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Rotational Dynamics of a grinding wheel

  1. Mar 31, 2009 #1
    Ok i am not good with physics and these are some review problems for my test coming up. Please help.

    At time t1 = 0 an electric grinding wheel has an angular velocity of 28.0 rad/s. It has a
    constant angular acceleration of 32.0 rad/s2. At time t2 = 2.20 s a circuit breaker trips, and the
    wheel then turns through another 350 rad as it coasts to a stop at constant angular acceleration.

    (i) Through what total angle did the wheel turn between t = 0 and the time it again
    stopped?




    ANSWER: _______________ Rad

    (ii) At what time did the wheel stop?





    ANSWER: ________________ ____

    (iii) What was the angular acceleration of the wheel as it slowed down?






    ANSWER: ________________ ____


    answers are A1) 489 rad A2) 9.31 sec A3) -13.83 rad/sec^2
    Please show work.
     
  2. jcsd
  3. Mar 31, 2009 #2

    rl.bhat

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    Please show work.
    First of all please show your attempts. What are relevant equations in rotational motion can be applied in this problem?
     
  4. Mar 31, 2009 #3
    Q1(i) Through what total angle did the wheel turn between t = 0 and the time it again
    stopped?

    My attempt was:

    28.0 + 32.0(2.20)^2= 182.88 + 350 = 532.88 rad

    The answer was suppose to be 489 Rad.

    (ii) At what time did the wheel stop?

    My attempt was

    2.20 + 350/? = ???

    I am not sure how i am suppose to find the decrease in acceleration of the angular velocity because i don;t know how long it took and its not given.


    (iii) What was the angular acceleration of the wheel as it slowed down?

    I can't solve for this until i know answer B.
     
  5. Mar 31, 2009 #4

    rl.bhat

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    28.0 + 32.0(2.20)^2
    For this which formula you are using?
     
  6. Mar 31, 2009 #5
    Well for that i was using the

    Change in distance = initial velocity + 1/2 a t^2

    Then i added that to the 350 Rad it said it traveled once the breaker was tripped.

    But i realized i forgot the 1/2 in my equation and end up getting 450.44. Which ends up being wrong again.
     
  7. Mar 31, 2009 #6

    rl.bhat

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    Change in distance = initial velocity + 1/2 a t^2
    It should be
    Change in distance = initial velocity*t + 1/2 a t^2
    Also find the velocity at 2.2 s. This is necessary for the second part.
     
  8. Mar 31, 2009 #7
    Ok The velocity at 2.2 sec is =732 rad/s^2

    I got that by

    Vf=V0 + at
    Vf = 28.0 + (32.0)(2.20)

    However i realized that I couldn't find an equation that didn't have both time and acceleration involved. What would i do to isolate just time while using the Initial velocity as 732 rad/sec^2 and the final velocity as 0 rad/sec^2. With the total distance traveled to be 350 rads?

    It doesn't give me a mass or a frictional force to work with.
     
  9. Mar 31, 2009 #8

    rl.bhat

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    Ok The velocity at 2.2 sec is =732 rad/s^2
    This is wrong. Is it the velocity or change in distance or some thing else?

    Find out three equations of motion.
     
  10. Mar 31, 2009 #9
    wow totally my bad. The answer was 98.4. entered the equation as 2.80+320(2.2)
     
  11. Mar 31, 2009 #10

    rl.bhat

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    Now you can use v^2 -u^2 = 2as to find a and t.
     
  12. Mar 31, 2009 #11
    OK i got the answers for both B and C. I had the right idea but i keep messing up with entering it into my calculator so i need to check that before i post stuff.

    However i got the answer to C first then B.

    c) Vf^2 = V0^2 + 2 a s

    0 = (98.4)^2 + 2(a)(350)
    a=-13.8

    B) t = (Vf-V0)/a

    t= (98.4-0)/13.8
    t= 7.13

    T(total) = 2.2 + 7.13 = 9.33



    If i were suppose to get it in the order of the test goes how would i go about that.

    Thank you for your patience with me and help
     
  13. Mar 31, 2009 #12
    v^2 -u^2 = 2as

    What is the u^2?
     
  14. Mar 31, 2009 #13

    rl.bhat

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    Initial velocity i.e vo.
    You can't find the answers in the given order.
     
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