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Non-uniform circular motion

  1. Oct 28, 2004 #1

    mad

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    Hello all,

    I need help to understand non-uniform circular motion. I know there's a centripetal acceleration and a tangential acceleration and the result of both are the acceleration. But what I dont understand is how can I find the tangential acceleration. I've been searching for a while now and some sites says with the slope of the acceleration (how can I find this if I dont know the total acceleration) and some says dv/dt. Please someone enlighten me on this with explanations and perhaps an example because I have an exam in 3 days and I must master this.

    Thanks a lot
     
  2. jcsd
  3. Oct 28, 2004 #2

    arildno

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    In a circular motion, the tangential acceleration is given as the derivative of the SPEED.
     
  4. Oct 28, 2004 #3

    Doc Al

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    The magnitude of the tangential acceleration can be computed using: [itex]a_t = dv/dt[/itex], where v is the speed ([itex]v = \omega r[/itex]).

    The magnitude of the radial acceleration is [itex]a_r = v^2/r[/itex].

    Try posting a specific problem and your solution if you need more help.
     
  5. Oct 28, 2004 #4

    mad

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    Thanks for the fast responses guys! Thats what I dont understand. If they say the car has a speed of say 20m/s, the diameter is 10 (r=5) then centripetal acceleration is v^2/r. -- that okay

    But the derivate of speed .. would that be 0? (derivate of 20) Or do I need to put it in a formula.. like v^2=Vo^2 + 2ax(x-xo) .. thats what I need help with.
    Thanks!
     
  6. Oct 28, 2004 #5

    arildno

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    You need to know how the speed changes with time in order to calculate the tangential acceleration
     
  7. Oct 28, 2004 #6

    Doc Al

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    You have to be told how the speed is changing. If it's not changing (uniform cirular motion) then the tangential acceleration would be zero. If it's uniformly accelerating, they have to give some info so you can figure it out.

    (I'll bet arildno will say the same thing! :smile: )
     
  8. Oct 28, 2004 #7

    arildno

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    You betted correctly, Doc Al..:smile:
     
  9. Oct 28, 2004 #8

    Doc Al

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    You're too fast for me, arildno. I'm getting old and slow... :biggrin:

    (And congrats on your new medal! :wink: )
     
  10. Oct 28, 2004 #9

    mad

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    Alright, here is a problem I tried to do.. but I really dont know how.

    A car approaches a turn with r= 40m. When its velocity is oriented to the north, its velocity changes at a rate of 2m/s^2 and the total acceleration is oriented 30 (degres) north to the west.
    What is the speed at that instant?

    also, if you guys could give me an example on how to calculate the tangential acceleration in that problem, I would appreciate

    Thanks a lot and sorry for my english
     
  11. Oct 28, 2004 #10

    Doc Al

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    I'm going to guess that when they say the "velocity" changes, what they really mean is that the "speed" changes. So they give you the tangential acceleration--it's 2 m/s^2. You have to figure out what the radial acceleration must be to have a net acceleration in the direction indicated. Once you know the radial acceleration, you can figure out the speed.
     
  12. Oct 28, 2004 #11

    mad

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    I made an illustration. I dont understand how the total acceleration can be like that. Please check the attachment
     

    Attached Files:

  13. Oct 28, 2004 #12

    Doc Al

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    When your attachment gets approved, I'll take a look. But I don't see the problem: I assume the car travels north and is turning west. The tangential acceleration points north and the radial acceleration points west. The total acceleration is the vector sum of those components.
     
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