Finding the maximum acceleration

[voko]

Oops, I made another slip. We are dealing with 3D, so the point is $(x, y, z)$, but I used $z$ to mean something else.

Pranav, you will have to fix my mistakes and introduce some consistent notation.

 

[Saitama]

Let's say one sphere is A, another is B. What is the potential of A at distance $z$ from its center if $z \ge R$? If $z < R$?

For $z\geq R$, it is simply $kQ/z^2$.

For $z<R$, the potential is $\displaystyle \frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)$

Now, let the centers of A and B be separated by $d$. Let $(x, y)$ be the coordinates of a point of B. What is its distance from the center of A? What is the potential of A at that point?

I take the sphere A to be centred at origin. The distance of $(x,y)$ from origin is $\sqrt{x^2+y^2}$. Potential at $(x,y)$ is calculated using the above formulas.

I was thinking that won't it be better to ask distance of $(x,y,z)$ from centre of A? We have a 3-dimensional ball, sorry if I am acting dumb.

EDIT: Okay, I see, we both posted at the same time. :)

So the distance from centre of A is $\sqrt{x^2+y^2+z^2}$.

 

[voko]

I would rather make the center of B the origin. I would further stack A on top (i.e., z-axis) of B and use spherical coordinates. Caveat: these suggestions are based solely on my gut feeling, so could be misleading.

Obviously, you will need to find a way to split the integration domain into two parts.

 

[Saitama]

I would rather make the center of B the origin. I would further stack A on top (i.e., z-axis) of B and use spherical coordinates. Caveat: these suggestions are based solely on my gut feeling, so could be misleading.

Obviously, you will need to find a way to split the integration domain into two parts.

I am not comfortable with spherical coordinates. I was trying something before you asked me to calculate the distance of $(x,y,z)$ from centre of A. Please look at the attachment. If my approach is too complicated, I would happily switch to spherical coordinates.

I consider a red sphere of radius r and thickness dr passing through the sphere B. This red sphere forms a spherical cap inside B. The area of the spherical cap is given $2\pi r^2(1-\cos\theta)$. I used the law of cosines and substituted $\cos\theta=\frac{r^2+d^2-R^2}{2rd}$. After some simplification, I get:

$$dA=\frac{\pi r}{d}\left(R^2-(r+d)^2\right)$$

Hence, the differential volume is $dV=(dA)dr$. Since I have the volume charge density, I can find the charge contained in it.

I would like to know if I am correct so far. :)

EDIT: I am not sure but I think dV should (dA)(rdθ) instead of (dA)dr.

 

[voko]

That looks correct to me, except that it should be $(r - d)^2$. And that should be just $A$, not $dA$, because the area is not a differential.

 

[Saitama]

That looks correct to me, except that it should be $(r - d)^2$. And that should be just $A$, not $dA$, because the area is not a differential.

Thank you voko for checking the work, yes it should be (r-d)^2. :)

Can you please tell what should be the differential volume? Should it be A(dr) or A(rdθ)? I am inclined towards the latter.

 

[voko]

The only differential thing in your approach is the thickness of the spherical surface $dr$. Obviously $r$ and $\theta$ are functionally interdependent, so you could find (in theory) $r = f(\theta)$, then you would be able to do $dV = A(r) dr = A(f(\theta)) f'(\theta) d\theta$, but I see no reason you would want to do that, as that only seems to complicate matters for no obvious benefit.

 

[Saitama]

The only differential thing in your approach is the thickness of the spherical surface $dr$. Obviously $r$ and $\theta$ are functionally interdependent, so you could find (in theory) $r = f(\theta)$, then you would be able to do $dV = A(r) dr = A(f(\theta)) f'(\theta) d\theta$, but I see no reason you would want to do that, as that only seems to complicate matters for no obvious benefit.

Yes, I am not interested in expressing r in terms of theta as it would, as you say, definitely complicate the things. But I can't see why it is Adr instead A(rdθ), can you please explain?

Using dV=Adr, the charge contained in this volume is:

$$dq=\frac{\rho\pi r}{d}\left(R^2-(r-d)^2\right)dr$$

I set up the potential energy integrals in the following manner:

$$U=\int_{d-R}^R \frac{kQ\, dq}{2R}\left(3-\frac{r^2}{R^2}\right)+\int_R^{d+R} \frac{kQ\, dq}{r}$$

I solve the integrals separately.

First integral is:
$$\int_{d-R}^R \frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)\frac{\rho\pi r}{d}\left(R^2-(r-d)^2\right)dr$$

I solved the integral using wolfram alpha and got:
$$-\frac{kQ\rho \pi}{2Rd}\frac{d(d-2R)^2(d^3+4d^2R-18dR^2-48R^3)}{60R^2}$$

Second integral is:
$$\int_R^{d+R} \frac{kQ\rho \pi}{d}\left(R^2-(r-d)^2\right)dr$$

Solving:
$$-\frac{kQ\rho \pi}{d}\frac{1}{3}d^2(d-3R)$$

These look dirty, am I going in the right direction?

 

[voko]

Yes, I am not interested in expressing r in terms of theta as it would, as you say, definitely complicate the things. But I can't see why it is Adr instead A(rdθ), can you please explain?

It is actually quite strange to me that you are asking for an explanation. The whole idea of your approach is that you "slice" sphere B with spherical sections of radius $r$. If the area of such a section is $A$, and the thickness is infinitesimal $dr$, then the volume of the "slice" can only be $A dr$. Why would it be $A r d\theta$ instead?

Using dV=Adr, the charge contained in this volume is:

$$dq=\frac{\rho\pi r}{d}\left(R^2-(r-d)^2\right)dr$$

You could use the fact that $\rho = \dfrac Q V = \dfrac Q {^4 / _3 \pi R^3}$, then $dq = \dfrac Q {^4 / _3 \pi R^3} \dfrac{\pi r}{d}\left(R^2-(r-d)^2\right)dr = \dfrac {3} {4d'} Q r' \left(1 - (r' - d')^2\right) dr'$, where $r = R r', \ d = R d'$, which would simplify your equations somewhat. But fundamentally, I believe you are on te right track.

The important question, which you have not answered, is what you are going to do with the expression for the potential energy.

 

[Saitama]

The important question, which you have not answered, is what you are going to do with the expression for the potential energy.

I am not sure, substituting d=0, I can have the potential energy at zero separation and equating with kinetic energy gives the mass of balls.

Do I find the force using the expression for potential energy?

 

[voko]

It is not the force you are looking for, it is the separation where the acceleration is at a maximum.

 

[rcgldr]

It is not the force you are looking for, it is the separation where the acceleration is at a maximum.

The problem asks for the magnitude of the maximum acceleration, not the magnitude of the separation of maximum acceleration.

The maximum speed is stated as 10 m /s, but is this the maximum speed of each sphere relative to the center of mass of both spheres or the maximum rate of closure between the centers of mass?

Observation, assuming Q1 = -Q2, as soon as the non-conducting spheres intersect, a "lens" of neutral charge is created, with a net force of zero.

If a formula for force versus separation F(s) can be determined (this is known for non-intersecting non-conducting spheres and needs to be determined for intersecting non-conducting spheres), then the potential energy difference between two points of separation equals the intergral of -F(s)ds from the first point to the second point.

 

[haruspex]

Observation, assuming Q1 = -Q2, as soon as the non-conducting spheres intersect, a "lens" of neutral charge is created, with a net force of zero.

That might be more confusing than helpful. I think we have to assume that each sphere behaves as a rigid body, its parts held together by some unspecified structural forces. Otherwise it gets really tough. So the repulsion between the portion of a sphere outside the lens and the portion of the same sphere inside the lens is canceled by these structural forces. Hence the attractive force between one charge in the lens and the other charge outside the lens is not cancelled.

 

[voko]

The problem asks for the magnitude of the maximum acceleration, not the magnitude of the separation of maximum acceleration.

Yes, that is correct. However, I do not think you can do that without first finding the separation where that happens.

Observation, assuming Q1 = -Q2, as soon as the non-conducting spheres intersect, a "lens" of neutral charge is created, with a net force of zero.

I had this idea initially, which I then rejected. So did you, if I remember correctly. It seems definitely true that we could ignore the overlapping parts of the spheres because the total charge there is zero. If we do however, we will have to deal with the potential of a spherical cap, which is hardly a simplification.

The maximum speed is stated as 10 m /s, but is this the maximum speed of each sphere relative to the center of mass of both spheres or the maximum rate of closure between the centers of mass?

That is a good question. I had doubts about that as well. The phrasing suggests to me that it is the speed of each sphere in the initial rest frame.

 

[rcgldr]

Observation, assuming Q1 = -Q2, as soon as the non-conducting spheres intersect, a "lens" of neutral charge is created, with a net force of zero.

That might be more confusing than helpful. I think we have to assume that each sphere behaves as a rigid body, its parts held together by some unspecified structural forces. Otherwise it gets really tough. So the repulsion between the portion of a sphere outside the lens and the portion of the same sphere inside the lens is canceled by these structural forces. Hence the attractive force between one charge in the lens and the other charge outside the lens is not cancelled.

I had this idea initially, which I then rejected. So did you, if I remember correctly. It seems definitely true that we could ignore the overlapping parts of the spheres because the total charge there is zero. If we do however, we will have to deal with the potential of a spherical cap, which is hardly a simplification.

Assuming that Q1 = -Q2 and that density is uniform, then the intersecting lens volume has equal distribution and densities of positive and negative charges. The total charge of that volume is zero. Any attractive force between a charged volume within the lens and a charged volume outside the lens is countered by an equal in magnitude but opposing repulsive force by the opposing charge that occupies the same volume within the lens and that charged volume outside the lens.

Instead of polar coordinates, it might be better to consider the center of mass of both spheres to be located on the x axis, separated by distance s, with the y and z components of force cancelling due to symmetry, leading to solving for force as a function of the separation distance as I mentioned before. It seems that the math for this would be complicated, far more than I would expect in an introductory physics class.

 

[Saitama]

I am still lost, can you please tell me what should be the next step?

Should I use conservation of energy?

 

[voko]

If you have a function of the potential energy vs the separation, can you deduce where the acceleration is maximal?

 

[Saitama]

If you have a function of the potential energy vs the separation, can you deduce where the acceleration is maximal?

I am not sure but $F=ma=-\frac{dU}{dx}$ where x is the separation between the centres of sphere. To find the maximum acceleration, I set $d^2U/dx^2$ equal to zero, right?

 

[voko]

Correct. Graphically, that would correspond to the inflection point of the potential energy.

 

[Saitama]

Correct. Graphically, that would correspond to the inflection point of the potential energy.

I plugged the expression for potential energy in wolfram alpha. (I did not enter the constant product $KQ\rho \pi$)

http://www.wolframalpha.com/input/?...4x^2b-18xb^2-48b^3))/(120b^3)+(1/3)x(x-3b))=0

I have replaced R with b and d with x.

From the given solutions, the only one which is possible is x=b or d=R. Have I done this correctly?

 

[voko]

Without re-doing all that algebra myself, I cannot say for sure. The result looks reasonable, however, and, frankly, I trust WA more at doing algebra that myself :)

Now that you know where the max acceleration is, what is it?

 

[Saitama]

Without re-doing all that algebra myself, I cannot say for sure. The result looks reasonable, however, and, frankly, I trust WA more at doing algebra that myself :)

Now that you know where the max acceleration is, what is it?

I tried it further but when I submit the answer, it is marked incorrect. :(

Here's what I did. At zero separation, velocity is max and equal to potential energy at the instant.

Hence,

$$\frac{1}{2}mv^2=\frac{kQ\rho \pi}{2R}\frac{4R^2\cdot 48R^3}{60R^2} \Rightarrow mv^2=\frac{16}{5}kQ\rho\pi R^2 \,\, (*)$$

Since $ma=-dU/dx$, I calculated dU/dx at x=R and comes out to be 5R/8.

http://www.wolframalpha.com/input/?...-18xb^2-48b^3))/(120b^3)+(1/3)x(x-3b))+at+x=b

Hence,
$$ma=\frac{5}{8}kQ\rho \pi R \,\, (**)$$

Dividing (**) by (*),

$$\frac{a}{v^2}=\frac{25}{16\cdot 8 R}$$

Since v=10 m/s and R=1m,
$$a=19.531 \,\,m/s^2$$

Unfortunately, this is marked incorrect. I don't think this answer is wrong, it could be an error. If its an error, it should be corrected shortly. :)

 

[voko]

You are using the signs in potential energy very liberally, but that does not seem to be a problem here.

Have you considered what is said in the final paragraph of #44?

 

[Saitama]

Have you considered what is said in the final paragraph of #44?

I did look at it but I don't know how to set up the integrals here. My method of using spherical caps doesn't seem to work here. :(

 

[voko]

The final paragraph of #44 was "That is a good question. I had doubts about that as well. The phrasing suggests to me that it is the speed of each sphere in the initial rest frame."

Is that how you interpret the problem? Is that reflected in your equation for the conservation of energy?

 

[Saitama]

The final paragraph of #44 was "That is a good question. I had doubts about that as well. The phrasing suggests to me that it is the speed of each sphere in the initial rest frame."

Sorry.

Is that how you interpret the problem? Is that reflected in your equation for the conservation of energy?

Yes, I interpret it that way and I think my energy equation is consistent with that.

I guess I have found an error in my working. I should add the kinetic energies of both the balls in my energy equation, right?

I took 1/2mv^2 as the kinetic energy of the system where as it should be 1/2mv^2+1/2mv^2=mv^2. Am I right?

EDIT: Thanks a lot voko! I corrected the error and got a=39.0625 m/s^2 and this is marked correct.

Thanks a lot for all the help! It was fun discussing the problem. I got to learn new things.

 

[voko]

Yes, that was an interesting problem, Pranav. I should say I like the method you came up with, it seems quite original and clever to me, as compared to the run of the mill methods employing spherical or cylindrical coordinates one would typically use in such problems.

 

[Saitama]

I should say I like the method you came up with, it seems quite original and clever to me, as compared to the run of the mill methods employing spherical or cylindrical coordinates one would typically use in such problems.

Thank you voko but honestly, its all because of PF heroes like you I can think of these methods, I owe a lot to you guys. ;)

I am interested in learning about the other coordinate systems but unfortunately, I will have to wait another 6 months for that.

 

[haruspex]

Any attractive force between a charged volume within the lens and a charged volume outside the lens is countered by an equal in magnitude but opposing repulsive force by the opposing charge that occupies the same volume within the lens and that charged volume outside the lens.

Sure, but if you are going to treat the four regions (+ and - inside the lens, + and - outside the lens) as independent entities then the spheres will not remain as spheres. The lenses will be repelled from their parent bodies. To maintain the integrity of the spheres, that repulsion must be countered by some unknown structural force.
For clarity, let me invent some notation. S⁺, S^- represent the spheres, and E⁺, L⁺ etc. their portions external to and within the lens. A:B represents the electrical force between entities A and B. The net force of attraction of the spheres is S⁺:S^- = (E⁺+L⁺):(E^-+L^-) = E⁺:E^- + E⁺:L^- + L⁺:E^- + L⁺:L^- = E⁺:E^- + E⁺:L^- + L⁺:E^-.

 

[rcgldr]

Any attractive force between a charged volume within the lens and a charged volume outside the lens is countered by an equal in magnitude but opposing repulsive force by the opposing charge that occupies the same volume within the lens and that charged volume outside the lens.

Sure, but if you are going to treat the four regions (+ and - inside the lens, + and - outside the lens) as independent entities then the spheres will not remain as spheres.

I only meant the forces related to the volume within the lens is zero, not that it has to be solved that way. To solve the problem, you can calculate the force as if the spheres didn't overlap, then subtract a force to compensate for the volume within the lens, which is what the OP ended up doing, although he was able to calculate directly for potential energy, while I was concerned he would need to create functions for force versus separation and integrate those to determine potential energy.

update - That should have been "... which is not what the OP ended up doing ..."