Classic Incline problem with cylinder

  • #1
ChiralSuperfields
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Homework Statement
Please seee below
Relevant Equations
Please seee below
FOr this,
Xg0AOpjUG5kSkda5tE2YaRRH9qpUvQmMNSROeGoy-iZg5eltVk.png

Use alternate coordinate system
ld6Bcxc8QBtLLX-byrt_J2HKSZLKZpLujuHHeo6PDe-CSUC4Pc.png

With ##ȳ##-axis parallel to incline and ##x̄##-axis parallel to the x-axis. Kinetic energy using this alternate coordinate system is ##T = \frac{1}{2}M\dot x_p^2 + \frac{1}{2}mR^2\dot \phi^2 + \frac{1}{2}m(\dot x̄^2 + \dot ȳ^2 + 2\dot x̄ \dot ȳ \cos\beta)## and potential energy is ##V = -mgȳ\sin \beta##. Note how potential energy of block was not included as we don’t know where the height of the COM is, however, it is a constant, so we can just all the high c, and it will add a ##+C## to the lagrangian

Now transform to original x-y coordinate system using transformations ##ȳ = -\frac{y}{sin \beta}## and ##x̄ = x + y\cot \beta## from decomposing ##ȳ## along the x and y axes. I think that ##\hat ȳ = \hat i - \hat j##. Then one would substitute the these transformations into the kinetic energy and potential energy expressions, however, how does one the ##x## and ##y## in these transformations in terms of ##x_p## and ##\phi##.

Thanks!
 
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  • #2
The cylinfer has velocity of
x direction : ##R\dot{\phi} \cos \beta + \dot{x}_P##
y direction : -##R\dot{\phi} \sin \beta ##
and potential energy of ##-mgR\phi \sin \beta## which is 0 for ##\phi##= 0
 
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  • #3
ChiralSuperfields said:
##\frac{1}{2}m(\dot {\bar x}^2 + \dot {\bar y}^2 + 2\dot {\bar x} \dot {\bar y} \cos\beta)##
Check: does that give the right answer when ##\beta=0##?
Btw, the latex works better using \dot{\bar x}.
ChiralSuperfields said:
I think that ##\hat ȳ = \hat i - \hat j##.
Very unlikely that ## \hat i - \hat j## would happen to have unit magnitude, but I'm not sure how you are defining them.
 
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  • #4
anuttarasammyak said:
The cylinfer has velocity of
x direction : ##R\dot{\phi} \cos \beta + \dot{x}_P##
y direction : -##R\dot{\phi} \sin \beta ##
and potential energy of ##-mgR\phi \sin \beta## which is 0 for ##\phi##= 0
haruspex said:
Check: does that give the right answer when ##\beta=0##?
Btw, the latex works better using \dot{\bar x}.

Very unlikely that ## \hat i - \hat j## would happen to have unit magnitude, but I'm not sure how you are defining them.
@anuttarasammyak @haruspex

Thank you both for your replies!

I see how you got the relation for y by using the rolling condition vector decomposition. However, sorry I am confused how you got the relation for x. I get everything apart from the + time derivative of x_p term. Do you please know why it is the time derivative of x_p not the time derivative of x (since it is along the x-axis)?

Also yes, I think that the expression I found the the cylinder velocity works for the special case of ##\beta = 0##. However, it is strange to think about since I already defined a non-orthogonal coordinate system to be parallel to the incline for the current ##\beta## shown in the diagram, so I assume the coordinate system would remain fixed if we change the beta. However, there is a potential option that we could redefine the coordinate system again. Do you please know whether you intended for the coordinate system to also be rotated too as ##\beta## approaches zero or is it just meant to be fixed?

Thanks!
 
  • #5
ChiralSuperfields said:
Also yes, I think that the expression I found the the cylinder velocity works for the special case of β=0. However, it is strange to think about since I already defined a non-orthogonal coordinate system to be parallel to the incline for the current β shown in the diagram, so I assume the coordinate system would remain fixed if we change the beta. However, there is a potential option that we could redefine the coordinate system again. Do you please know whether you intended for the coordinate system to also be rotated too as β approaches zero or is it just meant to be fixed?
Setting ##\beta =0## is just a way to verify that you have the correct form for the kinetic energy. If it is correct, it should work for any ##\beta.## Of course, the task in this case becomes a bit of a challenge to the unwary reader.. Of all the choices of symbols at your disposal you had to pick ##y## for the direction down the incline and ignore that ##x## and ##y## are already defined in the drawing as the standard Cartesian coordinates. So when ##\beta =0##, your ##y## becomes the already defined ##x## and is orthogonal to the already defined ##y##. That's confusing.

If I were doing this, I would pick ##s## as the generalized coordinate along the incline, with ##s=0##, say, at the upper corner of the wedge and proceed from there. Note that when ##\beta =0##, ##s\rightarrow x##.
 
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  • #6
You still have not said how you define ##\hat i, \hat j##. If those are ##\hat x, \hat y## then I get ##\hat{\bar y}=\cos(\beta)\hat i-\sin(\beta)\hat j##.
 
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  • #7
ChiralSuperfields said:
Do you please know why it is the time derivative of x_p not the time derivative of x (since it is along the x-axis)?
x_p is x coordinate of the rectangle corner (or any other part fixed point is OK) of the moving triangle. What is x you mentioned ?
 
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  • #8
anuttarasammyak said:
The cylinfer has velocity of
x direction : ##R\dot{\phi} \cos \beta + \dot{x}_P##
@ChiralSuperfields is using ##x## as the same as ##x_P## and ##\bar x = x_P+R{\phi} \cos \beta##.
 
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  • #9
kuruman said:
Setting ##\beta =0## is just a way to verify that you have the correct form for the kinetic energy. If it is correct, it should work for any ##\beta.## Of course, the task in this case becomes a bit of a challenge to the unwary reader.. Of all the choices of symbols at your disposal you had to pick ##y## for the direction down the incline and ignore that ##x## and ##y## are already defined in the drawing as the standard Cartesian coordinates. So when ##\beta =0##, your ##y## becomes the already defined ##x## and is orthogonal to the already defined ##y##. That's confusing.

If I were doing this, I would pick ##s## as the generalized coordinate along the incline, with ##s=0##, say, at the upper corner of the wedge and proceed from there. Note that when ##\beta =0##, ##s\rightarrow x##.
haruspex said:
You still have not said how you define ##\hat i, \hat j##. If those are ##\hat x, \hat y## then I get ##\hat{\bar y}=\cos(\beta)\hat i-\sin(\beta)\hat j##.
anuttarasammyak said:
x_p is x coordinate of the rectangle corner (or any other part fixed point is OK) of the moving triangle. What is x you mentioned ?
haruspex said:
@ChiralSuperfields is using ##x## as the same as ##x_P## and ##\bar x = x_P+R{\phi} \cos \beta##.
Thank you for your replies @kuruman, @haruspex and @anuttarasammyak!

Re: Coordinate system. I agree, sorry about that.

Re: Unit vector. Yes, sorry, that is my mistake, that makes sense the unit vector is that.

Re: x vs x_p coordinate. I don't think that x is the same as x_p since x_p ∈ x in the x-y coordinate system (Not x-bar, y-bar coordinate system). The x bar is in the different coordinate system with the x bar and y bar axes. The x that I for the x and y axes.

Thanks!
 
  • #10
Also does someone please know what the GPE of the block is? Is it ok if I just say the heigh of the COM is y = -k, where k is a constant?

For (c), I am trying to solve the equations of motion. In terms of integration constants, I solve the second order ODE coupled system and I have ##\phi = ht + k## and ##x_p = wt + c##. However, does someone please know what is good assumption to make to find Inital conditions?

I was thinking ##\phi(0) = \phi_{qi}##, ##\dot \phi(0) =\dot \phi_{qi}## ##x_p(0) = x_{2e}## ##\dot x_p(0) = \dot x_{2e}##

Thanks!
 
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  • #11
ChiralSuperfields said:
The x that I for the x and y axes.
Does your "x" or "time derivative of x" have something to do with the Lagrangian of
1714287201357.png

? I do not find it in parameter of the Lagrangian.
 
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  • #12
anuttarasammyak said:
Does your "x" have something to do with the Lagrangian of
View attachment 344176
? I do not find it in parameter of the Lagrangian.
Thank you for your reply @anuttarasammyak!

Sorry I am confused what you are saying. Do you please mean that I need ##\phi## and ##x_p## axes?

That is a interesting idea to consider. However, do you please know how one would draw them?
1714287591411.png

Not sure what one would call them but is is certainly not orthogonal. Maybe simi-cyclic coordinate system?
 
  • #13
Axes are x and y as I wrote x- y- directions of velocity in my post #2.
In xy coordinate, ##P(x_p,0)##. The contact point on the slope ##C(x_C,y_C)## in xy axes coordinate is
[tex]x_C=x_P+R\phi \cos \beta+x_{C0}[/tex]
[tex]y_C=-R\phi \sin \beta+y_{C0}[/tex]
where ##(x_{C0}, y_{C0})## is initial position of C when t=0, ##\phi=0##. Then the center of cylinder ##Q(x_Q,y_Q)## are given as
[tex]x_Q=x_C+ R\sin \beta[/tex]
[tex]y_Q=y_C+R\cos \beta[/tex]
by adding constants. All these things are in works of xy coordinates. You get Lagrangian from these Cartesian expressions. Once you got Lagrangian written by ##x_P## and ##\phi## , forget about Cartesian coordinates and go into analytics of ##x_P## and ##\phi## by Lagrange equation, which is the merit of analytical mechanics.
 
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  • #14
ChiralSuperfields said:
x vs x_p coordinate. I don't think that x is the same as x_p
You are right…
As I understand your working, (x, y) is the location of the centre of the cylinder in the coordinates as diagrammed.
If ##(\bar x, \bar y)## is the same point in your ##(\hat{\bar x}, \hat{\bar y})## coordinates then ##x\hat x+y\hat y=\bar x\hat{\bar x}+\bar y\hat{\bar y}##.
Also, ##\hat{\bar x}=\hat x##, ##\hat{\bar y}=\hat x\cos(\beta)-\hat y\sin(\beta)##, whence ##x=\bar x+\bar y\cos(\beta)##, ##y=-\bar y\sin(\beta)##.

This choice of coordinates effectively considers the motion of the cylinder as the same motion as the block (##\dot x_P=\dot{\bar x}##) plus a motion down the slope of the block (##\dot{\bar y}=R\dot\phi##).
 
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