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Another centripetal Acceleration problem

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data
    (a) What is the tangential acceleration of a bug on the rim of a 78 rpm record 10 in. in diameter if the record moves from rest to its final angular velocity in 3 s? (b) When the record is at its final speed, what is the tangential velocity of the bug? Its tangential acceleration? Its radial acceleration?


    2. Relevant equations
    ω = ωo + [itex]\alpha[/itex]t



    3. The attempt at a solution
    I have solved everything in the problem accept the second and third part of (b). I managed to calculate the tangential velocity of the bug just fine. The way I read this question is that once it hits 78 rpm it stops accelerating and that becomes the final velocity. This would cause the tangential acceleration in part (b) to be zero which my book agrees with. The third part of (b) I don't understand though. The book says that the radial acceleration at that point would be 8.39 m/s. If the tangential acceleration is zero, how could there be a radial acceleration and how would I calculate it?
     
  2. jcsd
  3. Jun 7, 2014 #2

    Nathanael

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    It's talking about the inward acceleration needed to move in a circle at a constant speed

    (The centripetal acceleration from "uniform circular motion")
     
  4. Jun 7, 2014 #3

    SammyS

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    Does an object undergoing uniform circular motion have a non-zero acceleration?
    If no: isn't its the direction of its velocity changing?

    If yes: What is that acceleration called, & what is its direction?​

    Darn !!
    Nathanael beat me !
     
  5. Jun 7, 2014 #4
    I wonder why they would call it the radial acceleration instead of the centripetal acceleration? I followed the formula for centripetal acceleration which is
    a = v2/r
    (1.04)2/.127 = 8.52 m/s which is not the correct answer the correct answer is 8.39. I am using the velocity that I solved for in the first part of (b) which the book says is correct. The radius I calculated by converting inches to meters
    10/39.37/2 = .127
     
    Last edited: Jun 7, 2014
  6. Jun 7, 2014 #5

    SammyS

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    It's the component that's along the radius. It's perpendicular to the tangential acceleration.
     
  7. Jun 7, 2014 #6

    Nathanael

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    It basically just means "along the radius" which is a way of saying "towards the center" which is what centripetal means.

    It's just terminology


    This time you beat me, Sammy! Hahah
     
  8. Jun 7, 2014 #7
    check out my edited post
     
  9. Jun 7, 2014 #8

    Nathanael

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    Weird. I would say the book made a mistake. I have no idea why it would be 8.39

    I used a more specific velocity (1.037) and the answer came out to 8.47 which is a little closer. I don't know why the book says that it's 8.39, I really think it's just a mistake.
     
  10. Jun 7, 2014 #9
    OK cool. Thanks!
     
  11. Jun 7, 2014 #10
    Just as an additional question if the centripetal force is the force pulling inward to the center what is the force pushing out that keeps it in equilibrium? The only other acceleration is the one tangent to the circle and its not pulling outward.
     
  12. Jun 8, 2014 #11

    Nathanael

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    If there was an equal force outward then it would just move in a straight path.

    The centripetal force is the required net force to move in a circle.

    (Centripetal force is not really a "force" in the sense that gravity is a force, it's just the net force that is required for uniform circular motion. It could be caused by tension, friction, gravity, etc.)

    (An example of tension causing it would be spinning an object around on a string. An example of friction causing it would be a car driving in a circle. An example of gravity causing it would be an object orbiting a planet.)
     
  13. Jun 8, 2014 #12

    SammyS

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    (78/60×2 π)^2×5×2.54 = 847.325...

    The book's answer must have some rounding at an intermediate step, or steps
     
  14. Jun 8, 2014 #13
    Oh I get it now. Thanks!
     
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