What minimum speed do we need to give a charged ball?

[Lotto]

Homework Statement

A small charged ball is attached to a string of a lenght $l=10 \, \mathrm{cm}$. The ball have a charge of $Q=10\,\mathrm {\mu C}$. The rope is attached to a point with the same charge $Q$. What minimum speed do we need to give the ball for it to move in a full circle?

Solve for $m=150\,\mathrm g$ and $m=50\,\mathrm g$.

Relevant Equations

$m\frac {{v_2}^2 }{l}=T_2+mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}$
$\frac 12 m{v_2}^2+2mlg=\frac 12 m{v_1}^2$

First, if we sign the speed in the highest point of the ball's trajectory to be $v_2$, we can write

$m\frac {{v_2}^2 }{l}=T_2+mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}$.

Now, depending on the ball's particular mass, the electrostatic force can be bigger of smaller than its gravity force. So:

1) $m=150\,\mathrm g$: $mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}>0$
2) $m=50\,\mathrm g$: $mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}<0$.

In the first case, the gravity force is bigger, so (from the non-inertial point of view) there has to be a centrifugal force that ballances them. But the tension force can be zero, so then we can calculate the speed $v_2$ and by using $\frac 12 m{v_2}^2+2mlg=\frac 12 m{v_1}^2$ we can determine $v_1$.

In the second case, the gravity force is smaller than the electrostatic force, so there has to be a tension force so that the ball can move. But I think that the centrifugal force can be zero, we don't need it. So $v_2=0$ and by using $\frac 12 m{v_2}^2+2mlg=\frac 12 m{v_1}^2$ we can determine $v_1$ again.

Are my thoughts correct?

 

[kuruman]

Homework Statement: A small charged ball is attached to a string of a lenght $l=10\,mathrm {cm}$. The ball have a charge of $Q=10\,\mathrm {\mu C}$. The rope is attached to a point with the same charge $Q$. What minimum speed do we need to give the ball for it to move in a full circle?

Solve for $m=150\,\mathrm g$ and $m=50\,\mathrm g$.
Relevant Equations: #m\frac {{v_2}^2 }{l}=T_2+mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}$\frac 12 m{v_2}^2+2mlg=\frac 12 m{v_1}^2$

First, if we sign the speed in the highest point of the ball's trajectory to be $v_2$, we can write

$m\frac {{v_2}^2 }{l}=T_2+mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}$.

Now, depending on the ball's particular mass, the electrostatic force can be bigger of smaller than its gravity force. So:

1) $m=150\,\mathrm g$: $mg--\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}>0$
2) $m=50\,\mathrm g$: $mg--\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}<0$.

In the first case, the gravity force is bigger, so (from the non-inertial point of view) there has to be a centrifugal force that ballances them. But the tension force can be zero, so then we can calculate the speed $v_2$ and by using $\frac 12 m{v_2}^2+2mlg=\frac 12 m{v_1}^2$ we can determine $v_1$.

In the second case, the gravity force is smaller than the electrostatic force, so there has to be a tension force so that the ball can move. But I think that the centrifugal force can be zero, we don't need it. So $v_2=0$ and by using $\frac 12 m{v_2}^2+2mlg=\frac 12 m{v_1}^2$ we can determine $v_1$ again.

Are my thoughts correct?

Please fix the LaTeX to make you post legible.

 

[kuruman]

Thank you for cleaning up the LaTeX. It's much better now. My reading of the problem is that gravity is out of the picture. Imagine the balls on a frictionless horizontal table. They are separated by distance $l$. The masses are going around each other. What is the radius of the circle that each mass describes? Hint 1: if it is $l$, then one of the masses must be at the axis of rotation. Hint 2: One mass is 3 times the other mass.

1) $m=150\,\mathrm g$: $mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}>0$
2) $m=50\,\mathrm g$: $mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}<0$.

This is utterly confusing. You cannot use the same symbol for different masses.

 

[Lotto]

Thank you for cleaning up the LaTeX. It's much better now. My reading of the problem is that gravity is out of the picture. Imagine the balls on a frictionless horizontal table. They are separated by distance $l$. What is the radius of the circle that each mass describes? Hint 1: if it is $l$, then one of the masses must be at the axis of rotation. Hint 2: One mass is 3 times the other mass.

This is utterly confusing. You cannot use the same symbol for different masses.

I have propably described the problem poorly. The ball is moving in a vertical plane. Here is the picture that will describe it better:

Sorry for the confusion.

 

[kuruman]

Three questions:
1. Is it given that the mass at point S is fixed? If so, why do we need to know what this mass is?
2. Which mass is at S and which mass is at A?
3. Why are points B, C and D relevant? Is there more to the problem that you did not mention?

Thanks.

 

[Lotto]

Three questions:
1. Is it given that the mass at point S is fixed? If so, why do we need to know what this mass is?
2. Which mass is at S and which mass is at A?
3. Why are points B, C and D relevant? Is there more to the problem that you did not mention?

Thanks.

The point S has no mass and is fixed, the only thing that has a mass is that big black point - that is the ball. We know its mass. In the fist case, it is 150 g, in the second case, it is 50 g. So we should solve the problem for two different masses.

The points A and C correspond to my points 1 and 2. The speed $v_0$ corresponds to my $v_1$.

The other points are relevant in the second problem, where I am to calculate tension forces acting in them. But that would be trivial after solving this problem.

 

[kuruman]

OK, I see now. Thanks for the clarification. I thought there were two balls each of charge Q. I modified my messages in order not to confuse anyone else.
How do you know that (1) is positive and (2) is negative? Did you put in the numbers?
1) $m=150\,\mathrm g$: $mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}>0$
2) $m=50\,\mathrm g$: $mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}<0$.

I get that both are negative, but I am not always right so please check the numbers.

This equation
$m\frac {{v_2}^2 }{l}=T_2+mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}~$ is good. You are looking for the minimum speed $v_0$ if the mass is to go around the circle. How can this equation help you?

 

[Lotto]

OK, I see now. Thanks for the clarification. I thought there were two balls each of charge Q. I modified my messages in order not to confuse anyone else.
How do you know that (1) is positive and (2) is negative? Did you put in the numbers?
1) $m=150\,\mathrm g$: $mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}>0$
2) $m=50\,\mathrm g$: $mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}<0$.

I get that both are negative, but I am not always right so please check the numbers.

This equation
$m\frac {{v_2}^2 }{l}=T_2+mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}~$ is good. You are looking for the minimum speed $v_0$ if the mass is to go around the circle. How can this equation help you?

For 150 g, I get circa 0,57 N and for 50 g I get -0,41 N.

I as wrote above, if the gravity force is bigger that the electrostatic force, then we can let the tension force to be zero. So from that equation I can calculate $v_2$ and then by using the law of conservation of energy I can determine $v_1$ ($v_0$).

 

[kuruman]

With
$k=9\times 10^9~(\text{N}\cdot \text{m/C})^2$, $Q=10~\mu\rm{C}=10^{-5}~\rm{C}$ and $l=10~\rm{cm}=10^{-1}~\rm{m}$,
the electric force is $$F_e=\frac{9\times 10^9\times (10^{-5})^2}{(10^{-1})^2}=\frac{9\times 10^9\times 10^{-10}}{10^{-2}}=90~\rm{N}.$$ The weights are
$m_1g=50\times 10^{-3}~(\rm{kg})\times 10~\rm{m/s}^2=0.5~\rm{N}$
$m_2g=3m_1g=1.5~\rm{N}$.

Both weights are less than the electric force so it appears that the tension will not go to zero even when the speed is zero.

 

[Lotto]

With
$k=9\times 10^9~(\text{N}\cdot \text{m/C})^2$, $Q=10~\mu\rm{C}=10^{-5}~\rm{C}$ and $l=10~\rm{cm}=10^{-1}~\rm{m}$,
the electric force is $$F_e=\frac{9\times 10^9\times (10^{-5})^2}{(10^{-1})^2}=\frac{9\times 10^9\times 10^{-10}}{10^{-2}}=90~\rm{N}.$$ The weights are
$m_1g=50\times 10^{-3}~(\rm{kg})\times 10~\rm{m/s}^2=0.5~\rm{N}$
$m_2g=3m_1g=1.5~\rm{N}$.

Both weights are less than the electric force so it appears that the tension will not go to zero even when the speed is zero. Your numbers are consistent with forgetting to square the denominator.

Ou... I am sorry again, the charge is $1.0 \,\mathrm{\mu C}$, not $10$. I am somehow scatterbrained today. But I calculated with the right value.

 

[kuruman]

Ou... I am sorry again, the charge is $1.0 \,\mathrm{\mu C}$, not $10$. I am somehow scatterbrained today. But I calculated with the right value.

That's OK. Can you finish the problem now? I suggest that for each mass you find an expression for $v_{\text{min}}$ and substitute the mass values at the very end. It will be easier for us to check your work.

 

[Lotto]

That's OK. Can you finish the problem now? I suggest that for each mass you find an expression for $v_{\text{min}}$ and substitute the mass values at the very end. It will be easier for us to check your work.

So, for 150 g:

$m\frac {{v_2}^2 }{l}=mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}$,

so ${v_2}^2=gl-\frac{1}{4\pi \epsilon}\frac{Q^2}{m l}$.

$\frac 12 m{v_2}^2+2mlg=\frac 12 m{v_1}^2$,

so

$v_1=\sqrt{5gl-\frac{1}{4\pi \epsilon}\frac{Q^2}{m l}}=2.1\,\mathrm {m\cdot s^{-1}}$.

For 50 g:

$0=T_2+mg-\frac{1}{4\pi \epsilon}\frac{Q^2}{l^2}$,

but $T_2$ is not important. So

$2mlg=\frac 12 m{v_1}^2$
$v_1=2\sqrt{lg}=2.0\,\mathrm {m\cdot s^{-1}}$.

Is it correct?

 

[kuruman]

Is it correct?

I didn't run the numbers, but the equations agree with mine. Good job.