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Circular motion and centripetal force

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m kg lies on the top of a smooth sphere of radius 2 m.
    The sphere is fixed on a horizontal table at P.
    The particle is slightly displaced and slides down the sphere.
    The particle leaves the sphere at B and strikes the table at Q.
    Find
    (i) the speed of the particle at B
    (ii) the speed of the particle on striking the table at Q.



    2. Relevant equations
    F = mv^2/r

    3. The attempt at a solution
    I had no way to go about this question (however part ii can be easily gotten from conversion from potential energy to kinetic), but it is from a section on circular motion, and in the marking scheme for this question, they write force equations for the mass as if it were undergoing circular motion ie they use the equation for centripetal force F = mv^2/r . The solution is at http://www.examinations.ie/archive/markingschemes/2010/LC020ALP000EV.pdf [Broken] for anyone interested, it's question 6 part a. I'm just questioning their use of the formula for circular motion - surely, the formula only holds if v is a constant, however, the mass is clearly accelerating so this is not the case, so how can they justify using this equation in their solution? Thanks.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 4, 2012 #2

    rock.freak667

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    The formula is being applied at point B, at that point, they are finding the value for v.

    However, the unwritten assumption is that the speed is constant. Otherwise you'd have an additional acceleration term.


    Remember you can travel in a circle at constant speed but still be accelerating because there is a change in velocity due to a change in direction.
     
  4. Apr 4, 2012 #3
    but it starts off at rest so to have any velocity it must have accelerated ie the magnitude of it's velocity increases?
     
  5. Apr 4, 2012 #4

    rock.freak667

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    It accelerated under the influence of gravity. Otherwise you'd have a radial component and a tangential component for acceleration. So it is more or less circular motion with constant speed.
     
  6. Apr 4, 2012 #5

    tms

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    Although the speed is not constant while the particle is on the sphere, the equation for circular motion holds instantaneously. Note that as the particle moves, the tangential and radial components of the force are also changing.
     
  7. Apr 4, 2012 #6
    The point at which the particle leaves the sphere is when it starts free-fall.
    Before this point the particle 'sits' on the sphere because the R>0, =>mgCosθ >mv^2/r.

    Yes, the particle were undergoing circular motion.
     
  8. Apr 6, 2012 #7
    ok i kind of see what you mean, but would it would be right to say that this is not uniform circular motion because the angular velocity is not constant?
     
  9. Apr 6, 2012 #8

    ehild

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    It is not uniform circular motion, but it is circular motion while the particle is in the sphere (point B). While moving on he sphere, its radial component of acceleration is v^2/R. That means the resultant force on the particle must have mv^2/R radial component. What forces act on the particle?

    ehild
     
  10. Apr 6, 2012 #9

    ehild

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    It is not uniform circular motion, but it is circular motion while the particle is on the sphere (point B). While moving on the sphere, the radial component of acceleration is v^2/R. That means the resultant force on the particle must have mv^2/R radial component. What forces act on the particle?

    ehild
     
  11. Apr 6, 2012 #10
    ah ok. gravity and reaction?
     
  12. Apr 6, 2012 #11

    ehild

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    Yes. Note that the reaction force can not pull the particle, it pushes only.

    Apply conservation of energy to get the speed of the particle at a certain position. Check if the normal component of gravity is sufficient to supply the centripetal force.

    ehild
     
  13. Apr 6, 2012 #12
    at B mgcos(theta)=mv^2/r
     
    Last edited by a moderator: May 5, 2017
  14. Apr 6, 2012 #13
    Thanks for the replies, the bit i dont understand is the use of the centripetal force equation though. In the derivation of the centripetal force f = mv^2/r, it was taken that the mass is moving at a uniform speed and it's acceleration vector was towards the centre of the circle, however if the mass is not moving with uniform speed, then the acceleration vector is not towards the centre of the circle, and in this case how can you use the centripetal force equation? Or can the formula be used no matter whether or not the acceleration/force is pointed towards the centre? Or are we just saying the component of force that is directed towards the centre of the circle is mv^2/r but that's only a component of the entire force? Sorry I'm so confused haha thanks.
     
  15. Apr 6, 2012 #14
    ah ok nevermind i think i get it thanks
     
  16. Apr 6, 2012 #15

    cepheid

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    I think we have a winner
     
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