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I Circular motion in a cone

  1. Jan 11, 2017 #1
    Hello
    I am a little confused by the following problem: Mass in cone: A particle of mass m slides without friction on the inside of a cone. The axis of the cone is vertical, and gravity is directed downward. The apex half-angle of the cone is θ, as shown. The path of the particle happens to be a circle in a horizontal plane. The speed of the particle is v0. Draw a force diagram and find the radius of the circular path in terms of v0, g, and θ. I have arrived at the following solution which I assume is correct r = v^2 * tan (θ) / g. I arrive at my solution by using the equation: N(normal force)* sin(θ)=mg (since there is no vertical accelaration). But this means that the normal force is larger than mg. This intuitively seems strange to me, isn't it mg that creates the normal force in the first place? I am probably missing some very basic detail..
     
  2. jcsd
  3. Jan 11, 2017 #2

    BruceW

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    Homework Helper

    Hi,
    your solution looks good to me !

    It's true that the Normal force is a reaction to the mass being pushed into the cone, but there is no reason for the normal force to be less than the gravitational force on the block. The normal force will be whatever it needs to be, to stop the mass from squeezing through the hard cone.
     
  4. Jan 11, 2017 #3

    Doc Al

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    Staff: Mentor

    The mass is accelerating and the normal force will adjust accordingly.

    Here's an example that might be clearer: Imagine yourself on a scale in an elevator. At constant speed, the normal force equals your weight. But what about when the elevator begins accelerating upward?
     
  5. Jan 11, 2017 #4
    Thanks for the replies, I am still a little confused though. In my free body diagramm I have the gravitational force mg pointing down. I then decompose mg into 2 components: one perpedicular to the cone and one paralell. Isn't it only the perpendicular component that causes N?
     
  6. Jan 11, 2017 #5

    Doc Al

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    Staff: Mentor

    You could analyze things that way, but realize that the acceleration has a component perpendicular to the surface.

    Easier to use vertical and horizontal components. (Decompose the normal force into components, not the weight.)
     
  7. Jan 11, 2017 #6
    But why isn't the normal force equal in magnitude to the perpendicular component of the weight?
     
  8. Jan 11, 2017 #7

    Doc Al

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    Staff: Mentor

    Because there's a component of acceleration in that direction.

    For example: A block sliding down an incline. In that case, the acceleration is parallel to the surface thus ΣF = 0 perpendicular to the surface. So N - mgcosθ = 0.

    But in your situation, perpendicular to the surface ΣF ≠ 0.
     
  9. Jan 11, 2017 #8
    I think it's starting to make sense now (thanks for being patient). The problem was that I was thinking of the normal force as a reaction to the weight so it didn't make sense to me that it could be larger. I still don't really understand what is "causing" the normal force to be so large when the only other force is gravitation.
     
  10. Jan 11, 2017 #9

    jbriggs444

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    Science Advisor

    The contact force of ball on cone is equal and opposite to the contact force of cone on ball. The gravitational force of earth on ball is equal and opposite to the gravitational force of ball on earth. That's Newton's third law.

    There is no similar relationship between the gravitational force of earth on ball and the contact force of ball on cone.

    Newton's second law comes close. ΣF = ma. If the ball were not accelerating then "ma" would be zero. In that case, the contact force of cone on ball and the gravitational force of earth on ball would have to add to zero. It is easy to use this to build a mistaken intuition that things can only push when they are themselves being pushed. But it just ain't so. If you are accelerating, you can push on one thing without being pushed by something else.
     
  11. Jan 11, 2017 #10
    Ok, it just clicked. I had a lot of things mixed up. I think that mulling this over was very useful, thanks to everyone who helped.
     
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