I just don't understand this Spring - ElectroStatic problem

[flyingpig]

Homework Statement

[PLAIN]http://img832.imageshack.us/img832/17/unledqp.png

Above is the problem and the solutions

The Attempt at a Solution

Now I really have problems with how the factors of the radius is in there. I did it differently

This is how I did it

$$\sum W_{ab} = \Delta K$$

$$\int \vec{F_{e}}\cdot \vec{ds} + \int \vec{F_{s}} \cdot \vec{ds} = \Delta K$$

$$\int \frac{-k_{e}qq}{r^2}\cdot \vec{dr} + \int -k_{s}\vec{x} \cdot \vec{dx} = \Delta K$$

Here is where it got difficult. I have three questions first of all I have

$$\int_{L+l}^{L} \frac{-k_{e}qq}{r^2}\cdot \vec{dr} + \int_{x=l}^{x=0} -k_{s}\vec{x} \cdot \vec{dx} = \Delta K$$

Now I know that it is wrong (because the solution is right...), but I just want to know (for future references) for the limits of integration on the springs, is it right to say x= 0 when I want to say that the object comes back to the equilibrium point? Or should do this instead?

$$\int_{L+l}^{L} \frac{-k_{e}qq}{r^2}\cdot \vec{dr} + \int_{x=l}^{x=L} -k_{s}\vec{x} \cdot \vec{dx} = \Delta K$$

Where it comes back to the distance L, the natural length of the spring? These plus and minus signs are killing me in physics...

Now for the final question. Just why do they have the radius in there? It messes things up!

The spring is attached to the outside of the sphere. In other words

[PLAIN]http://img508.imageshack.us/img508/2757/unledhvs.png

Is it attached to the centre of the sphere??

 

[jhae2.718]

Now for the final question. Just why do they have the radius in there? It messes things up!

Charged spheres act like point charges at the center of the sphere. L+l+R₁+R₂ is the center to center distance at the initial state. That's in the expression for the electric potential energy, not the spring energy, so how the spring is attached is irrelevant.

The limits on the integral should be from L+l+R₁+R₂ to L+R₁+R₂ for the EPE, and from l to 0 for the spring PE.

 

[flyingpig]

Also, I used $$-kx$$ instead of $$kx$$, the solution used $$kx$$, why? My calculus textbook also used $$kx$$ but I assumed it is wrong because the writers are mathematicans (no offense intended...)

 

[flyingpig]

Charged spheres act like point charges at the center of the sphere. That's in the expression for the electric potential energy, not the spring energy, so how the spring is attached is irrelevant.

Oh wow that is just tricky lol

But could you answer the question about the bounds of integration for the spring? We nevre covered SHM in mechanics

 

[jhae2.718]

I edited my first post with the bounds. I was originally mistaken for the spring, your limits look correct. (The x is the distance from the equilibrium point, after all...I forgot that originally.)

 

[jhae2.718]

Also, I used $$-kx$$ instead of $$kx$$, the solution used $$kx$$, why? My calculus textbook also used $$kx$$ but I assumed it is wrong because the writers are mathematicans (no offense intended...)

The minus indicates that the direction of the force is opposite the displacement from the equilibrium position. I find it easier to ignore the minus and determine the sign based on what direction the force should act.

 

[flyingpig]

I edited my first post with the bounds. I was originally mistaken for the spring, your limits look correct. (The x is the distance from the equilibrium point, after all...I forgot that originally.)

What about having x = L instead of x = 0? What does x = L mean then?

Also, what if it was somewhere "behind" the equilibrium point? My intuition is that it would be

$$\int_{x = l}^{x = -d_1} kx \cdot \vec{dx}$$

Noticed +kx

[PLAIN]http://img844.imageshack.us/img844/103/unleddw.png

 

[jhae2.718]

What about having x = L instead of x = 0? What does x = L mean then?

Also, what if it was somewhere "behind" the equilibrium point? My intuition is that it would be

$$\int_{x = l}^{x = -d_1} kx \cdot \vec{dx}$$

Noticed +kx

...img...

Actually, you were right originally, if we take positive x to the right. Then the net force is $F_{net}=-\frac{kqq}{r^2}-kx$, and using the Work-Energy Theorem, the integral of these forces is equal to $\Delta K$.

For your question here, using the same coordinate system the kx term would be positive (i.e. to the right) and then you would do the integral accordingly.

 

[jhae2.718]

E.g.:
$$\begin{align*} \Delta K &= -\int\limits_{L+l+R_1+R_2}^{L+R_1+R_2} \frac{k_eq^2}{r^2}dr-\int_{l}^0 kxdx \\ &= \left[\frac{k_eq^2}{r}\right]_{L+l+R_1+R_2}^{L+R_1+R_2} - \left[\frac{kx^2}{2}\right]_{l}^0\\ \frac{mv^2}{2} &= \frac{k_eq^2}{L+R_1+R_2}-\frac{k_eq^2}{L+l+R_1+R_2}-\left[0-\frac{kl^2}{2}\right] \\ \frac{mv^2}{2} &= \frac{k_eq^2}{L+R_1+R_2}-\frac{k_eq^2}{L+l+R_1+R_2}+\frac{kl^2}{2}\\ v^2 &= \frac{2}{m}\left(\frac{k_eq^2}{L+R_1+R_2}-\frac{k_eq^2}{L+l+R_1+R_2}+\frac{kl^2}{2}\right)\\ v &= \sqrt{\frac{2}{m}\left(\frac{k_eq^2}{L+R_1+R_2}-\frac{k_eq^2}{L+l+R_1+R_2}+\frac{kl^2}{2}\right)} \end{align*}$$

 

[flyingpig]

Why would you take x = -l?

 

[jhae2.718]

Just the way I defined the coordinate system for the spring. (The sign of l doesn't affect anything here.)

Edit: Actually I'm wrong here. I'm leaving it for the thread record, but see my post below.

 

[flyingpig]

Was I right about x = -d₁

 

[jhae2.718]

Assuming you're taking x to the right as positive and the equilibrium point as x=0, yes.

 

[jhae2.718]

Why would you take x = -l?

Actually, you caught me here. It should be positive l. (See how easy it is to make sign errors? I'm kind of busy, so if I do something stupid like that point it out!)

 

[flyingpig]

Actually, you caught me here. It should be positive l. (See how easy it is to make sign errors? I'm kind of busy, so if I do something stupid like that point it out!)

When I said x=d₁, did you noticed that I said it was +kx not - kx...?

 

[flyingpig]

Sammy are you reading this?

 

[SammyS]

Sammy are you reading this?

I am now, but after all of this dialogue, I think I will have to word any reply very carefully so you know just which aspect of the question is being addressed.

 

[SammyS]

...

The Attempt at a Solution

... I did it differently.
This is how I did it:

$$\sum W_{ab} = \Delta K$$

$$\int \vec{F_{e}}\cdot \vec{ds} + \int \vec{F_{s}} \cdot \vec{ds} = \Delta K$$

$$\int \frac{-k_{e}qq}{r^2}\cdot \vec{dr} + \int -k_{s}\vec{x} \cdot \vec{dx} = \Delta K$$

Here is where it got difficult. I have three questions first of all I have

$$\int_{L+l}^{L} \frac{-k_{e}qq}{r^2}\cdot \vec{dr} + \int_{x=l}^{x=0} -k_{s}\vec{x} \cdot \vec{dx} = \Delta K$$

Now I know that it is wrong (because the solution is right...), but I just want to know (for future references) for the limits of integration on the springs, is it right to say x= 0 when I want to say that the object comes back to the equilibrium point?

First comment: It's very important to define your variables clearly.

The form of the electrostatic force tells me that the r vector goes from the center of the large sphere to the center of the small sphere.

It appears to me that x represents the amount the spring is stretched from its equilibrium length.​

Once you set things up you can incorporate the effect of the dot product into your expressions, and drop the vector symbols, so r is the magnitude of the r vector.

Your expression should then be:

$$\int_{L+l+R_1+R_2}^{L+R_1+R_2} \frac{-k_{e}qq}{r^2}\ dr\ +\ \int_{x=l}^{x=0} -k_{s}x \ dx = \Delta K\ .$$

...Or should I do this instead?

$$\int_{L+l}^{L} \frac{-k_{e}qq}{r^2}\cdot \vec{dr} + \int_{x=l}^{x=L} -k_{s}\vec{x} \cdot \vec{dx} = \Delta K\$$

Where it comes back to the distance L, the natural length of the spring? These plus and minus signs are killing me in physics...

If you do it this way, you're saying that x represents the total length of the spring. It's equilibrium length is L, so the force exerted by the spring is -k_s(x-L). Your equation would then be:

$$\int_{L+l+R_1+R_2}^{L+R_1+R_2} \frac{-k_{e}qq}{r^2}\,dr\ +\ \int_{x=L+l}^{x=L} -k_{s}(x-L) \,dx = \Delta K\ .$$

Either way, ΔK = (1/2)mv².

...

[PLAIN]http://img508.imageshack.us/img508/2757/unledhvs.png

 

[flyingpig]

Sammy, did I get the -d₁ right as well? I will read detaily of your post once I finish supper (which I just spilled as I am typing this...)

 

[SammyS]

Sammy, did I get the -d₁ right as well? I will read details of your post once I finish supper (which I just spilled as I am typing this...)

If you mean in the following:

$$\int_{x = l}^{x = -d_1} kx \cdot \vec{dx}$$

then yes, you you used d₁ correctly. However, as noted elsewhere, there should be a negative sign on the spring force.

$$\int_{x = l}^{x = -d_1}{-kx}\ dx$$

 

[flyingpig]

First comment: It's very important to define your variables clearly.

The form of the electrostatic force tells me that the r vector goes from the center of the large sphere to the center of the small sphere.

It appears to me that x represents the amount the spring is stretched from its equilibrium length.​

Once you set things up you can incorporate the effect of the dot product into your expressions, and drop the vector symbols, so r is the magnitude of the r vector.

Your expression should then be:

$$\int_{L+l+R_1+R_2}^{L+R_1+R_2} \frac{-k_{e}qq}{r^2}\ dr\ +\ \int_{x=l}^{x=0} -k_{s}x \ dx = \Delta K\ .$$

If you do it this way, you're saying that x represents the total length of the spring. It's equilibrium length is L, so the force exerted by the spring is -k_s(x-L). Your equation would then be:

$$\int_{L+l+R_1+R_2}^{L+R_1+R_2} \frac{-k_{e}qq}{r^2}\,dr\ +\ \int_{x=L+l}^{x=L} -k_{s}(x-L) \,dx = \Delta K\ .$$

Either way, ΔK = (1/2)mv².

Oh okay, I understand now! Thanks!

 

[sagardip]

To flyingpig.
What the solution requires is the use of Conservation of energy which states that if no external force acts on the given system then the total energy of the system remains the same. Here by total energy we mean the kinetic and the potential energies of the system.
Now in case of a spring the potential energy is always positive irrespective of the fact whether it is compressed or stretched.Can you figure this out?You will understand the mistake in your solution if you understand this.

 

[sagardip]

Potential energy is the work done by an external agent in moving a body slowly from the point of zero potential to the given point.Lets consider the two cases of compression and extension of a spring.
In case of compression the spring exerts an outward force on the body so as to push it towards the equilibrium point.Now in order to compress the spring slowly an external agent has to apply a force equal to the spring force at any instant but in opposite direction that is in the direction of the displacement of the body.so the potential energy stored in the spring after compressing it through L metres would be equal to the work done by force Kx from the equilibrium point to the final point i.e L.Here while integrating the expression for work done we take Kx dx and not -Kx dx because it is not the work done by the spring force but the work done by the external agent in moving the body slowly from the initial to the final position and hence the direction of the force is along that of the displacement of the body. Hence we get (0.5)KL^2 as the potential energy of the body in the compressed state of the spring.

Similarly in case of extention the force due to spring is opposite to the direction of displacement of the body and hence the force applied by the external agent is along the direction of motion and so dW= Kx dx and not -Kx dx.

 

[sagardip]

and the limits are from 0 to L because the x in the integration is just the extention or the compression in the spring.Here x=0 represents the equilibrium position where no external force is required to keep the body in a state of rest.

 

[flyingpig]

I justaI aws doing this problem again and I found some problems, different ones.

1. How come for the work done by electric force's distance is $$r = R_1 + R_2 + L$$, I thought the distance, once it hits x = 0 (equilibrium point), would be $$r = R_1 + L$$