Path of the particle on inclined plane

[Vibhor]

The important thing is to find the direction of the friction force just before slipping.

If particle sits at rest in third quadrant i.e when T = 0 , then friction acts up the slope .

If $\phi$ is the angle that the string makes to the (negative) y-axis, try to find the angle between the string and the friction force in terms of $\phi$.

The angle would be $\phi$ .

Also think about which way the particle is going to move when it slips.

The particle will try to move at an angle $180° - 2\phi$ with the -y axis measured CW .

 

[TSny]

OK, your directions look good. But it seems to me that the arrows you added should be drawn at the particle rather than the hole.
[EDIT: I noticed that you did not draw the direction of $f$ very accurately. Your value of $180^o - 2\phi$ is correct. But if you draw this angle more accurately, I think you will see that $f$ points in a different direction.]

So I think you have the correct angle for the direction of motion of the particle. Let $r$ be the length of the string between the particle and the hole. You want to find the path of the particle. In polar coordinates, this would be $r$ as a function $\phi$.

Suppose the particle undergoes a small displacement $ds$. Can you find expressions for $dr$ and $d\phi$ in terms of $ds$, $r$, and $\phi$?

 

[Vibhor]

Let $r$ be the length of the string between the particle and the hole. You want to find the path of the particle. In polar coordinates, this would be $r$ as a function $\phi$.

Suppose the particle undergoes a small displacement $ds$. Can you find expressions for $dr$ and $d\phi$ in terms of $ds$, $r$, and $\phi$?

$d\vec{s} = dr\hat{r} + rd\phi \hat{\phi}$

 

[TSny]

$d\vec{s} = dr\hat{r} + rd\phi \hat{\phi}$

What if you project this equation onto the direction $\hat{r}$? onto the direction $\hat{\phi}$?

This will give you two equations to work with.

 

[Vibhor]

What if you project this equation onto the direction $\hat{r}$? onto the direction $\hat{\phi}$?

You mean I need to take dot product of $d\vec{s}$ with $\hat{r}$ and with $\hat{\phi}$ separately ?

 

[TSny]

Yes.

 

[Vibhor]

$d\vec{s} \cdot \hat{r}= dr$

$d\vec{s} \cdot \hat{\phi}= rd\phi$

 

[TSny]

Yes, but the left hand sides can be expressed explicitly in terms of the magnitude of the vector $\vec{ds}$ and the angle between $\vec{ds}$ and the unit vectors.

Thus, you will need to know the angle between $\vec{ds}$ and $\hat{r}$. Use your diagram (drawn carefully).

 

[Vibhor]

The angle between $\vec{ds}$ and $\hat{r}$ is $180° - \phi$ and that between $\vec{ds}$ and $\hat{\phi}$ is $270° - \phi$ .

Is that right ?

 

[TSny]

The angle between $\vec{ds}$ and $\hat{r}$ is $180° - \phi$ and that between $\vec{ds}$ and $\hat{\phi}$ is $270° - \phi$ .

Is that right ?

No. Can you show a diagram with the forces acting on the particle and also show the vector $\vec{ds}$?

Note that in the figure, I made $\phi$ roughly 30^o. So, what is the approximate value of $180 - 2\phi$ ? Make sure you draw $f$ approximately in this direction. Then draw $\vec{ds}$ in the appropriate direction.

[EDIT: Sorry, Vibhor, your values for the angles are correct! When I worked through this part of the problem, I did not use the unit vectors $\hat{r}$ and $\hat{\phi}$. When I read your last thread, I had the vectors pictured in my head in their opposite directions.]

 

[TSny]

So, since you got the angles correct, your diagram hopefully looks something like this:

 

[TSny]

So, using your expressions for the angles between $\vec{ds}$ and the unit vectors, try expressing the left sides of the two equations below in terms of the magnitude of the displacement $ds$ and the angle $\phi$.

$\vec{ds} \cdot \hat{r} = dr$
$\vec{ds} \cdot \hat{\phi} = rd\phi$

 

[Vibhor]

[EDIT: Sorry, Vibhor, your values for the angles are correct! When I worked through this part of the problem, I did not use the unit vectors $\hat{r}$ and $\hat{\phi}$. When I read your last thread, I had the vectors pictured in my head in their opposite directions.]

Ok

I got really discouraged reading your initial reply in #70 . But now I feel better

 

[Vibhor]

$-dscos\phi = dr$

$-dssin\phi = rd\phi$

From this I get $\frac{dr}{r} = \frac{d\phi}{tan\phi}$

 

[TSny]

$-dscos\phi = dr$

$-dssin\phi = rd\phi$

From this I get $\frac{dr}{r} = \frac{d\phi}{tan\phi}$

Great! The first two equations above can pretty much be read directly off the diagram. But using the unit vectors is good, too.

 

[Vibhor]

Integrating I get $r=Csin\phi$ , where $C$ is a constant .

 

[TSny]

Yep.

 

[Vibhor]

Does that mean distance of the particle from the hole is oscillating (alternately increasing and decreasing ) like a sine curve ?

 

[TSny]

I don't think so.
Perhaps the curve will be more recognizable when expressed in terms of the Cartesian coordinates of the particle (x, y).

 

[Vibhor]

Perhaps the curve will be more recognizable when expressed in terms of the Cartesian coordinates of the particle (x, y).

Should I replace r =$\sqrt{x^2+y^2}$ and $\sin\phi = \frac{x}{r}$ to express curve in cartesian coordinates ?

 

[TSny]

Almost. There's a sign error. (Of course you now know not to believe anything I say, but I really do think there is a sign error.)

 

[Vibhor]

r =$\sqrt{x^2+y^2}$ and $\sin\phi = -\frac{x}{r}$

 

[TSny]

Yes.

 

[Vibhor]

Ok .

This gives $\left( x + \frac{c}{2} \right)^2 + y^2 = (\frac{c}{2})^2$ . This is a circle centered at $(-\frac{c}{2} , 0)$ and radius $\frac{c}{2}$ .

 

[TSny]

I believe that's correct.

 

[Vibhor]

Ok .

In polar coordinates we measure angles anti clockwise from x - axis but in this problem we measured it clockwise from -y axis .Doesn't this make a difference ?

It is a very naive question but I am not too familiar with polar coordinates .

 

[TSny]

It doesn't make any difference as long as you make your definitions clear.

Does the particle end up where you expect it would?

 

[Vibhor]

It doesn't make any difference as long as you make your definitions clear.

Ok . Good point .

Does the particle end up where you expect it would?

I don't know . Even though I have got the equation of the curve , I can't imagine how the particle is actually moving (except it is moving in a circular fashion not centered at the hole) . I am still not sure how to determine constant $C$ .

 

[TSny]

Can you relate C to the place where the particle crossed the x-axis in going from quadrant II to quadrant III?

 

[Vibhor]

Can you relate C to the place where the particle crossed the x-axis in going from quadrant II to quadrant III?

But that depends on where the particle was initially ( at rest ) in the second quadrant . As it moves vertically down slope in the second quadrant .