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Homework Help: Circular Motion problem

  1. Jun 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Please see attachment my problem.

    2. Relevant equations

    Law of conservation of energy
    Newton's Second law for circular motion

    3. The attempt at a solution

    Please see attachment.
    I didn't get the answer same as solution. Please help me.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 9, 2010 #2

    ehild

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    R2 and R1 are vectors, of opposite directions. You calculated their magnitude. Determine the vector difference.

    ehild
     
  4. Jun 9, 2010 #3
    Thank you for quick reply. This is the old exam question. But I need magnitutude of R2-R1. Please see the exam solution.
     
  5. Jun 9, 2010 #4

    ehild

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    Oops. Take care of the direction of R1 and the condition that theta should be less than 33.7 degrees. Is then cosθ lower or higher than 2/3? Just think of the case when θ=0. What is the direction of R1?

    ehild
     
  6. Jun 9, 2010 #5
    If Θ=0, direction of R1 is toward the centre at any pointl. I didn't get idea as well.
     
  7. Jun 10, 2010 #6
    I confused one thing is in solution,
    R1=mgcos(theta)-mv[2]/a

    I got
    R1=mv[2]/a-mgcos(theta)

    Could you explain for this? Thanks a lot.
     
  8. Jun 10, 2010 #7

    ehild

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    It is

    R1=mgcos(theta)-mv^2/a,

    because R1 points upward, away from the centre. The direction of the force R1 is wrong as shown in the figure. Just think: What is the direction of R1 at the top, so as the bead do not fall down?

    As R1 is the magnitude of the force, it can not be negative. But your formula for R1 results in a negative number.



    ehild
     
    Last edited: Jun 10, 2010
  9. Jun 10, 2010 #8
    No. This is vertical circular motion. For bead threaded onto the wire , direction of normal force is toward the centre at any point. If the particle is moving outside the loop, normal force is outward.
     
  10. Jun 10, 2010 #9

    ehild

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    It is a bead on a wire. The normal force can be both outward and inward. How can the bead be in rest at the top, if two downward forces act on it?

    Imagine you have this loop of wire, and you hold a bead to it at the top, just below the wire so the normal force acts downward. Release the bead, what happens?

    ehild
     
  11. Jun 10, 2010 #10
    According to the mechanics text books, only we consider one normal force. Not two normal force. Please see Application of Mathematics or Mechanics about vertical circular motion.
     
  12. Jun 10, 2010 #11
    At the top two down forces acting on it. Normal force and weight. For roller coaster, at the top , normal force and weight directions are downward as well.
     
  13. Jun 10, 2010 #12

    ehild

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    Yes, but the roller coaster has some speed at the top. If it does not move, it falls down.

    The centripetal force is the resultant of normal force and gravity.

    mv^2/R = N +G. N=mv^2/R-G, which can be either positive or negative, depending on v.

    Why don't you calculate the value R1 with your formula for a small angle, for example 10 degrees, instead of sticking to the statement that "For bead threaded onto ("on" in the original text) the wire , direction of normal force is toward the centre at any point"? The bead has a hole, and the wire goes through it. The wire can exert force both toward and away the centre, as it can be in contact with opposite walls of that hole in the bead.

    I have to leave now. Think about what I have said.

    ehild
     
    Last edited: Jun 10, 2010
  14. Jun 10, 2010 #13

    Dear ehild,
    Thanks a lot for explaining patiently. When is velocity negative? I didn't see any velocity is negative in my text book. We consider force only. If the force is same as acceletatin direction, usually we take positive.
     
  15. Jun 10, 2010 #14

    ehild

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    Velocity is a vector, it can have any directions. It is speed, the magnitude of velocity, that can not be negative.

    You are right, the force is considered positive with respect to the acceleration if it has the same direction as that of the acceleration.

    Now, about normal force: The (centripetal) acceleration points toward the centre. If the bead is at the top of the circle, gravity G=mg points also towards the centre so it is positive. The centripetal force is the sum of gravity G and the normal force R1. All these forces are vectors, they act along the vertical direction now. You found that R1=mv2/a-mgcos(theta). Calculate, R1 for theta =0, 30° and 40°, please, and tell me what you got.


    ehild
     
  16. Jun 11, 2010 #15
    I got these values.
    R(40)>R(30)>R(0)
     
  17. Jun 11, 2010 #16

    ehild

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    It is all right that R1 increases with the angle, but is it positive or negative?

    You wrote that

    R1= mv12/a-mgcos(θ),

    and

    1/2 mv12=mg(1-cos(θ)),

    right?

    So

    R1=mg(2-3 cos(θ)).

    If θ=0, R1=mg(2-3)=-mg negative.
    If θ=30°, R1=mg(2-2.598)= -0.598 mg, negative.
    If θ=40°, R1=mg(2-2.298)= -0.298 mg, negative.

    You need the magnitude of the normal force. It can not be negative. Your formula is wrong.
    The magnitude of the normal force is the negative of the values above. The correct form is

    R1=|mg(2-3 cos(θ))| =mg(3 cos(θ)-2)

    ehild
     
  18. Jun 11, 2010 #17
    If R is negative, the particle will fall out from the circular loop.

    Thank you very much ehild. Now I got a lot of knowledge. Really thanks.
     
  19. Jun 11, 2010 #18

    ehild

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    Well, this one will not fall down, as it is a bead, (a small ball with a hole through the middle) and the wire goes through the hole.

    ehild
     
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