Path of the particle on inclined plane

[Vibhor]

I understand what you are saying but , direction of $\vec{f}$ has to be in opposite direction to that of the resultant vector of $\vec{T} + \vec{W}$ . I need to first make $\vec{W}$ and $\vec{T}$ and then draw $\vec{f}$ such that it acts opposite to the resultant of the other two vectors .

If f = W , then forces cannot add to zero in second quadrant .

 

[Nidum]

Interesting to note that since it is the resultant of the two driving forces that is trying to overcome friction enough to cause motion and that the resultant has different values according to where the particle is on the plane then the particle could depending on location be stopped , accelerating or moving at constant velocity .

To establish slow stable motion in all locations the source of the string tension would have to have intelligence and vary the tension value according to particle location .

 

[TSny]

If f = W , then forces cannot add to zero in second quadrant .

There is one situation where they can add to zero. What would T need to be?

 

[Vibhor]

There is one situation where they can add to zero. What would T need to be?

T needs to be zero .

 

[TSny]

Right. So, if the slightest tension is applied to the string, what will happen?

 

[Vibhor]

The particle will move radially inwards towards the hole .

 

[TSny]

The particle will move radially inwards towards the hole .

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?

 

[Vibhor]

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

The particle starts its motion in first or second quadrant ??

 

[TSny]

Sorry again! We're talking second quadrant!

 

[Vibhor]

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?

In +y direction .

If you know the direction of the kinetic friction, what can you say about the direction of the motion?

In -y direction i.e down the slope .

 

[TSny]

Yes.

 

[Vibhor]

But it cannot move down the slope , as the string holds it up . It has to move in +x direction as well . Is it making circular motion in second quadrant ??

 

[TSny]

If I'm working it correctly, it does not move in a circle in the second quadrant. The string does not need to hold it up. It can slide down the slope while you take up slack in the string with T = 0.

 

[Vibhor]

In the question it states " The string is pulled so slowly " . Doesn't it mean the length of the string on the plane decreases continuously ( distance between hole and particle decreases ) ? If it is allowed to slide down the slope exclusively in -y direction , it is possible only if length of the string from the hole to the particle increases

 

[TSny]

What happens to distance between the particle and the hole as it slides down in the second quadrant?

 

[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.]