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Circular Motion - particle in a bowl

  1. Sep 1, 2008 #1
    For the above problem i got the answer. But i got another question

    I really apprecaite if any one can give an answer for this question
    This is the question

    A particle of mass 0.3kg moves with an angular velocity of 10rad per second in a horizontal circle 20cm inside a smooth hemispherical bowl. FInd the reaction of the bowl on the particle and the radius of the bowl.

    Here i found the centripetal of acting on the object
    F=0.3*10^2*20/100
    F=6N

    Is this the reaction?
    After this i don't know how to find the radius
     
  2. jcsd
  3. Sep 1, 2008 #2

    Doc Al

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    Staff: Mentor

    You found the centripetal force acting on the particle. That's only one component of the reaction force.

    Draw a diagram of the particle showing all the forces acting on it. Hint: What can you deduce from the fact that the bowl is smooth?
     
  4. Sep 1, 2008 #3
    ohhh yes if it is smooth centripetal force is the only componant of the reaction force as there is no friction
    But how can i find the radius of the bowl??
     
  5. Sep 1, 2008 #4

    Doc Al

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    Staff: Mentor

    True, there is no friction, but that does not mean that the reaction force is horizontal. Hint: Consider the vertical forces acting on the particle.
    First find the direction of the reaction force, then use some trig.
     
  6. Sep 1, 2008 #5
    I think the centripetal force we found should be normal to the inner surface of the bowl
    as centripetal acceleration always goes through the center of the circle ,,,i got a triangle which one side is 20cm and its hypotenus is the centripetal force which is 6N
    After that i got stucked..
     
  7. Sep 1, 2008 #6
    ohh sorry its not the centripetal acceleration its the centriprtal force that goes through the center of the bowl
     
  8. Sep 1, 2008 #7

    Doc Al

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    Staff: Mentor

    But the circle is horizontal, thus the centripetal force and acceleration will be horizontal.

    The force that the bowl exerts on the particle is normal to the surface of the bowl. What component of that force equals the centripetal force?
     
  9. Sep 1, 2008 #8
    I think we have to resolve the normal reaction force excert by the bowl on the particle like "Rcos theta"
    then it is equal to the 6N which is centripetal force
    I don't know ,,,is it right??
     
  10. Sep 1, 2008 #9

    Doc Al

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    Staff: Mentor

    Good. Now analyze the vertical components.
     
  11. Sep 1, 2008 #10
    i think i got now
    Rcos theta= 6N as centripetal force is 6N
    Rsin theta = 3N as mass of the particle is 0.3kg
    therefore:
    tan theta = 1/2
    theta = 26 deg and 34 mintues
    then
    from the triangle
    cos theta =20/radius
    we know theta therefore we can find cos theta
    cos theta = 0.8954
    0.8954 = 20/radius
    radius= 20/0.8954
    By solving this you can get the lenght of radius
    I hope this is right
    Is it?????
     
  12. Sep 1, 2008 #11

    Doc Al

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    I didn't check your calculations, but that's exactly the right idea. Good!
     
  13. Sep 1, 2008 #12
    thank you for your help
    I appreaciate it
     
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