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How long the ball remains in contact with the ground when dropped

  1. Jul 8, 2014 #1
    1. The problem statement, all variables and given/known data
    A ball of mass 1kg is dropped from a height 1m . Estimate how long the ball is in contact with the ground in seconds ?

    Details and assumption

    Simplify the question by modeling the ball as an ideal spring of spring constant 100Nm .

    Take gravitational acceleration if necessary as 9.8m/s2

    2. Relevant equations
    Work done by spring = 1/2kx^2.
    Change in kinetic energy = Work done.


    3. The attempt at a solution
    Let the spring stops(ball) after it is bring displaced from Xm from its mean position.
    Using the work energy theorem-
    mgX-[itex]\frac{kx^{2}}{2}[/itex]=0-m/2*2gh
    where m is the mass of spring.I got square of velocity before impact as 2gh(h is the height mentioned above).
    So I got the displacement of spring from its mean position to stop.
    I tried to use equations of motion but acceleration is not constant.So I tried to use calculus.
    I got Acceleration as a function of x =g-kx/m .Could you please help me in solving this question further.
     
  2. jcsd
  3. Jul 8, 2014 #2
    The answer is 0.341 seconds.It can be wrong.I am not sure about it.
     
  4. Jul 8, 2014 #3

    ehild

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    Is not it 0.314 s?

    ehild
     
  5. Jul 8, 2014 #4

    ZetaOfThree

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    Here's a hint: remember that the period of oscillation of a spring is [itex]2\pi \sqrt{\frac{m}{k}}[/itex] where [itex] k [/itex] is the spring constant and [itex] m [/itex] is the mass. See if you can use that fact to solve this problem.
     
  6. Jul 8, 2014 #5
    Yes you are right sir.It was my mistake.
     
  7. Jul 8, 2014 #6
    On putting values in this equation I got t=0.628 seconds.But this value is twice of the answer(0.314) above.
     
  8. Jul 8, 2014 #7

    ZetaOfThree

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    Think about the motion of the ball when it's in contact with the ground. It doesn't oscillate through a full period, does it?
     
  9. Jul 8, 2014 #8
    When the spring touches the ground it is then compressed but after some displacement from the mean position it is not possible to compress it more then spring relax itself.Is this not the full oscillation?
     
  10. Jul 8, 2014 #9

    ehild

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    Not. Imagine a sine function. It starts from zero - goes to maximum- decreases to zero - turns to negative - reaches minimum-goes to zero. That is a full period. The ball can only be compressed, the ground can not stretch it. There is no "negative compression".

    ehild
     
  11. Jul 9, 2014 #10
    Before colliding with the the ground spring was displaced 0cm from its mean position.It is then compressed and it get displaced by maximum displacement in negative direction.It then relax itself and reaches the mean position again.So this oscillation was from zero to minimum and again to zero.So this is only half oscillation.So the required time is half of the total time period.I understood this.Is this right?
     
  12. Jul 9, 2014 #11

    ehild

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    Yes, it is right.

    ehild
     
  13. Jul 9, 2014 #12
    What is the proof of T=2∏√(m/k).
    I have not studied Harmonic motion and oscillation yet.It would be nice if somebody could send me link from where I could study about this.Link of a video will be appreciated.
     
  14. Jul 10, 2014 #13

    ehild

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    You can find lot of videos and pages browsing "simple harmonic motion".

    The equation for the time period T=2∏√(m/k) is equivalent to ω=√(k/m), as the angular frequency is ω=2π/T. The derivation of the equation for ω can be found here, for example.

    Or see http://en.wikipedia.org/wiki/Simple_harmonic_motion

    ehild
     
    Last edited by a moderator: Sep 25, 2014
  15. Jul 10, 2014 #14
    Thank you all of you for helping me.
     
  16. Jul 10, 2014 #15

    ehild

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    Funny things happen to a ball dropped on the ground. See . So the approximation that it is in contact with the ground for half period is very rude. The ball really vibrates like a spring when it is in the air again, but during the time it is in contact with the ground it gets very much deformed and it does not look like performing SHM.

    ehild
     
    Last edited by a moderator: Sep 25, 2014
  17. Jul 11, 2014 #16
    Thank you
     
    Last edited by a moderator: Sep 25, 2014
  18. Jul 11, 2014 #17
    Hello ehild

    Somehow I am not convinced that the ball undergoes half the oscillation , for the reason that ball just touching the floor is not the equilibrium position . I tried a little differently .Not sure if I have got it right .

    Let ##v_0## be the speed of the ball when it comes in contact with the ground . ##v_0 = \sqrt{2gh}## . x axis points vertically downwards with x=0 is the location of CM when the ball first comes in contact with the floor.

    Applying conservation of energy

    $$\frac{1}{2}m{v_0}^2 = -mgx + \frac{1}{2}mv^2+ \frac{1}{2}kx^2$$ . Here 'x' is the displacement of the CM .

    When the ball is fully compressed , v=0 and let x_0 be the maximum displacement of the CM .

    $$\frac{1}{2}m{v_0}^2 = -mgx_0 + \frac{1}{2}kx_0^2$$

    $$ x_0 = \frac{mg}{k}+ \sqrt{\frac{{v_0}^2mk}{k^2}+\frac{m^2g^2}{k^2}}$$

    $$ x_0 - \frac{mg}{k} = \sqrt{\frac{{v_0}^2mk}{k^2}+\frac{m^2g^2}{k^2}}$$

    From the first equation , $$ v = (\sqrt{\frac{k}{m}})\sqrt{\frac{{v_0}^2mk}{k^2}+\frac{m^2g^2}{k^2}-(x-\frac{mg}{k})^2}$$

    $$ v = (\sqrt{\frac{k}{m}})\sqrt{(x_0-\frac{mg}{k})^2-(x-\frac{mg}{k})^2}$$

    Now writing v=dx/dt and integrating with x(0→x_0) and t(0→T/2) doesn't give the time period as ##\frac{\pi}{10}## .
     
  19. Jul 11, 2014 #18
    This means my first step was right.I did it in the question (post#1).I learned it in doing 'Mechanics problem -- 2 masses & spring on a surface' from you and partially by sankalpmittal on this forum.
     
  20. Jul 11, 2014 #19
    I understood it till till eq(2).How did you get eq(3).Have I mentioned the first equation correctly?Is this the equation you were talking about.How did you eliminate x[itex]_{0}[/itex] from eq(2) and get it in terms of v.I am really confused .
     
  21. Jul 11, 2014 #20

    ehild

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    And what is v?

    Remember the problem says that the ball is like a spring. It is not a rigid body. Its parts move with different velocities with respect to the CM. The kinetic energy has contribution from that internal motion.

    If the problem was " a small ball of mass m falls onto the top of a vertical spring from high h, how long is the ball in contact with the spring" then it could be solved by this method.

    ehild
     
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