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Homework Help: Weird circular motion problem

  1. Nov 19, 2005 #1
    A little block with mass m lies still on the top of a frictionless sphere with radius R. Somebody gently hits the block, causing it to slide down the sphere. The mass loses contact with the sphere when the angle between the positionvector and the vertical equals 0c. The drawing below shows the position of the block on two moments in the movement. The velocity caused by the hit in the beginning is negligible.

    http://img453.imageshack.us/img453/6496/sphericalsurface0qu.gif [Broken]

    With this problem I found a couple of answers and I'd like to know if I'm doing it correctly. There are a few questions I don't know how to answer so help would be appreciated very very much!
    a) Express the distance s in R and 0.

    s= R * 0 right? That is when 0= in rad. If 0 is in degree it should be s= R (2pi *0 / 360) correct?

    b) Which forces work on the block if it's sliding down the sphere? Draw them.

    I guess that'll be the gravitationforce and the normal force, drawn below:

    http://img173.imageshack.us/img173/4374/sphericalsurface25zz.gif [Broken]

    c) Split the forces in tangential components and perpendicular components on the spherical surface.

    I did it this way, could somebody please tell me if I put the angle 0 between the right vectors?:

    http://img467.imageshack.us/img467/2815/sphericalsurface35tt.gif [Broken]

    d) Express the centripetal acceleration Ac and the tangential acceleration At in the forces found by b.

    If 0 is chosen well in the x-direction (the direction of the tangential acceleration) Fres= Fz,//= Fz sin 0 = m * At. Therefore At= (Fz sin (0))/m

    In the y-direction (the direction of the centripetal acceleration) Fres= Fz,|- Fn= Fz cos 0 - Fn= m* Ac. Therefore Ac= (Fz cos(0) -Fn)/m. Correct?

    From here I don't get the problem any longer:

    e) At the point where the mass begins losing contact with the surface there are two conditions:

    Condition 1: |Ac|= g cos(0)
    Condition 2: |Ac|= v^2/R where v= the speed of the block.

    Explain why these conditions are set here.

    f) Vind the angle 0c where the block begins losing contact with the surface.

    Please help! I'd appreciate it very much! Thanks for your effort!
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 19, 2005 #2


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    a) correct
    b) correct
    c) correct
    You place theta right as:
    [tex]\theta = \frac{\pi}{2} - \theta _c = \theta _1[/tex]
    Where [itex]\theta _1[/itex] is the angle at the point of the block.
    d) looks right
    e) This is the essential part of the problem. Give it some more thought. When does the block lose contact?
    f) See conditions from e). How can you determine the speed of the block at a certain point?
  4. Nov 19, 2005 #3


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    e) The block loses contact when the normal force is equal to the centripal force, then there is nothing no keep the block on the sphere. Or in other words, when the 2 accelerations are equal, that explains the two conditions.
  5. Nov 19, 2005 #4
    Thank you! I just don't get what you did with theta. I see you used pi/2, but the picture of 1/4 of a circle I drew doesn't mean the block will go off the circle at pi/2....

    Concerning e) I know the block loses contact when Fn =0 (do you mean that?), therefore Ac= (Fz cos(0) -Fn)/m becomes Fz cos(0)/m= mg cos(0)/m= g cos(0) and condition 2... well all I know is that that's the formula for the centripetal acceleration :grumpy: But is that all I needed to answer there? I mean at any point on this movement on the sphere there are these two conditions, only g cos(0) would be (Fz cos(0) -Fn)/m ....

    About f) well I know v= 2pi s/T, so v= 2pi R 0/T. Substitution in condition 2 gives Ac= (4 pi^2 R 0^2)/T^2 but I don't know what to do now....
  6. Nov 19, 2005 #5
    Oh thanks Daniel! I missed your post when I posted mine :) I think I understood e) now....I'm just figuring it out for myself...
  7. Nov 19, 2005 #6


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    That isn't precise, the normal force will always be bigger than 0 as long as the theta_c is bigger than zero. But even if the normal force isn't zero, the block will lose contact when the sum of the forces on the block is zero.
    EDIT: Sorry Lisa, I just missed your post as I was posting mine:rofl:
  8. Nov 19, 2005 #7
    Use conservation of energy. The potential energy at the top is mgR, right? Now find the potential energy at the point it loses contact, as well as the kinetic energy. It will be useful to note that:


    ...now do a little algebra and you're done.
  9. Nov 19, 2005 #8
    That's a bit of a problem.... I also thought of that way, but my teacher told me I need to do it without using any energy laws, because he wants me to know to work with other laws... I know it REALLY sucks....
  10. Nov 19, 2005 #9


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    I just showed that your thinking in placing theta where you placed it was correct. pi/2 has nothing to do with the point where there's no longer contact. It's merely a right angle.

    Although correct, I don't quite appreciate you cramming in and ruining the thought process for the OP. This is especially irritating as you posted before the OP had responded to my reply.
  11. Nov 19, 2005 #10


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    Sorry Päällikkö.
  12. Nov 19, 2005 #11
    So if I understood it correctly the answer to e is:

    'The block gets off the sphere when Fn= Fz,|=Fmpz therefore if Fn= mg cos 0 = mv^2/r and (according to F=m*a) when a= mg cos 0/m= g cos 0 needs to equal a= (mv^2/r)/m= v^2/r'
  13. Nov 19, 2005 #12
    I think I still don't get it, when Fn= Fz,| there is no Fmpz and ac=0 therefore ? Why does it need to equal g cos(0) and v^2/R ???
  14. Nov 19, 2005 #13
    And what about f... I still don't know from where to start :S....
  15. Nov 19, 2005 #14


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    What is Fmpz?

    Fc = Fz,| - Fn
    c indicating centripetal.

    When is there no contact?
  16. Nov 19, 2005 #15
    Oops sorry :D Fmpz= Fc :D My book is Dutch and calls the centripetal force 'middelpuntzoekende kracht' (Fmpz)
  17. Nov 19, 2005 #16
    What I ment is when there is no contact Fz,| = Fn and Fc= 0 therefore... but why is ac= g cos 0 and v^2/r ? If Fc=0 ac needs to be 0 too right?
  18. Nov 19, 2005 #17


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    When there's no contact, there's no supporting force from the surface.
    Thus Fn = 0.

    Ok :smile:. Maybe I should add some Finnish terms to add to the confusion :smile:.

    EDIT: As a sidenote, using (i)tex-tags or sub and sup -tags clarifies the notation quite a bit. I now know what you mean by different markings, but just a hint for future posts, as I usually rather reply to posts where the problem and work are clearly stated.
    Last edited: Nov 19, 2005
  19. Nov 19, 2005 #18
    Ah well in that case my original thought was correct right?:
  20. Nov 19, 2005 #19
    Thanx for the tip! I didn't know it was possible! I'll surely use them in the future!
  21. Nov 19, 2005 #20


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    I'm not quite sure what you meant here, though.
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