- #51

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

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- #51

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

- #52

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

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

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. :)

- #53

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

- #54

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

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- #55

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

- #56

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Sorry.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."

Yes, I interpret it that way and I think my energy equation is consistent with that.Is that how you interpret the problem? Is that reflected in your equation for the conservation of energy?

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.

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- #57

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- #58

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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 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.

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

- #59

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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.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.

For clarity, let me invent some notation. S

- #60

rcgldr

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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.

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.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.

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- #61

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Well, it wasn't quite that simple. Treating them as though they don't overlap means taking them to be point masses. The trouble with a portion within the lens is not only that there is no net force between the +ve and -ve elements of the lens, but that there is a reduced force between those elements and the portions outside the lens. Pranav-arora quite correctly ran separate integrals for d-R to R and R to d+R. In the first, only the portion of the other sphere at radius < r contributes to the attraction, so it gets a factor (r/R)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,

Yes, the integrals could have dealt with forces instead of potentials. Not sure which is simpler algebraically.

- #62

rcgldr

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You're correct. This is how I initially thought it could be done, but the OP's method turned out to be bettter.Well, it wasn't quite that simple. Pranav-arora quite correctly ran separate integrals for d-R to R and R to d+R.

That was where I was getting it wrong. I kept thinking the volume within the lens would not contribute to any attractive force outside the lens, because I wasn't taking into account that the repulsive force is countered by that unknown structural force. You mentioned this earlier, but I didn't understand this until you mentioned it again.To maintain the integrity of the spheres, that repulsion must be countered by some unknown structural force.

and this was an intro physics class problem?

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- #63

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I am not sure how you classify the problems into "intro" and "advanced" but when I looked at the problem, I did not know I will have to do those integrals. Anyways, it was more of a Math problem than Physics. :Pand this was an intro physics class problem?

- #64

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However, it is very physical in the sense that it examines your knowledge of reference frames, energy conservation, and the relationships between kinetic energy and velocity, potential energy and force, force and acceleration. That was the important part, and I think you struggled with that more than with its mathematical content.

- #65

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Yes, you are right, I have myself noticed that I can do the Math and its the Physics part I have trouble figuring out.However, it is very physical in the sense that it examines your knowledge of reference frames, energy conservation, and the relationships between kinetic energy and velocity, potential energy and force, force and acceleration. That was the important part, and I think you struggled with that more than with its mathematical content.

- #66

rcgldr

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- #67

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I am not sure what x and z are. I do hope that Pranav can address that :)

- #68

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The potential over a sphere of radius r is constant, so I can take spherical caps and consider them as dq instead of selecting smaller point charges. Is this what you ask?

I found something: http://arxiv.org/pdf/0903.3304.pdf

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