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Body sliding down an accelerating hemisphere

  1. Dec 21, 2013 #1
    We have a frictionless hemisphere M,of a radius R and a body m on the top of it.We give to the hemisphere a horizontal constant acceleration W and consequently the body begins to slide down the accelerating hemisphere.What is the relative velocity of the body in respect to the hemisphere when the body leaves the hemisphere?And what is the angle θ that has made the body when it leaves the hemisphere,applying for W=g?
    I thought that the faster the hemisphere moves forward, the faster the height of the body in y-axis reduces,so W=Ay,where Ay is the acceleration of the body in y-axis.The only acceleration of the body is the centripetal An,which breaks down to Ax and Ay for my convenience.So:Ay=W=>An*cosθ=W (1). Also: ΣFr=mAn=>mgcosθ-N=mAn (where N the normal reaction)<=>An=(mgcosθ-N)/m (2).From (1),(2):mgcosθ2-Ncosθ=mW.When the body leaves the hemisphere:N=0,so cosθ2=W/g=>cosθ=[itex]\sqrt{W/g}[/itex].But I also have:
    An=u2/R =>W/cosθ=u2/R =>u2=gR[itex]\sqrt{W/g}[/itex] <=>
    u=[itex]\sqrt{gR\sqrt{W/g}}[/itex].But that u is the velocity in respect to some motionless observer outside the system.To find the relative velocity we have to find and the velocity of the hemisphere the time the body leaves it.For the hemisphere:Uhsph=Wt and for the body:uy=Ay*t=Wt.But uy=ucos(π/2-θ)=>uy=sinθ*u.So:
    uhsph=usinθ.
    urltv=u-uhsph(vectors) => urltv=sqrt{u2+uhsph2}=>...=>urltv=[itex]\sqrt{R*(2g-W)\sqrt{W/g}}[/itex].
    But the values for θ are just not normal,so I guess that the assumption Ay=W is wrong,right?
    Can you tell me the principles that rule over the phenomenon (and the relative equations)?Any idea what to do?
     
    Last edited: Dec 21, 2013
  2. jcsd
  3. Dec 21, 2013 #2

    TSny

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    In the inertial "lab" frame, the body does not move in a circle. So, I don't think you can say that the only acceleration the body has is centripetal acceleration in this frame of reference.

    It might be much easier to switch to the frame of reference of the hemisphere. Are you familiar with the idea that in a reference frame that has uniform acceleration in a straight line, you can take into account the effect of the acceleration by just using a modified acceleration of gravity? That is, you can treat the reference frame of the hemisphere as an inertial reference frame if you modify the magnitude and direction of the acceleration of gravity.
     
  4. Dec 21, 2013 #3
    No,I haven't learnt anything like that.Can you explain in further detail?And what is the orbit of the body m,why isn't it a part of a circle?
     
  5. Dec 21, 2013 #4

    TSny

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    Suppose you are riding in a train that has a constant acceleration along a straight track.
    In the inertial earth frame you can apply Newton’s second law to the motion of a body:

    ##\sum{\vec{F}} = m\vec{a}##

    You can write the acceleration of the body relative to the earth as a sum of the acceleration of the body relative to the train and the acceleration of the train relative to the earth:

    ##\vec{a} = \vec{a}_{b/t} + \vec{a}_{t/e}##

    So, ##\sum{\vec{F}} = m( \vec{a}_{b/t} + \vec{a}_{t/e})##

    Rearrange:

    ##\sum{\vec{F}} - m\vec{a}_{t/e} = m\vec{a}_{b/t} ##

    So, the acceleration of the body relative to the train obeys the 2nd law if you include an extra “fictitious†force ##- m\vec{a}_{t/e}## acting on the body.

    If you combine this extra force with the true force of gravity acing on the body, then you have

    ##m\vec{g}- m\vec{a}_{t/e} = m(\vec{g}- \vec{a}_{t/e} ) \equiv m\vec{g}_{\rm eff}##

    So, Newton’s 2nd law will still hold in the train if you replace the true acceleration of gravity vector by a new effective acceleration of gravity vector

    ##\vec{g}_{\rm eff} \equiv \vec{g} - \vec{a}_{t/e}##
     
    Last edited: Dec 21, 2013
  6. Dec 21, 2013 #5

    TSny

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    In the frame of reference of the hemisphere, the body moves along a circular arc with increasing speed. But relative to the earth, the body has an additional horizontally accelerated motion. So, relative to the earth, the body does not move along a circle.
     
  7. Dec 21, 2013 #6
    But how does it get this horizontal acceleration,since no force acts towards the direction of the hemisphere on the body(no friction)?Is it from the normal force?Is it towards where the hemisphere is moving or from the opposite?And can you describe me precisely its exact orbit?Is it something like an ellipse?I just can't visualize it.
     
  8. Dec 21, 2013 #7

    TSny

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    Yes, in the earth frame the body gets it's horizontal acceleration, ##a_x##, solely from the x-component of the normal force. By thinking about the direction of the normal force, you should be able to decide on the direction of the x-component of acceleration of the body relative to the earth.

    I don't believe the precise trajectory relative to the earth will be a simple curve like an ellipse or parabola. But if you draw a few sketches at different moments you should be able to see the rough shape of the trajectory relative to the earth.
     
  9. Dec 21, 2013 #8
    If I consider the hemisphere reference frame as you said then we have only centripetal acceleration for the body.So how do I combine this with what you mentioned above about the effective acceleration,since there is no horizontal acceleration for the body?The only thing that changes for my first post if we consider this frame of reference isn't that u is the relative velocity?
     
  10. Dec 21, 2013 #9

    TSny

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    In the hemisphere frame, there will be both centripetal acceleration and tangential acceleration. It will be similar to the more familiar problem where you let a block slide down a hemisphere that is kept at rest. But, now you will need to use an effective acceleration of gravity ##\vec{g}_{\rm eff} \equiv \vec{g} - \vec{a}_{hemisphere/earth}##

    This means that the "force of gravity" ##m\vec{g}_{\rm eff}## will not be vertical and it will have a magnitude that differs from normal mg.
     
    Last edited: Dec 21, 2013
  11. Dec 21, 2013 #10
    By that you mean that wherever I have g I use instead of it the geff?And wherever I find u is the relative velocity?And I justify the use of geff by the explanation you said,despite the fact we haven't learnt such a thing?
     
  12. Dec 21, 2013 #11

    TSny

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    Not just change the magnitude of g, but also the direction. Note that geff is found by vector addition of the true gravitational acceleration vector and the negative of the acceleration vector of the hemisphere (relative to the earth).

    Yes, the velocity of the body in the hemisphere frame is the velocity of the body relative to the hemisphere.

    Since I don't know what you have covered in your course, I don't know what method you are expected to use. Can you relate this problem to what you are currently studying in the course?
     
    Last edited: Dec 21, 2013
  13. Dec 21, 2013 #12

    TSny

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    Note. The nice thing about using the effective g is that you can still use energy concepts just as you would do if the hemisphere is not allowed to move relative to the earth. But you do need to take into account the complication that ##\vec{g}_{\rm eff}## is no longer perpendicular to the ground.
     
  14. Dec 22, 2013 #13
    In our course we have just "touched" the relative motions,nothing any further in detail...So the body has two accelerations in this reference frame,centripetal and effective (with its direction) and using the equations for the centripetal acceleration and the conservation of energy we solve the problem in that way?
     
  15. Dec 22, 2013 #14

    TSny

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    Yes. Relative to the hemisphere, the acceleration of the body can be broken down into centripetal and tangential components. As you say, you can use the centripetal component of F = ma and conservation of energy to solve the problem.
     
  16. Dec 22, 2013 #15
    Because I am confused about how to use practically geff,can you write me a few equations about the body,showing the usage of geff?
     
  17. Dec 22, 2013 #16

    TSny

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    Recall ##\vec{g}_{\rm eff} = \vec{g} - \vec{a}_{hemisphere/earth}##.

    See if you can write expressions for the magnitude and direction of ##\vec{g}_{\rm eff}## in terms of g and W.
     
  18. Dec 22, 2013 #17
    I think I got it.The relative velocity will be a function of the angle,right?Here is what I have found:
    urelative=[itex]\sqrt{}(R(gcosθ-Wsinθ))[/itex] and
    cosθ(g+2sqrt{g2+W2})-Wsinθ-2sqrt{g2+W2}=0. If the latter one is right,how do I solve this? But thankfully,our professor said that the angle should be around 17 degrees for W=g,so I plugged this in the angle equation and I got 0.5...=0,which leads me to consider the equations to be right,because he said around 17 degrees,so we are almost there!What do you think?
     
    Last edited: Dec 22, 2013
  19. Dec 22, 2013 #18

    TSny

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    Yes.

    Check to see if these two equations make sense at θ = 0?
     
  20. Dec 22, 2013 #19
    What,do you think they are wrong?
     
  21. Dec 22, 2013 #20

    TSny

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    Take your equation urelative=[itex]\sqrt{}(R(gcosθ-Wsinθ))[/itex].

    What does this give you for urelative when θ = 0?
     
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