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Weight after sliding off a ramp

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  1. Jan 25, 2017 #1
    1. The problem statement, all variables and given/known data
    A ball with mass of m starts sliding off a ramp with height h = 3R and when it slides off the ball does a "loop" once. ( Just see the picture , it's hard to explain :biggrin:). I have to find the weight of the ball at the bottom and at the top of the circle.

    h = 3R

    2. Relevant equations
    Ek = mv2 / 2
    Ep = mgh
    Work = F * h
    3. The attempt at a solution

    I had found the weight of the ball at the top of the circle, although not sure if correct:

    I found the potential energy (or work) at the top of the circle, which is 2Rmg
    Weight = Work / height = 2Rmg / 2R = mg

    And then I have to find the weight at the bottom of the circle, before doing the first loop, but I am not exactly sure how. A good hint would be awesome :woot:
     

    Attached Files:

  2. jcsd
  3. Jan 25, 2017 #2

    BvU

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    Hi,

    Your relevant equations don't look as if they have anything with weight in them ? Also, ##h## appears but is never used after that.

    Please inform us why the weight of the ball should be different at the top and at the bottom ? What is the definition of weight you intend to use ?
    what is the reference point for your 'Work' ?

    To me, weight is ##mg## just like what you end up with (by coincidence ?) However, if we agree that weight is mass times vertical acceleration,
    ( [edit] Not good. If it lies still on the table, the weight isn't zero -- you' ll have to help me there :smile:)
    then we might get
    answers for top and bottom while in the loop that differ from ##mg##.

    What kind of trajectory is the loop ? Any equations for the acceleration ?

    You also have to make an approximation: the size of the ball isn't given, so the best you can do is assume it can be ignored.

    PS IMHO this is a rather badly thought out exercise; we'll see how far we can get.
     
  4. Jan 25, 2017 #3
    My guess is that Weight is different at the top and at the bottom because of different value of acceleration. I have correct answers, but I don't really know how to get the bottom weight. Bottom is 6mg, top is mg.

    For Work, I did : Work = 2Rmg ( the work of force of gravity).

    The loop has a form of a simple circle, as seen in the picture. No given equations or acceleration.
     
  5. Jan 25, 2017 #4

    BvU

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    You will need something. If it isn't given, you're supposed to know !
    Let's try to simplify a little first: what is the (expression for the) horizontal velocity at the end of the ramp when it enters the loop ? (hint: you can use your first two relevant equations for that)
     
  6. Jan 25, 2017 #5
    v = sqrt ( 2g * 3R) ? Got from kinetic and potential energy formula.
     
  7. Jan 25, 2017 #6

    BvU

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    Excellent. Now, at that moment it is describing a loop with radius R. What is the expression for the acceleration needed to describe a circular loop ?
     
  8. Jan 25, 2017 #7
    a = v2 / R ? :-p
     
  9. Jan 25, 2017 #8

    BvU

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    The smile indicates you start to see the light !
    Now what about the top of the loop ?
     
  10. Jan 25, 2017 #9
    Ok, I seem to have got correct answer for the bottom, 7mg ! ( I accidently typed 6mg before :P) Trying to get for the top now.
     
  11. Jan 25, 2017 #10
    Velocity at the top is zero, because potential energy is at maximum ? In that case, I get the answer mg, right? :rolleyes:
     
  12. Jan 25, 2017 #11

    BvU

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    Is that so ? What about ##E_{\rm kin} = \Delta E_{\rm pot}## ?
     
  13. Jan 25, 2017 #12
    I'm a little bit confused. Velocity is still sqrt( 6Rg ) at the top then ? Or rather, sqrt ( 4Rg ), because of 2R height ?
     
  14. Jan 25, 2017 #13

    BvU

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    One way is to ask, the other is to calculate. You found ##{1\over 2} mv^2 = mg\; 3R## for a height difference of 3R. What is the height difference when it's at the top of the loop ? So what's ##v^2## there ? So what's ##a## there ? So what do you get for the 'weight' ?
     
  15. Jan 25, 2017 #14
    The height difference is 1R, v is sqrt ( 2Rg ), and weight is 3mg, is that right ? :-p
     
  16. Jan 25, 2017 #15

    BvU

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    No; work a bit more meticulous. At the bottom you added 6 and 1. Which way were they pointing ?
     
  17. Jan 25, 2017 #16
    1 was pointing downwards, while 6 upwards. So do you mean that when the ball is at the top, 6 is pointing downwards as well ? Sorry, I'm a little lost right now :biggrin:
     
  18. Jan 25, 2017 #17

    BvU

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    At the top it's not 6
     
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