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Tangential accn.

  1. May 30, 2006 #1
    A particle starts to slide down a fixed smooth sphere from its top.When it breaks off the sphere,what is its tangential acceleration?
     
  2. jcsd
  3. May 31, 2006 #2

    mukundpa

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    The tangential acceleration is given by the component of its weight in tangential direction, because the only other force acting on the particle is the normal reaction.

    For this first you have to calculate the angle of the radius vector at which the particle leaves the sphere. (cos q = 2/3 ?)

    MP
     
  4. Jun 1, 2006 #3
    Before you start thinking about the tangential acceleration, you will need to figure out when the particle breaks off the sphere. Under what condition does the particle "fall off" the sphere? That is why mukundpa stated that [tex]\cos{q}=\frac{2}{3}[/tex]
     
  5. Jun 1, 2006 #4
    How did you get [itex] cos \theta = \frac{2}{3} [/itex]? The condition which you speak of is that the reaction of the particle to the surface of the sphere is zero?
     
  6. Jun 2, 2006 #5
    Doesn't matter, I realised after posting that that I did get [itex] cos \theta = \frac{2}{3} [/itex], its just I was working it out using a different angle, and so kept on getting [itex] sin \theta = \frac{2}{3} [/itex].
     
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