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Need help calculating angular velocity

  1. Nov 4, 2014 #1
    1. The problem statement, all variables and given/known data

    The average car today has a mass of 1100 kg, and when accelerating from rest, covers 0.25 miles in 15 seconds. Each rim and tire together has a diameter of 46 cm and a mass of 9.1kg. If we agree the rim and tire have the shape of a solid disk that rotates through its geometric center, what would be the kinetic energy of one of the tires, in Joules, at the end of the run?

    2. Relevant equations

    ke = 1/2 * I * w^2, I=1/2*r^2*m

    3. The attempt at a solution

    i calculated the moment of inertia to be .24 by I= 9.1kg*.23m^2*.5

    I calculated the angular velocity to be 116.59 rad/s by finding the rev/sec and then converting to rad/sec

    circum = .23*2*pi = 1.4451m

    and the rev: 402.25m/1.4451m = 278 rev. i got 402.25 by converting .25 miles to meters

    278/15= 18.5rev/sec

    18.5rev/sec* 2*pi = 116.59rad/ sec

    in the end i got KE= 1635 j

    i dont know if thats right. i also need help applying significant numbers. i dont know whether to apply them towards the final calculation or throughout the conversions
     
    Last edited: Nov 4, 2014
  2. jcsd
  3. Nov 4, 2014 #2

    mfb

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    The car is not moving with a constant velocity, that formula does not apply here.
    Round the final result to the right number digits and keep more digits in between, otherwise you can get too large rounding errors.

    There are various units missing in the equations.
     
  4. Nov 4, 2014 #3
    how would i find angular velocity and what units am i missing?
     
  5. Nov 4, 2014 #4

    mfb

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    Find the final speed of the car first.

    Missing units:
     
  6. Nov 4, 2014 #5
    okay. so using vf= 2d-vi/t

    2*402.25m/15s= 53.63m/s

    relating linear motion with rotational motion

    v/r=w

    53.63m/s/.23m =233.17

    is that right?
     
  7. Nov 4, 2014 #6

    haruspex

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    Yes, but it's usually a good idea to do all the working symbolically first, only plugging in numbers at the end. It makes it much easier for others to follow, makes it easier to spot mistakes, and minimises accumulation of rounding errors. In the present case, you'll find you don't need to calculate w at all. Indeed two of the given numbers are irrelevant.
     
  8. Nov 4, 2014 #7
    but the question is asking for rotational KE. dont i need to calculate w for that?
     
  9. Nov 4, 2014 #8

    haruspex

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    No, it asks for KE. And no, you will find you do not need to calculate w even if it did only ask for rotational KE. Just try it.
     
  10. Nov 4, 2014 #9
    i got it. thanks a lot.

    if you dont mind could you help me with applying significant figures?

    i have trouble with problems that need a lot conversions. so for example, 12 miles to m. i see 2 in miles so would that mean i have to round m to 2 significant figures as well?
     
  11. Nov 4, 2014 #10

    haruspex

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    As I said, you should work symbolically, only plugging in numbers at the final step. When you do so, use a couple more figures than in the provided data where you can, e.g. in the conversion factors. Only right at the end round to the number of figures implied by the given data.
     
  12. Nov 5, 2014 #11
    Got it. Thanks a lot. You helped me out a lot.
     
  13. Nov 5, 2014 #12
    Ive done the maths, you may as well have it.
    Moment of inertia = 0.240 kg-m²
    Rotation rate at final velocity = v / r = 233.24 rad / sec
    KE of 1 wheel = 6,546 Joules
     
  14. Nov 5, 2014 #13

    haruspex

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    No. Read post #8.
     
  15. Oct 24, 2015 #14
    Sorry, actually the final velocity assuming constant acceleration = 53.645 m/s resulting in a rotation rate of 53.645 / 0.23 = 233.238 rad/sec
    the resulting (rotational ) KE = 6,546.9 Joules, but that's only the rotational KE, it also has linear KE also......
     
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