight down. $\endgroup$ – istigatrice May 1 '15 at 12:47
  • $\begingroup$ The velocity vector describes a circle - when the dipole is aligned with the electric field, the instantaneous velocity is straight down. That is independent of the path it took to get there - what matters is what the velocity is at the moment we consider it. What do you mean by "apparently this isn't the right way to do it"? What new information do you have? $\endgroup$ – Floris May 1 '15 at 14:33
  • $\begingroup$ @WenqiZhang - I added the "complete" calculation. See if this clears it up for you - and if you have independent confirmation of the correctness of this answer it would be appreciated. $\endgroup$ – Floris May 1 '15 at 15:03
  • $\begingroup$ Hi Floris, Your answer is the same as what we came up with, just we got there with a different method. Our homework is online and thus we can check our answers instantly so when I said "this isn't the right way to do it" I probably should have said we found an alternative way of doing it which yielded the same answer (sorry about not being clear). However, I am still confused as to what displacement you used for the work calculation, was it the horizontal? Why? I thought the displacement had to be in the same direction as the work done? $\endgroup$ – istigatrice May 2 '15 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.