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What is the tension just before and just after touching the nail?

  1. Jun 10, 2013 #1
    1. The problem statement, all variables and given/known data
    In the given system(see attachment), when the ball of mass 'm' is released from A, it will swing down the dotted arc as shown. A nail is located at a distance 'd' below the point of suspension (O) at M.
    (a) What is the tension just before and just after touching the nail?





    2. My Attempt
    (a)See attachment
    Attachment 1 has the situation.
    Attachment 2 has the FBD when the sphere is just about to touch the nail.
    Attachment 3 has the FBD just after when the sphere has touched the nail.
    From attachment 2, equation become as:
    Net force towards the centre= Centripetal Force = (mV^{2}_{B})/l = T[itex]_{i}[/itex]-mg
    →T[itex]_{i}[/itex]-mg = (mV^[itex]^{2}_{B}[/itex])/l (Equation 1 )
    As we have V^[itex]^{2}_{B}[/itex]=2gl( By conservation of energy from position A to position B )
    so we get T[itex]_{i}[/itex]=3mg

    3. My Doubt
    My doubt was that according to equation 1, the net vertical force is the centripetal force which is radially inwards. So why doesn't the particle go radially upwards to the the centre of the circle's path it is follwing? The same doubt arises for the equations of "FBD of sphere just after touching the nail".. Please reply soon..
     

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    Last edited: Jun 10, 2013
  2. jcsd
  3. Jun 10, 2013 #2

    mfb

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    It is accelerated upwards, but it has a horizontal velocity component, so it does not go radially inward.
     
  4. Jun 10, 2013 #3
    What is the horizontal component of acceleration? At the lowermost position? Tangential? Ok, so what would be its magnitude? Maybe I know it but just conforming..would it be the rate of change of velocity at B? That is d([itex]\sqrt{2gl}[/itex]/dt? And the direction would be same as that of velocity, that is tangentially?
     
  5. Jun 11, 2013 #4

    mfb

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    Zero. There is a horizontal component of the velocity, that is important for the trajectory.
    I don't understand what you are comparing here. A velocity is not the same as an acceleration.
     
  6. Jun 11, 2013 #5
    Sorry, I should have been more precise..I mean "rate of change of velocity with respect to time"
    would that give the horizontal acceleration at the lowermost point?
     
  7. Jun 11, 2013 #6

    mfb

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    Do you expect a non-zero rate of change there?
     
  8. Jun 12, 2013 #7
    Yes because it is a vertical circle and velocity decreases as the the sphere goes up due Loss in KE= Gain in PE
    So dv/dt would be negative means the sphere would be retarding...I guess this acceleration would be the acceleration caused due to one of the components of gravity when resolved along tangential direction..is that correct..hmmm and what you said is true..there won't be any horizontal acceleration to the ball at the lowermost point the ball goes tangential but the radial acceleration pulls it back..
     
    Last edited: Jun 12, 2013
  9. Jun 12, 2013 #8
    But my question is what makes the sphere go along the circle everytime even though there is a centripetal force present..why doesn't particle go towards the centre of the circle its following because "centripetal" force in centre seeking so why doesn't particle want to seek the centre? What stops it from doing so?
     
  10. Jun 12, 2013 #9

    mfb

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    It has no reason why it should go towards the center. As soon as it would come closer to the center, the wire cannot not provide the required tension any more. Neglecting vibrations, the tension in the wire is always just enough to keep it on its circular track.
     
  11. Jun 14, 2013 #10
    If there is tension in the wire that keeps the sphere then there has to be some outward force on the sphere also so that the inward (tension) force is balanced out..what is that force?
     
  12. Jun 14, 2013 #11

    mfb

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    It has not, the ball is not in an equilibrium - it has a non-zero acceleration and changes its direction of motion according to that.
     
  13. Jun 14, 2013 #12
    So if the ball is in its lowermost point and then you say the ball has a non zero acceleration then the acceleration should be in a direction which is inlclined tothe horizontal. Right? I mean to say the ball has to go up from the left point of the lowermost point. So how do we determine that the acceleration of the ball is along that direction?
     
  14. Jun 14, 2013 #13

    mfb

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    At the lowest point, the acceleration is upwards. It has to be upwards (or downwards) - everything else would change its speed without changing its height, violating energy conservation.
    It is in the direction of the string, if string+gravity are aligned.
     
  15. Jun 14, 2013 #14

    462chevelle

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    centrifugal force would be what would you would call the opposite of centripetal. if you were swinging the same string and ball around in a circle. centripetal force is what makes it go in a circle and centrifugal force is what will make it stay in a horizontal line if your swinging it fast enough. this doesn't answer your question. just a comment.
     
  16. Jun 14, 2013 #15
    But centrifugal force is used only when we are working with a non inertial frame of reference..So as we are taking the observations from the ground or say we ourselves, both being inertial frames of reference so we can't include a pseudo/fictitious force that is the centrifugal force..Thanks still for raising this point.
     
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