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Multiple Choice Kinematics Problem

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=71151&stc=1&d=1404893091.gif

    2. Relevant equations



    3. The attempt at a solution


    ## α = v\frac{dv}{ds} ## and K is the kinetic energy .


    In all the four cases ## α = v\frac{dv}{ds} = mgsinθ ## where θ is the angle which the normal to the body makes with the vertical .

    Fig 1) K increases and tangential acceleration α decreases ,so matches with P)

    Fig 2) K decreases and tangential acceleration α also decreases ,so matches with S)

    Fig 3) K increases and tangential acceleration α also increases ,so matches with R)

    Fig 4) K decreases and tangential acceleration α increases ,so matches with Q)

    According to me the correct option is (C)

    Is this the correct option ?

    Many thanks
     

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  3. Jul 9, 2014 #2

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    I think you accidentally added an [itex] m [/itex] in there that doesn't belong.

    Yes, your choice looks correct to me. :approve:

    In case you missed it though, there is another way to express the [itex] \alpha = v \frac{dv}{ds} [/itex]. Can you represent [itex] v [/itex] in terms of [itex] ds [/itex] and [itex] dt [/itex]? if so, replace [itex] v [/itex] with that in your equation and see what happens. It should then be quite clear why the [itex] g \sin \theta [/itex] makes perfect sense. :wink:
     
  4. Jul 9, 2014 #3
    Thanks :)

    [itex] v = \frac{ds} {dt}[/itex] and [itex]α = \frac{dv}{dt}[/itex] which is the tangential acceleration .

    Is this what you are suggesting ?
     
  5. Jul 9, 2014 #4

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    That's it. :smile: It was just to point out that [itex] \alpha [/itex] is the ball's acceleration.
     
  6. Jul 9, 2014 #5

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    And yes, just to avoid ambiguity, [itex] \alpha [/itex] is the tangential component of acceleration, as you have correctly suggested. :smile: (There's also a normal component, but that's not relevant for this problem.)
     
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