Path of the particle on inclined plane

[Vibhor]

Sorry.I was thinking about third quadrant . So the particle moves in a straight line with constant speed down the slope in the second quadrant ??

 

[TSny]

I think so. The speed can be as slow as you want.

 

[Vibhor]

Ok . Now when the particle enters third quadrant , tension is non zero ,which means the forces cannot add up to zero .Or can they ?

 

[TSny]

Start with the particle sitting at rest at some point in the third quadrant with T = 0. Think about what happens as you slowly add tension to the string. Can you still have static equilibrium conditions for a finite amount of tension?

 

[Vibhor]

Start with the particle sitting at rest at some point in the third quadrant with T = 0. Think about what happens as you slowly add tension to the string. Can you still have static equilibrium conditions for a finite amount of tension?

As tension increases , the particle moves in +x as well as -y direction . Not sure whether we can definitely say it is a circular motion . For a finite amount of tension , the particle cannot be in static equilibrium.

 

[TSny]

As tension is applied, the friction force can change direction in an attempt to keep the particle in equilibrium.

 

[Vibhor]

Sure . But its magnitude will always be less than the resultant of the other two forces . The particle cannot be in static equilibrium .

 

[TSny]

Sure . But its magnitude will always be less than the resultant of the other two forces . The particle cannot be in static equilibrium .

I don't think that's true. It's easy to draw a force diagram for an arbitrary location in the 3rd quadrant where the three forces give equilibrium and T ≠ 0.

Actually, I think your equations in #15 [EDIT: #22] pretty much correspond to this situation. However, I think that it might be more convenient to let $\phi$ be the angle between the string and the negative y-axis for this quadrant.

 

[Vibhor]

I don't think that's true. It's easy to draw a force diagram for an arbitrary location in the 3rd quadrant where the three forces give equilibrium and T ≠ 0.

.You are right. Can I say path traversed would be circular .

Actually, I think your equations in #15 [EDIT: #22] pretty much correspond to this situation. However, I think that it might be more convenient to let $\phi$ be the angle between the string and the negative y-axis for this quadrant.

How can I find the path equation ? Equations in #22 do not yield anything useful.

 

[TSny]

.You are right. Can I say path traversed would be circular .

You'll need to find out.

How can I find the path equation ? Equations in #22 do not yield anything useful.

OK, you don't really need these equations. Just the picture you used to set up the equations. The important thing is to find the direction of the friction force just before slipping. 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 $\phi$. Also think about which way the particle is going to move when it slips.

Edit: For clarity, I added a picture below to show $\phi$.

 

[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 .

 

[TSny]

Exactly. You have to start the particle somewhere. So, C is determined by the initial conditions.

 

[Vibhor]

Back to second quadrant .

In the first second quadrant an infinitesimal amount of tension will set the particle in motion. If the tension is kept essentially at zero as the particle moves at a slow constant speed in the first second quadrant, what direction does the kinetic friction need to act so that the net force is zero? If you know the direction of the kinetic friction, what can you say about the direction of the motion?

I am still bemused by the fact that even if the string is pulled at some angle to the +y direction ( even though it is a very small magnitude Tension force ) at t=0 when the particle was at rest in second quadrant , the particle instead of going radially inwards towards the hole starts moving down slope .

Please explain it a bit more .

 

[TSny]

Back to second quadrant .
I am still bemused by the fact that even if the string is pulled at some angle to the +y direction ( even though it is a very small magnitude Tension force ) at t=0 when the particle was at rest in second quadrant , the particle instead of going radially inwards towards the hole starts moving down slope .

Please explain it a bit more .

I had to think about this, too. I attempted an explanation in post #42. I'll let you think about it some more. I'm shutting down for the night. Good work!

 

[haruspex]

Ok . Good point .
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 $C$ .

With a bit of geometry, you can get the curve fairly easily from TSny's excellent diagram in post #71.
The string makes the same angle to the 'vertical' through the hole as it does to the direction of travel of the mass. If you take the normal to the direction of travel, at the particle, and a horizontal line through the hole, these intersect at a point O. The string forms a chord of the circle centred at O and passing through the hole and the mass. The direction of travel is a tangent. So as the mass moves a short distance, its distance from O does not change. Having moved that distance, all of the above statements still apply, and with the same point O as centre.

Thanks TSny for taking over on this thread. I only have net access briefly each day at the moment (cycling Berlin to Copenhagen), and I'm sure I would not have found such a clear route to the answer.

 

[TSny]

With a bit of geometry, you can get the curve fairly easily ...

Nice.

(cycling Berlin to Copenhagen)

Wow . Enjoy and be safe.

 

[Vibhor]

I had to think about this, too. I attempted an explanation in post #42. I'll let you think about it some more.

Couple of thoughts .

a) As soon as tension is applied on the particle the ,the direction of friction changes ( from up slope ) , but the direction of net force on the particle has to be downward such that the particle starts moving downward .

b) As soon as tension exists in the string , the equilibrium of particle is disturbed which causes it to move down slope . But why down slope ?

 

[haruspex]

b) As soon as tension exists in the string , the equilibrium of particle is disturbed which causes it to move down slope . But why down slope ?

I assume you are referring to the case where the mass starts at a level above the hole. For any actual nonzero tension in the string, the motion of the mass will have a horizontal component towards the hole. But in the limit, as the tension tends to zero, that component tends to zero.

 

[Vibhor]

And why not vertical component also tend to zero and particle stays at rest instead of moving down slope ?

 

[Vibhor]

In the first second quadrant an infinitesimal amount of tension will set the particle in motion.

Could you please explain how does tension (acting at an angle) causes the particle to move down slope . I have thought about it but I am still unsure .

Thanks

 

[haruspex]

And why not vertical component also tend to zero and particle stays at rest instead of moving down slope ?

That can be answered by going back to the equation that allowed for arbitrary tension and acceleration you had before TSny joined the thread and generalising it to the case where the string is at some angle to the horizontal. You can take the limit as acceleration tends to zero and see that (where the particle is further up the plane than the hole) the path tends to the 'vertical'.
I don't know if there is another way to demonstrate it convincingly.