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Bar on inclined slot on one end

  1. Jan 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Here, have a diagram I made!

    Sg8Rgkw.png

    The problem is to find the moment about the midpoint of the bar when it is horizontal, given that the right end is moving in towards the frictionless slot at a velocity v. Gravity and friction are neglected in this system. If you want, assume you're looking down on this problem from above, and it is happening on a table.

    2. Relevant equations

    Force is the time derivative of momentum. Moment is time derivative of angular momentum.

    3. The attempt at a solution

    My problem stems from trying to figure out the geometry of this. I am not sure how to even approach this. See, if the slot was vertical instead of inclined, this would feel like a much simpler problem to me.

    I know that what I want to do is find an equation for angular momentum about the midpoint. This requires finding the x and y velocity of the midpoint, as well as an equation for the rate of change of the angle of the bar.

    I just do not know what to do about the geometry. Do I use the law of sines? Something is not clicking here.
     
  2. jcsd
  3. Jan 17, 2014 #2
    Nice diagram!

    I am slightly confused about your statement of the problem though. You say your finding the moment and you define the moment as the time derivative of angular momentum, which is torque.

    Anyways, I think you know where your going with he problem and your just curious how you find the angular velocity, right? On the ramp, you can set this problem up similar to something sliding down. (i.e. orient your coordinate system so x-axis is flat) If the velocity is traveling into your coordinate system, what is the magnitude of the y-component of the velocity?

    With that component, you know how fast the that end of the bar is moving in a particular direction and therefore you know the angular velocity. Sounds like you can handle it from there.
     
  4. Jan 18, 2014 #3
    Thanks @ the diagram comment!

    Hmm, this actually raises a few questions. What is the difference between torque and moment? As an engineer, I have used them interchangeably. I mean the "moment of force" or $$\vec{M}_G=\vec{r}\times\vec{F}=\vec{r}\times{\operatorname{d}\over \operatorname{d}\!t}m\dot{\vec{r}}$$. G is the center of mass of the bar.

    Also, I do not quite understand what you mean. Are you saying to set the X axis along the incline? I still do not see how I would work out the geometry. But, yes, it is precisely angular velocity of the bar, with respect to its center of mass, G, that I wish to find.
     
  5. Jan 18, 2014 #4
    Ohhh, I see. It seems the moment is just the torque about the center of mass. That make sense. Never seen that terminology, thanks for sharing.

    Instead of looking at you coordinate system where x-axis is parallel to the bar and y-axis perpendicular to it try this instead:

    • Orient your x-axis to be parallel to the inclined plane. Also your y-axis will be perpendicular to it as well. This is literally saying tilt the entire coordinate system by 30 degrees.
    • Using your velocity vector you can get an x and a y component. This is how fast that end of the bar is moving up the ramp and how fast it is moving perpendicular to the ramp. The perpendicular is the velocity that you would get your angular velocity from.

    I hope this is clear, as trying to explain this without picture is proving to be difficult. If you find that this is not that clear, let me know. I will just insert a picture.
     
  6. Jan 18, 2014 #5
    It is not clear to me how the slot and the bar are oriented. Please post the description of the problem exactly as it was given.
     
  7. Jan 18, 2014 #6
    The left part of the bar is allowed to slide freely (via a frictionless bearing) in the slot. The slot has an angle of 30 degrees from horizontal. The right part of the bar is allowed to slide freely across the horizontal surface (which it does, at velocity v). The bar is not restricted in angle at all, other than both ends are always touching their respective surfaces.
     
  8. Jan 18, 2014 #7
    Alright, so, this is what I ended up doing to approach a solution. I figured that the only thing that I need to know is the rate of change of ω and the inertia of the bar. Knowing both, the moment of force imparted on the bar, about the bar's center, is solved for.

    I used the law of sines. As the bar continues to get pushed in, it rises up and creates a triangle, with its large angle being 150 degrees.

    I kept the coordinate system the same, defining x to be along horizontal.

    uXT8z9r.png

    I determined [itex]\varphi[/itex] from the law of sines.

    $$\frac{\sin (\phi )}{L-v t}=\frac{\sin (150)}{L}$$

    From this, I was able to determine phi.
    $$\phi=\sin ^{-1}\left(\frac{L-v t}{2 L}\right)$$
    Using the special condition for bending moment, where the center of mass is the origin,
    $$M_G=I_G \alpha=I_G {\operatorname{d}\over \operatorname{d}\!t}\left({\operatorname{d}\over \operatorname{d}\!t}\phi\right)$$

    I was able to determine that the instantaneous moment about G (the center of mass), when the bar is horizontal, is..
    $$M_G=\frac{mv^2}{36\sqrt{3}}$$

    How does that look?
     
  9. Jan 19, 2014 #8
    It was not clear that the bar, the slot and the bar's velocity are all in one plane. If that is the case, you approach seems good, except that I am not sure how you got the final equation. Did you have a numeric value of ##L##?
     
  10. Jan 20, 2014 #9
    The L's ended up canceling out. No numeric value. It looks like it solely depended on the mass. Any length consideration got absorbed into the denominator.
     
  11. Jan 20, 2014 #10
    Ah, indeed. I made a mistake in my previous calculation. Now I get the same result. Well done!
     
  12. Jan 20, 2014 #11
    Phew! That's the first confirmation this semester that I have any clue what I'm doing. Thanks for the help!
     
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