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Kinematics angular speed

  1. Aug 15, 2016 #1
    1. The problem statement, all variables and given/known data
    I have chosen my axis as you can see on the picture. To the left, it is the positive 'r'-axis, up is the positive 'y'-axis. I have to calculate the angular speed. My solution is correct, but i'm getting into trouble at the end. I have to take the roots of the last equation to become the angular speed. Because my acceleration 'ar' is positive, I become a negatif number so I can't take the root. What am I doing wrong?


    2. Relevant equations


    3. The attempt at a solution
    YwxHKYw.png
    JfOC0sf.png

    Thank you!
     
  2. jcsd
  3. Aug 15, 2016 #2

    Doc Al

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    Can you provide a statement of the problem?
     
  4. Aug 15, 2016 #3

    jtbell

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    If the original statement is not in English, I suggest that you tell us both the original statement (in whatever language it's in), and then your attempt at translating it. That would be useful if someone reads this, who knows your language.
     
  5. Aug 15, 2016 #4
    Sorry thought the question was in the picture.

    The question is, calculate the angular speed so that r = 0,72 m. Everything else that you need is in the picture.
    The spring is 0,4 m long in rest. Spring constant is in the picture.

    Thanks guys!
     
  6. Aug 15, 2016 #5

    Doc Al

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    Shouldn't we have the mass?
     
  7. Aug 15, 2016 #6

    Doc Al

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    It's difficult to see what you're doing that leads you to try to take the square root of a negative number.

    The force and the acceleration all point to the left (in your diagram). No need for any negative numbers. (Or if you use them, they will cancel out.)
     
  8. Aug 15, 2016 #7
    http://blob:http%3A//imgur.com/fe849a41-c1e6-4557-98ab-d2fc40a81c18 42OFqZj.png

    I have to put my acceleration, calculated out of the free body diagram, in the ar = ... equation. Since my ar is positive and r(double point) = 0, I can't take the root to find my angular acceleration. If something is not clear enough, just say me. Thanks!

    EDIT: The mass is 3 kg.
     
  9. Aug 15, 2016 #8

    Doc Al

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    The negative sign in the equation ##a_r = - r \dot{\theta}^2## just indicates the direction of the acceleration, which is centripetal. When doing your calculation, you'll set the magnitude of the acceleration equal to ## r \dot{\theta}^2## and leave off the minus sign.
     
  10. Aug 15, 2016 #9
    Thank you. But how do I know this? Is 'r' a vector? I see the formulas just as a formula where I have to fill in my value of something without changing signs in the formula. How do I know when I can change a value? Maybe this is a weird question, but I want to get the bigger view. Thank you
     
  11. Aug 15, 2016 #10

    Doc Al

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    You should not just plug numbers into a formula blindly without first understanding what it means. ar is a component, which can be negative or positive depending on the direction. In this case, the negative sign tells you that the direction is negative (which is toward the axis).
     
  12. Aug 15, 2016 #11
    Yes indeed, I know that the acceleration sign is - or + if a result is negatif, it is poiting in the other direction. But in this case, my ar was positive. How did I know that the right part of the equation to calculate the angular speed must be positive and NOT negative? This is not the acceleration ... Do you understand me?

    Or on another view. Because my ar is positive, the right part SHOULD be also positive so I can remove the minus sign? It sound weird to me.
     
  13. Aug 15, 2016 #12

    Doc Al

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    The way I look at it is this. You solved for the acceleration using Newton's 2nd law and a force analysis. Then you used an expression for centripetal acceleration. I would have used ##a_r = \omega^2 r##. The only reason your expression has a minus sign is that you are "borrowing" from a more elaborate treatment of cylindrical coordinates and the signs indicate direction. No real need for that if you know the formula for centripetal acceleration in terms of ω.

    Whenever a minus sign appears in a formula, you need to understand what it means. Here it just shows direction, since we're discussing components of vectors. (By the way, if you consistently used a minus sign to indicate "left" then your value for force and acceleration would also have been negative, and thus the negative signs would have canceled out.)
     
  14. Aug 15, 2016 #13

    Doc Al

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    Realize (to repeat myself a bit more clearly) that your ar is positive means that you are using a sign convention where left is positive. But the formula you grabbed for the centripetal acceleration uses the opposite sign convention.
     
  15. Aug 15, 2016 #14

    David Lewis

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    Plus or minus sign gives you the directional sense (or sense) of the vector. The direction of the vector's line of action can be spelled out in a diagram or represented by a unit vector in a Cartesian coordinate system.
     
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