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How to find angular acceleration

  1. Apr 11, 2013 #1
    a)t=0 t=7
    190/60*2pi=19.8 2950/60*2pi=308
    308/7=41

    I dont know what to do with acceleration with t=0.

    b)for tangental I used the equation alpha*R

    for centripetal I used w^2*R

    c) I think the equation might be applied
    atan(atan/acen)
     

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  2. jcsd
  3. Apr 11, 2013 #2

    haruspex

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    You are told the acceleration is constant over the 7 seconds. The question asks for the acceleration at t=0+ε.
     
  4. Apr 11, 2013 #3
    what do you mean? what does the E stand for?
     
  5. Apr 11, 2013 #4

    haruspex

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    It's epsilon, a fairly standard notation for an arbitrarily small (usually positive) quantity. In other words, it isn't asking for the acceleration exactly at t=0 (which would be impossible to determine) but the acceleration just a fraction later.
    You are told the acceleration is constant over the 7 seconds, but that could mean over the interval [0, 7] or [0, 7) or (0, 7] or (0, 7). The point is that 0+ε lies in all of them, whereas 0 only lies in the first two.
     
  6. Apr 11, 2013 #5
    so can I calculate the acceleration at 1 sec?
     
  7. Apr 11, 2013 #6

    haruspex

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    What's to calculate? The acceleration is constant, and you already know what it is .
     
  8. Apr 11, 2013 #7
    so acceleration is the same being 41.28?
     
  9. Apr 11, 2013 #8

    haruspex

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    That's it.
     
  10. Apr 11, 2013 #9
    I have part a and b
    for part c I used atan(atan/acen) to find angle but its incorrect.
     
  11. Apr 11, 2013 #10

    haruspex

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    That's the right principle. Maybe you made an arithmetic error, or perhaps you need to consider that the radius, as a vector, points outwards, whereas the acceleration points somewhat inwards. (I.e., the centripetal acceleration is negative).
     
  12. Apr 11, 2013 #11
    for t=0 I did 7.4(7.4/71.2)=0.76
    t=7 7.4(7.4/17177)=0.0031
     
  13. Apr 11, 2013 #12

    haruspex

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    You seem to have confused yourself by an overloaded abbreviation. You wrote previously "atan(atan/acen)", which I interpreted as arctan(atan/acen), since that is correct, but in your numerical working you have interpreted it as atan(atan/acen)
     
  14. Apr 12, 2013 #13
    so im suppose tan-1(7.4/71.2) ?
     
    Last edited: Apr 12, 2013
  15. Apr 12, 2013 #14

    haruspex

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    Yes. Do you see why? Draw a right-angled triangle with the side opposite angle theta representing the tangential acceleration, 7.4, and the (non-hypotenuse) side adjacent representing the centripetal acceleration, 71.2. What would be the value of theta?
     
  16. Apr 12, 2013 #15
    yes i do ,aslo my calculator was in radians at first so I got confused, Thanks for the help!
     
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