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Rigid body kinetic energy+ constraints (upper level classical mechanics)

  1. Jul 3, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Using the corresponding constraints conditions, calculate the kinetic energy of
    1)A homogeneous cylinder of radius a that rolls inside a cylindrical surface of radius R>a.


    2. Relevant equations
    My toughts: I hope they meants "roll without slipping". Let's consider this case!
    [itex]T=\frac{mv ^2_{CM}}{2}+\frac{I_1\omega ^2 _x+I_2\omega ^2 _y+I_3 \omega ^2 _z}{2}[/itex].


    3. The attempt at a solution
    Ok I already found out that [itex]I_1=I_2=\frac{m}{4} \left ( \frac{h^2}{3}+a ^2 \right )[/itex] and [itex]I_3=\frac{ma^2}{2}[/itex].
    The constraint is that the line of the rolling cylinder in contact with the cylindrical surface does not slide. Mathematically I'm not really sure how to write this down. First, I'd set my coordinate system in the center of the rolling cylinder. (I remember in introductory physics that we dealt with this by setting the relative speed of the cylinders to be 0 m/s, but here I'm not sure)
    Considering the axis along the height of the rolling cylinder as the z-axis, I think that [itex]\omega _x =\omega _y =\vec 0[/itex]. While [itex]\omega _z \neq 0 \neq \text{constant}[/itex]. It would be worth [itex]\omega _z (t)= \frac{d \theta (t)}{dt}[/itex] but I'm not sure I should use "theta" as angle. I've read in Goldstein that I should use Euler angles for this, but I have absolutely no idea how.
    If the cylinder starts rolling at [itex]t=0[/itex] from its maximum height, [itex]\theta (0)=0[/itex], which is an initial condition for the motion equation of the cylinder.

    Any help is appreciated, as usual. :smile:

    Edit: Just checked out the picture of Euler angles in wikipedia and I think that instead of theta I should use beta.
     
    Last edited: Jul 3, 2011
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  3. Jul 7, 2011 #2

    lanedance

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    using the circumference of the circular cross section of each cyclinder (outer & rolling) you should be able to relate the angular velocity of the centre of mass to to the angular velocity of the rotation of the inner cylinder itself
     
  4. Jul 7, 2011 #3

    fluidistic

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    I appreciate your help very much. However I do not know how to do so.
    I realize that the center of mass' motion is a circle of radius R-a. I know that the circumferences of the outer and inner cylinders are [itex]2 \pi R[/itex] and [itex]2\pi a[/itex] a respectively.
    I know that the instant velocity of the inner cylinder surface with respect to the outer one is 0m/s since I considered no rolling.
     
  5. Jul 7, 2011 #4

    lanedance

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    It will help to consider 2 angular velocities, that of the CoM [itex] \omega_{CM}[/itex] and that of the smaller cylinder itself [itex] \omega_{c}[/itex]

    Now, how about first considering the case where the smaller cylinder is rigidly rotating about at the centre of the larger cylinder. In this case the surface slides. Imagine a rigid arm bolted across the smaller cylinder on a pin at the centre of the large cylinder.

    Then if the centre of mass velocity is [itex] v_{CM} = \omega_{CM}(r-a)[/itex] then velocity of the small cylinder scales with distance form the inner cylinder centre and at the surface of the large cylinder it is [itex] v_R = v_{CM} \frac{R}{R-a}[/itex]. This is the relative sliding velocity.

    What is the angular velocity of the cylinder [itex] \omega_{c}[/itex]
    and what is the total energy of this system?


    Now when you move to the no sliding condition you have
    [itex] v_R = 0[/itex]

    So there is an extra angular rotation of the body we must also account for . We want to relate the angular velocity of the CoM to the angular velocity of the small body itself.

    2 equivalent ways to relate these
    - use the zero velocity constraint to solve for the required body angular velocity
    - use the ratio of circumferences to find out how many times the inner cylinder rotates in a single revolution of the larger cylinder
     
    Last edited: Jul 7, 2011
  6. Jul 7, 2011 #5

    fluidistic

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    Thanks for your patience and time!
    Ok, perfect.

    I've no problem with this, but here the English language is also a problem for me. I don't really understand the situation. Do you mean that a=R/2 (condition to be fixed on the center of the outer cylinder) and also that it's forced to slide over the outer cylinder's internal surface? Do you consider the outer cylinder to "swing" with respect to an external reference frame?

    Ok I understand this
    Hmm, I don't know how to reach this.

    Not sure. Maybe [itex]\omega _c =v_R/a[/itex]?
    I hope the outer cylinder isn't swinging. In this case it would be [itex]\frac{mv^2_{CM} }{2}+\frac{I \omega _c }{2}+mg(h-a)[/itex] where h is the height of the center of mass with respect to the lowest line in the outer cylinder.


    Fine.

    I don't understand why.
    In other words [itex]\omega _{CM}[/itex] in function of [itex]\omega _c[/itex]?

    Ok. I'll try once I understand better all the previous things.
     
  7. Jul 8, 2011 #6

    lanedance

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    ok, that confused things more than i would have liked... lets just deal with the problem at hand... where the cylinder does not slide

    lets start by working in the absolute reference frame and assuming we know [itex] \omega_{CM} [/itex],
    R is radius of large cylinder
    a that of small

    then the velocity of the centre of mass of the small cylinder is
    [tex] v_{CM} = \omega_{CM} (r-a)[/tex]
    the velocity of the surface (at r= a) is zero as it does not slip
    [tex] v_{a} = 0[/tex]

    Now consider the dashed reference frame moving with the centre of mass,
    [tex] v_{CM}' = 0[/tex]
    [tex] v_{a}' = -\omega_{CM} (r-a) [/tex]

    Can you find the angular velocity?
     
  8. Jul 11, 2011 #7

    fluidistic

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    I'm trying hard but can't make it.
    I know that the small cylinder can make R/a rotations within the big cylinder before returning back to its original position.
    So we have [itex]v_a'=-\omega _{CM}(R-a)[/itex]. I'm not understanding very well the minus sign here though. Anyway I think that [itex]v_a'=\omega _c a \Rightarrow \omega _{CM}=\omega _c \frac{a}{a-R}[/itex]. I'm totally stuck here.
    P.S.:I realize my last equation isn't true. The modulus of [itex]\omega _{CM}[/itex] is greater than the one of [itex]\omega _c[/itex] which is senseless. By total intuition I would write [itex]\omega _{CM}=\frac {\omega _ca}{R}[/itex] but I've no idea if it's right nor how to derive it.

    Edit: My intuition was wrong.
     
    Last edited: Jul 11, 2011
  9. Jul 12, 2011 #8

    lanedance

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    I think you are on the right track

    the minus sign they are in different direction hopefully you have drawn a good picture

    first of the 2 cylinders together, working in the global ref frame (large cylinder is stationary)
    - make the contact point at the far right (positive x direction)
    - say [itex] \omega_{CM} [/itex] is CCW
    - this makes [itex]v_{CM}=\omega_{CM}(R-a)[/itex] in the positive y direction
    - mark the contact point as v_a=0

    now draw the single cylinder alone, still in the same ref frame, and mark on the same velocities

    finally draw the the single cylinder alone, now in the ref frame of the small cylinders CoM velocity, but not rotating
    - [itex]v_{CM}'=0[/itex] by def'n
    - [itex]v_{a}'=v_a-v_{CM} = 0-\omega_{CM}(R-a)[/itex]
    so writing the
    [tex]v_{a}'=\omega_c a = -\omega_{CM}(R-a)[/tex]

    you get as you had (i prefer considering the little cylinder)
    [tex]\omega_c = \omega_{CM}\frac{(R-a)}{a} [/tex]

    now to convince yourself it is a reasonable answer

    first imagine the large cylinder was instantaneously removed
    - this says the small cylinder would move off at speed [itex]v_{CM =}\omega_{CM}(R-a)[/itex] and rotating at [itex]\omega_c = \omega_{CM}\frac{(R-a)}{a} [/itex]
    seem reasonable

    examine the limiting behaviour
    [tex]\omega_c = \omega_{CM}\frac{(R-a)}{a} [/tex]

    when [itex]a<<R [/tex] the little cylinder cylinder spins really fast
    when [itex]a \approx R [/tex] the little cylinder cylinder barely rotates

    try and convince yourself this all makes sense (and feel free to question, though it look good to me everyone makes mistakes no and then;)
     
    Last edited: Jul 12, 2011
  10. Jul 12, 2011 #9

    lanedance

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    the other thing to consider as mentioned before is the ratio of circumference

    imagine the large cylinder is unrolled flat, it has a length of [itex] 2 \pi R[/itex]

    if you rolled the small cylinder along the unrolled length without slipping it would rotate [itex] 2 \pi \frac{R}{a}[/itex] radians

    when you roll the cylinder back up, this is not quite true and i think you actually rotate only [itex] 2 \pi (\frac{R}{a}-1)[/itex] radians, as you remove whole rotation as you go round the cylinder, but you may want to check that (consider the limiting case where R~a)
     
  11. Jul 13, 2011 #10

    fluidistic

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    Thanks once again lanedance.
    Actually I didn't understand how it's possible to get [itex]\omega _c =\frac{\omega _{CM} (R-a) }{a}[/itex]. I get [itex]\omega _c =\frac{\omega _{CM} (a-R) }{a}[/itex]. Since a-R<0, it means that my answer says that if the cylinder making a clock-wise swing inside the big cylinder, then it is rotating on itself counter-clowisely. While your answer implies that [itex]\omega _c[/itex] and [itex]\omega _{CM}[/itex] have the same direction and hence if the cylinder swings clock-wise then it also rotate on itself clockwise. Am I right on this?

    Also I didn't get the point of the ratio circumference. It's supposed to help me to find [itex]\omega _c[/itex], thus [itex]\omega _{CM}[/itex] and thus [itex]V _{CM}[/itex] I guess... but how?
     
  12. Jul 14, 2011 #11

    lanedance

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    you're right, whilst i mentioned the rotation directions i dropped a minus sign - good pick up

    the ratio of circumferences was (for me) an interesting insight to the problem, however if its not helping or confusing the issue, don't worry about it...
     
  13. Jul 14, 2011 #12

    fluidistic

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    Ok :)
    Now how do I continue?
    Once I've [itex]\omega _c[/itex], [itex]\omega {CM}[/itex] or [itex]V_{CM}[/itex] I'm done.
     
  14. Jul 14, 2011 #13

    first of all i must say the question is askin energy not momentum or velocity so you need not go into vectors and coordinate system as energy is scaler...yes u can do it with cooerdinate but it will complicate solution

    i will try to do it with superposition principle

    "cylinder is moving inside the big cyllinder that means the work done by the normal force is zero":- implies KE is constant:- again IMPLIES :if we remove big cyllinder KE remains same(as force was perpendicular to motion and the friction doesnt work in rolling)

    after removing big cyll small cyll moves lenearly and also rotates

    now we have to find 'w' (angular vel about its axis afterwards)

    and lenear velocity 'v' (afterwards)


    NOW the solution:
    lets us cosider small cylinders angular velocity(abouts its own axis)= 'w' {remains same}
    its speed in circular motion be 'v'= w*(R-a) {after removing big Cyll it is its lenear velocity}
    'a' is the radius of small cylinder

    moment of inertia along small cyllinders centre axis=1/2(m*a^2)


    now total energy= translation KE+ rotational KE
    =1/2(m*v^2) + 1/2(I*w^2)

    this is the correct solution

    if any doubts plz ask
     
  15. Jul 14, 2011 #14

    fluidistic

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    Thanks for the reply :D Well, I don't really understand why you consider the case when we remove the big cylinder... I mean it's not stated in the problem.
    Shouldn't the solution be =1/2(m*v'^2) + 1/2(I*w'^2)? (notice the primes)


    Edit: Also I don't agree with you when you say that the kinetic energy remains constant. When the small cylinder is at its max height, its kinetic energy is null.
     
    Last edited: Jul 14, 2011
  16. Jul 14, 2011 #15

    i did not consider gravity....it is no where mentioned in the question

    you shud post full question then
     
  17. Jul 14, 2011 #16

    lanedance

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    should be pretty much there - can you take it from here?
     
  18. Jul 14, 2011 #17

    fluidistic

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    My bad, for some reason I thought there was gravity in this problem. You are right, there is none. However I didn't really understand the point of removing the big cylinder. I understand that we can do it since there's no external force "eating" the kinetic energy of the small cylinder, but I don't understand the point of doing it. Where does this simplify the equations?

    My bad... I had a big mental block lasting for days. I think I get it now.
    [itex]T=\frac{ma^2 \omega _c ^2 }{2} \left ( \frac{1}{(a-R)^2 }+1 \right )[/itex].
    I thought that [itex]\omega _c[/itex] was unknown and was a property of the cylinders, which is totally false. It's a "given" value. I hope I didn't make any algebra error.
     
  19. Jul 14, 2011 #18
    what does the big cyllindrical surface doing?? :-it is providing 1) normal force 2)friction to support rolling.

    for me:i dont know the big cyllinder
    i only know the two forces

    now situation: "a cyllinder being acted upon by two forces which dont work"

    now as both forces do not do any work on the little cyllinder....removing these forces would
    not change KE of the cyllinder

    so as soon as we remove these forces the motion of cyllinder is linear as well as rotating about its axis this

    so the solutin remains same (provided there is no gravity)
     
  20. Jul 14, 2011 #19

    lanedance

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    yeah so as darkxponent has said, the energy of the little cylinder is conserved

    now that you know how the angular velocities, you should be able to write the total energgy in terms of only the mass, radii & angular rotational speed of the CoM

    if required incorporating gravity would be a simple exercise, just adding a gravitational
    potential term
     
  21. Jul 14, 2011 #20

    fluidistic

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    Ah thanks... I think I understand. So in the situation of the little cylinder alone with linear motion, it would rotate on itself, not because of friction (since you removed it) but from its inertia.
     
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