How to find equation of motion for object leaving a curved ramp?

[Tanya Sharma]

I can't get this latex work

Right click on the equation in post#32 and view Tex Commands . You need to have $$ or ### before and after the code.

But wait a minute, did I put the friction force in the right sense? Shouldn't it be pointing toward the opposite direction of the tendency of movement?

Yes , friction opposes the motion .This is why we have a negative sign in front of frictional force T .

Or do we assume that the object moves only counterclockwise?

We aren't assuming . Isn't the object moving counterclockwise in the picture you have attached ?

Why haven't you given the exact problem statement as given to you in post#1 ? What are the given parameters in the problem ?

 

[cantleave]

Yes, it does move counterclockwise. So what do we do next?

 

[cantleave]

Why haven't you given the exact problem statement as given to you in post#1 ? What are the given parameters in the problem ?

I wasn't given exact values, it was just stated that those are known parameters, could be any number.

 

[verty]

Well it seems part 1 is more difficult than I even thought, it has no elementary solution. To give a formula will need unsolvable integrals or the Jacobi amplitude function.

 

[cantleave]

I understand. What about part 2,3 and 4? How do I proceed further?

 

[Tanya Sharma]

What is the general expression of acceleration in polar coordinates ?

Have a look at https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=dy&chap_sec=01.6&page=theory .

 

[cantleave]

a = (r'' - rΘ'$^{2}$)$\hat{r}$ + (2r'Θ' + rΘ'')$\hat{Θ}$

 

[Tanya Sharma]

Yes .

$$ \vec{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\theta} $$

Do you understand what these quantities represent ?

Can you calculate $\dot{r}$ and $\ddot{r}$ ?

 

[cantleave]

$$ \dot{r} $$ is the first derivative of the position function with respect to time = velocity, and $$ \ddot{r} $$ is the second derivative of the position = acceleration.

I suppose the thetas and funky stuff comes from the fact that we convert from Cartesian to polar coordinates. I'm familiar with calculus, I can calculate $$ \dot{r} $$ and $$ \ddot{r} $$.

 

[Tanya Sharma]

In the equation I have written in post#43 $r$ is not a vector .It is the distance of object from the origin ( center of the circle ) .

What is the value of $r$ when the object is moving along the curve ? Does $r$ depend on $\theta$ ?

 

[cantleave]

I thought r was the position vector that starts from the center of the circle.

r = 2pi/θ? or my other guess would be

the integral of velocity?

 

[Tanya Sharma]

I thought r was the position vector that starts from the center of the circle.

$\vec{r}$ is position vector whereas $r$ is the distance of the object from the origin .

r = 2pi/θ? or my other guess would be

the integral of velocity?

Please explain this .What is the reasoning behind this expression ?

 

[cantleave]

The distance traveled by the object on the path is the full length of the circle divided by the angle θ. As the object moves, θ becomes greater thus the formula r = 2piR/θ (I forgot the R, the radius of the circle)

By distance did you mean the shortest way to the object from the origin which is a straight line? In that case I'd use the law of cosines.

 

[Tanya Sharma]

The distanced traveled by the object on the path is the full length of the circle divided by the angle θ. As the object moves, θ becomes greater thus the formula r = 2piR/θ (I forgot the R, the radius of the circle)

Nonsense . Can you quote a reference ?

By distance did you mean the shortest way to the object from the origin which is a straight line? In that case I'd use the law of cosines.

I meant magnitude of the displacement of the object from the origin i.e shortest distance . Does this shortest distance depend on θ ?

Please note that origin is at the center of the circle.

 

[cantleave]

Ok. I think got it now.

r doesn't depend on Θ. r is a constant

 

[Tanya Sharma]

r doesn't depend on Θ. r is a constant

So what is $\dot{r}$ and $\ddot{r}$ ?

 

[cantleave]

$\dot{r}$ = r$\hat{r}$' = $r\dot{Θ}$$\hat{Θ}$
$\ddot{r}$= r($\dot{Θ}$$\hat{Θ}$)' = r($\ddot{\theta}$$\hat{\theta}$-$\dot{\theta}^{2}$$\hat{r}$)

Could you tell me please in a few words what are we trying to achieve? I mean, what is our strategy?

 

[Redbelly98]

I'll leave the helping to others who are already doing so, but I will just point out that I do not see an actual statement of the problem anywhere. As in, a word-for-word restatement, exactly as the problem was written out, posted here.

 

[Tanya Sharma]

I guess we have put in far more effort than required .

Part 1) is asking for equation of motion for AB .

For that we need to have two equations

1. ∑F_n = ma_n
2. ∑F_t = ma_t

Here ∑F_n is net force along normal direction (towards the center )
a_n is normal acceleration
∑F_t is net force along tangential direction
a_t is tangential acceleration

From post #32 :

$$N-mgcosθ = ma_n$$
$$-mgsinθ-\mu N = ma_t$$

The above two equations are your answer to part 1) .

Now ,For part 3) you need to solve the two equations of motions by putting $a_n = \frac{v^2}{r}$ and $a_t = \frac{dv}{dt}$ .This will give you a DE which I think is easy to solve .

 

[ehild]

Well it seems part 1 is more difficult than I even thought, it has no elementary solution. To give a formula will need unsolvable integrals or the Jacobi amplitude function.

Yes, part 1 is difficult enough, as N depends both on the position and speed, and so does the force of friction, but it has elementary solution. Using that T= μN and N= mv²/r +mgcosθ, we get from the Work-Energy Theorem that ΔKE=ΔW=-{mgsinθ+μ(mgcosθ+mv²/r)rΔθ

Use z=v² as variable : 0.5 dz/dθ = -gr(sinθ +μcosθ)-μz - a linear differential equation for z in terms of θ, the angle enclosed with the vertical.

ehild

 

[ehild]

cantleave,
you wrote the acceleration in the form
$\vec a$= (r'' - rΘ'2)$\hat r$ + (2r'Θ' + rΘ'')$\hatΘ$
which was correct and I presume you knew the meaning of the symbols. r is the radius of the circle the object moves on, Θ is the angle the line from the centre to the object encloses with the vertical, and ' means derivative with respect time.

As r is constant r' and r" are zero, you certainly knew.
Find the forces acting on the object and write out the radial and tangential components of the resultant force.

ehild

 

[Redbelly98]

I am temporarily closing this thread. I think we are wasting our time without the actual problem statement. Just for example, the list of "known" parameters provided in Post #1 had nothing about friction, and yet that is present in the attempt at a solution. So, does friction belong here or not? Without the actual problem wording, it is confusing for those trying to help.

cantleave, if you can provide the actual wording of the problem statement, we will re-open it.

To provide us with the wording, click the "Report" button on this post and type the problem statement into the text box that appears. No need to include any figures, we just want the wording of the problem statement.

Regards,

Redbelly98

 

[micromass]

Thread is now reopened. But I would still like to see the exact problem statement. So OP, if you're still with us, could you please do that? Thanks!

 

[cantleave]

Thanks for reopenning the thread.
There was no problem statement thus there was no mentioning of friction, neither that it should be ignored nor that it should be taken into consideration. The professor gave me the figure, listed the parameters that are known (by "known" I mean they could be any number) and the 4 questions.

The original figure didn't include any forces, only the velocity vector was shown as it is placed on the attached image. I was supposed to draw the forces that act in the system by myself and I assumed that there must be some kind of friction and that it is the opposite direction of the movement of the object. We usually use the T= μN (or T=< μN) statement in class when we deal with friction.

I would like to continue trying to solve this problem without taking into consideration friction. I apologize for my mistake.

Also I have been told that part 1.) can be easily solved using nothing more than conservation of energy. Could we possibly head this way? I'm familiar with calculus and differential equations but don't feel yet comfortable enough to put them into such a practical context. Of course if the solution requires using such advanced mathematics I'm open for the challenge.

Thank you

 

[ehild]

Also I have been told that part 1.) can be easily solved using nothing more than conservation of energy. Could we possibly head this way?

OK, there is no friction, energy is conserved. State the law of conservation of energy. What are the kinetic and potential energies of the object at the bottom of the slope and what are they at the top, when the radius drawn to the object makes the angle α to the vertical?

ehild

 

[cantleave]

I'm having an exam in 2 days, after that I will be able to concentrate more on this problem. Please don't close the thread. I'll be back as soon as I can. Thank you

 

[Redbelly98]

I'm having an exam in 2 days, after that I will be able to concentrate more on this problem. Please don't close the thread. I'll be back as soon as I can. Thank you

No problem, hope there are no hard feelings and good luck on your exam. Thank you for clarifying things.

Regards,

Redbelly98

 

[BiGyElLoWhAt]

Sorry cantleave, I didn't mean to bail on you -.-

I got busy.

I'm glad to see you made some steady progress.

P.S.
good luck on the exam ;-)