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Reaction on the particle in terms of angular velocity

  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A particle is attached by means of a light inextensible string to a point 0.4 m above a smooth horizontal table. The particle moves on the table in a circle of radius 0.3 m with angular velocity ω. Find the reaction on the particle in terms of ω. Hence find the maximum angular velocity for which the particle can remain on the table

    Answers: m (g - 0.4 ω2), √5g/2

    2. The attempt at a solution
    At first we find the hypotenuse: 0.42 + 0.32 = 0.52 m. Then we find sin α = 0.6 and cos α = 0.8.

    The maximum angular velocity is:
    R cos α = mg (vertical)
    R sin α = mω2r (horizontal)

    R = mg / 0.8 and R = mω2r / 0.6
    mg / 0.8 = mω2r / 0.6
    0.6 mg = 0.8 mω2r
    g = 0.4 ω2
    ω = √5g/2 or 5 rad s-1 (if g = 10).

    Though I don't know where to start with "Find the reaction on the particle in terms of ω." Isn't it R sin α = mω2r → R = 0.5 mω2?

    Update:
    We have R = 0.5 mω2 and R = mg / 0.8
    0.5 mω2 = mg / 0.8
    0.4 mω2 = mg
    0 = mg - 0.4 mω2
    0 = m (g - 0.4 ω2)

    It does fit the answer, but I don't quite understand how zero (0) represents "the reaction on the particle". Any ideas please?
     
  2. jcsd
  3. Sep 15, 2016 #2
    Start by drawing a diagram of the situation labeling all the forces.
     
  4. Sep 15, 2016 #3
    Did that. This is what I have:
    416fdbf474c1.jpg
     
  5. Sep 15, 2016 #4

    kuruman

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    I am not sure what "reaction" means in this context. From the wording, I guess that it is the normal force exerted by the table. It goes to zero when the mass is just about ready to lift off the surface. You need to write Newton's 2nd law in the horizontal and vertical direction and include the normal force in the vertical direction and the components of the tension in the vertical and radial directions.

    On edit: When I posted, the drawing was not available. You need to include the normal force in the vertical direction, not just the component of the tension.
     
  6. Sep 15, 2016 #5
    If it goes to zero isn't this correct then:
    We have R = 0.5 mω2 and R = mg / 0.8
    0.5 mω2 = mg / 0.8
    0.4 mω2 = mg
    0 = mg - 0.4 mω2
    0 = m (g - 0.4 ω2)
    ?

    You mean this kind of normal force?
     
  7. Sep 15, 2016 #6
    What is the meaning of your last equation? for the record.
     
  8. Sep 15, 2016 #7
    The vertical and horizontal components of force are equally strong and therefore are equal to zero. But again, I am not sure about this part, though it does fit the answer.
     
  9. Sep 15, 2016 #8
    Which means?
     
  10. Sep 15, 2016 #9
    This (0 = m (g - 0.4 ω2)) is the reaction on the particle in terms of ω?
     
  11. Sep 15, 2016 #10
    What I would say is that the vertical component of the tension in the string (i.e. the lifting component of the tension) is equal to the force of gravity resulting in no reaction of the table on the mass (i.e. the mass merely touches the table.

    It is the condition giving the angular frequency of rotation in which the mass just lifts off from the table.
     
  12. Sep 15, 2016 #11
    To finish up how would you write the expression for the reaction force of the table on the mass as a function of ω?
     
  13. Sep 15, 2016 #12
    F = mg
    F = mω2r

    mg = mω2r → mg - mω2r
    ?
     
  14. Sep 15, 2016 #13

    kuruman

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    Look at your drawing. Is the entire force F opposing the weight mg? If not, how much of F opposes mg? Same thing in the horizontal direction. What piece of F provides the centripetal acceleration?
     
  15. Sep 15, 2016 #14
    Just a note about notation that can help a bit. You used the same symbol in F in two different expressions. This is true only for the situation where the rotation is just fast enough to lift the mass off the table. You should first write the general case in which you would distinguish between the weight and the lifting forces of the rotation.

    Ie. Fr = mrω2 and

    Fw = mg

    Then apply the condition that the rotation is just fast enough to lift the mass of the table i.e. Fw = Fr
    Make sure you distinguish the forces first.
     
  16. Sep 16, 2016 #15
    F cos α opposes mg (vertical) and F sin α opposes mω2r.

    Where Fw is weight and Fr = radius / radian?
     
  17. Sep 16, 2016 #16

    kuruman

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    As gleem wrote, it is better to think of it as Fw = mg which is equal to (as you said) F cosα. If you put these two together what do you get? Likewise, put together Fr = mω2r and Fr = F sin α.
     
  18. Sep 16, 2016 #17
    If we put Fr = mω2r and Fw = mg together we have F cos α = mg and F sin α = mω2r
    mg / cos α = mω2r / sin α
    mg sin α = mω2r cos α
    g sin α = ω2r cos α
     
  19. Sep 16, 2016 #18

    kuruman

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    Can you solve this for ω which is one of the questions in the problem?

    Knowing ω, can you find F, which is the other question?
     
  20. Sep 16, 2016 #19
    g sin α = ω2r cos α
    ω = √(g sin α) / (r cos α)

    Isn't it what've I found in the first place?
     
  21. Sep 16, 2016 #20

    kuruman

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    Right. Just to make things neater what is sin α / cos α ? Also, what about F as a function of ω?
     
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