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Rotational Energy conservation

  1. Nov 28, 2006 #1
    The four wheels of a car have each a mass of 25kg and a radius of 30cm. The car's mass is 1000kg. We neglect the losses due to friction. We assimilate the wheels to homogeneous cylinders.

    a) What is the total kinetic energy of the car and the wheels if the speed of the car is 30m/s.

    b) What distance will the car travel till it stops going up an incline of 10 degrees if the initial speed is of 30m/s?

    For a I did,

    Ktot = rotational kinetic energy + linear kinetic energy
    = 1/2lw + 1/2mv^2
    = (1/2)(4.5)(100)^2 + (1/2)(1100)(30)^2
    = 518kJ

    For b I thought I could just use "gravitational potential = rotational kinetic energy + linear kinetic energy"

    h = xsin10
    So mgx = rotational kinetic energy + linear kinetic energy
    mgx = 518000
    x = -89m

    Obviously I did something wrong because I shouldn't be getting a minus...and on top of that I know the answer is 276m.

    Any ideas on what I did wrong..or what I might of forgotten to do?
     
  2. jcsd
  3. Nov 28, 2006 #2

    berkeman

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    Staff: Mentor

    How do you go from mgx = 518000 to a negative answer for x? You are approaching it correctly using energies....
     
  4. Nov 28, 2006 #3
    Sorry, that should read mgxsin10 = 518000

    sin10 gives -0.544 therefore giving me a negative answer.

    Any idea on what I might of done wrong?!
     
  5. Nov 28, 2006 #4

    berkeman

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    Staff: Mentor

    The sin of 10 degrees is not negative. Is your calculator in radian mode maybe?
     
  6. Nov 28, 2006 #5
    Oh geez...lol..I didn't realise I was in radians...

    Thanks
     
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