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Angular kinematics question

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    The angular position of a point on the rim of a 18.7 cm rotating wheel is given by θ(t) = 4.7 t2 − 7.2 t +9.7, where θ is measured in radians and t is measured in seconds.

    What is the instantaneous angular acceleration α of the point at time t = 6 s?
    What is the instantaneous tangential (not radial!) acceleration a of the point at time t =6 s?
    What is the instantaneous angular velocity ω of the point at time t = 9 s?
    What is the instantaneous speed v of the point at time t = 9 s?
    What is the average angular speed ωav of the point over the time interval starting time t = 6 s and ending at the time t = 9 s?
    Through what angular displacement Δθ does the wheel turn during this time?


    3. The attempt at a solution
    I get the angular position of 135.7 radian at 6 seconds but I do not understand where to go from there. I cannot just use those for angular velocity/acceleration right?

    Thanks.
     
  2. jcsd
  3. Sep 19, 2009 #2

    rl.bhat

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    If θ is given, how to find the angular velocity and angular acceleration.
    Can you find them by differentiating θ(t)
    What is relation between linear velocity and angular velocity?
     
  4. Sep 20, 2009 #3
    Sorry, I was just introduced to this and I am super confused.
     
  5. Sep 21, 2009 #4

    rl.bhat

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    Angular position of the wheel is given as θ(t) = 4.7 t2 − 7.2 t +9.7
    The angular velocity = ω = d[θ(t)] /dt
    The angular acceleration = α = d(ω)/dt
    linear velocity v = ω*R
    linear acceleration = a = Rα
     
  6. Sep 21, 2009 #5
    I am able to get the last two questions involving average angular speed and displacement because I can simply plug in the numbers but I don't know how to find the instantaneous values. I know how to find the speed up to that point but not specifically at that instant.
     
  7. Sep 21, 2009 #6
    hi, i too have a similar question to that, i was able to get the instantaneous angular velocity but doing derivatives then just plugin in the point, but i dont understand how to get the instantaneous angular acceleration and instantaneous tangential acceleration.
     
  8. Sep 21, 2009 #7

    rl.bhat

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    [tex]Angular acceleration \alpha = \frac{d^{2}\theta}{dt^{2}}[/tex]

    Tangential acceleration = Rα.
     
  9. Sep 21, 2009 #8
    sorry to be a bother, but i still dont quite understand that formula. is it another derivative?
     
  10. Sep 21, 2009 #9

    rl.bhat

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    Instantaneous angular velocity ω = dθ/dt
    Instantaneous angular acceleration = α = dω/dt
     
  11. Sep 22, 2009 #10
    I am also still confused about that formula. (surprise, surprise)
     
  12. Sep 22, 2009 #11

    rl.bhat

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    In he problem θ(t) = 4.7*t^2 - 7.2*t + 9.7
    Can you find the derivative of θ with respect to t?
     
  13. Sep 22, 2009 #12
    I am still awaiting the derivative lecture in calculus. I expect this is why I am unable to continue in physics.
     
  14. Sep 22, 2009 #13

    rl.bhat

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    Sorry. Without the knowledge of derivative you cannot solve this problem.
     
  15. Sep 22, 2009 #14
    Thanks for your help, this makes much more sense to me now. I guess I cant do my physics assignment though. haha
     
  16. Sep 22, 2009 #15

    rl.bhat

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    If you are very much particular about the physics assignment, why can't you open your maths book and go through basic rules of derivative?
    Just one rule is sufficient.
    If y = x^2, then dy/dx = 2x.
    And if y = x, then dy/dx = 1.
     
  17. Sep 22, 2009 #16
    so whats the rule for d^2y/dx^2?

    as in [tex]Angular acceleration \alpha = \frac{d^{2}\theta}{dt^{2}}[/tex]
     
  18. Sep 22, 2009 #17

    rl.bhat

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    If θ = t^2
    then dθ/dt = 2t
    and d^2(θ)/dt^2 = 2
     
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