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Two wheels rotating connected by a rod

  1. May 24, 2017 #1
    1. The problem statement, all variables and given/known data
    A cart has two cylindrical wheels connected by a weightless horizontal rod using weightless spokes and friction- less axis as shown in the figure. Each of the wheels is made of a homogeneous disc of radius R, and has a cylidrical hole of radius R/2 drilled coaxially at the distance R/3 from the centre of the wheel. The wheels are turned so that the holes point towards each other, and the cart is put into motion on a horizontal floor. What is the critical speed v by which the wheels start jumping?
    Answer: ##v=3\sqrt{gR}##

    IMG_0126.jpg
    2. Relevant equations


    3. The attempt at a solution
    I have no idea where to start! I have to find the centre of mass of each wheel. Then I must find the forces acting on it (only gravitational attraction) and then?
     
  2. jcsd
  3. May 24, 2017 #2

    BvU

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    Not good enough according to the PF guidelines !
    You could do that. Or you can work with R/2 disks with a negative mass ....
    There's the normal force too! Once that goes zero at some moment in a revolution the jumping isn't far away :rolleyes:

    PS is the connecting rod essential for answering the question / Or is it just to give the contraption a semblance of realism ?
     
  4. May 24, 2017 #3
    You gave me an idea! It's correct to think (in the reference frame of the wheel) that when centrifugal force equals gravitational force, then normal force goes zero and the wheel jumps?

    I don't know. I think that I have to use this (but I have no idea how to):cry:
     
  5. May 24, 2017 #4

    BvU

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    My intake is to forget the rod and look at the vertical acceleration of the c.o.m. because of the rolling motion with speed ##v##. If that offsets g you're at the critical speed.
     
  6. May 24, 2017 #5
    After some counts I came to this: the c.o.m. is ##\frac{R}{9}## distant from the centre of the cylinder, its angular speed ##\omega## is equal to the angular speed of the wheel, so ##v_{c.o.m}=\frac{v}{9}##.
    How can I find the vertical acceleration of the c.o.m.?
     
  7. May 24, 2017 #6

    BvU

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    Really ? So the center of mass stays behind and after a short while it is outside of the wheel ?
     
  8. May 24, 2017 #7
    Are both disks and the rod all in the same plane, or is this a 3D system with the rod normal to the two disks which are parallel to each other (like a rail road truck)?
     
  9. May 24, 2017 #8

    haruspex

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    I think the description is quite clear. Two parallel cylinders, axles linked by a horizontal rod normal to the axles. The complication is that the masses of the cylinders are not uniformly distributed, this being represented by holes drilled parallel to their axes (not, as the text says, coaxially).
     
  10. May 24, 2017 #9

    haruspex

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    I suggest it is so that the system would keep moving the same direction at any speed. The mass centre of the whole system has constant height. Tension and compression in the horizontal rod alternates: as the mass centre of the leading cylinder descends that cylinder pulls the other along.
    But since, as you say, "jumping" would mean the normal force from the ground vanishes, it seems that the forces in the rod are not relevant for that aspect.
     
  11. May 24, 2017 #10

    haruspex

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    Yes, that's the way to go.
    Don't deal with the c.o.m., that's unnecessarily complicated. As BvU wrote, treat the hole as a cylinder of negative mass superimposed on a solid cylinder. The solid cylinder has no centrifugal/centripetal force.
     
  12. May 25, 2017 #11
    How can I deal with this idea of superimposed cylinder? Because maybe it's useful to find the moment of inertia but I don't think I must use this to solve the problem.
     
  13. May 25, 2017 #12

    haruspex

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    No, you don't need to worry about moments of inertia here, just centripetal force.
    Think of the cylinder as consisting of a uniform solid cylinder with the hole as a cylinder of equal negative density superimposed on it.
    The centripetal force required is the sum of the two centripetal forces. The solid cylinder has no acceleration about the axis, so needs no centripetal force. So the centripetal force needed is for the negative cylinder only.
    Now, it might not be obvious, but it turns out that the centripetal force required for a solid body is the same as for a point mass at the same mass centre. So it is very easy to find the centripetal force.
    For the cylinder to become airborne, the max downward centripetal force equals the weight of the system.
     
  14. May 25, 2017 #13
    What is the radius to put in the centripetal formula? Is ##\frac{R}{3}## (but the answer is not correct) (or ##\frac{R}{3} + \frac{R}{9}##
     
  15. May 25, 2017 #14

    BvU

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    Why don't you do some work an write it down in a post ? If you derail, someone will help you back onto the right track.
     
  16. May 25, 2017 #15
    Ok I'm sorry. My work is this IMG_4427.JPG

    No, I'm wrong. Instead of M I have to put ##(M-m)##. So the final equation becomes ##\frac{R^2}{4} \frac{3v^2}{R} \geq \frac{3}{4} R^2 g##
    And I think that the radius for the centripetal force I'm taking is wrong.
     
  17. May 25, 2017 #16

    haruspex

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    Is that v the speed of the system or the instantaneous speed of the centre of a hole?
     
  18. May 25, 2017 #17
    I assumed v as the speed of the system.
     
  19. May 25, 2017 #18
    I would like to add a comment which does not have direct relation to the problem under consideration but it seems that it would also be useful here. For simplicity assume that a rigid body rotates about a fixed axis with constant angular velocity. What can one say about centrifugal forces relative to a body fixed frame? Very common wrong answer is as follows: the centrifugal force is applied to the center of mass.
     
  20. May 25, 2017 #19
    I can state that here the centre of rotation is not the centre of mass. Is it possible to use this fact to get the solution?
     
  21. May 25, 2017 #20

    haruspex

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    But in your centripetal force calculation you seem to have taken it as the tangential speed of the centre of the hole relative to the system.
     
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