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Find Speed in Centripetal Motion

  1. Nov 25, 2007 #1
    The radius of curvature of a loop-to-loop roller coaster is 12 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.4mg. Find the speed of the roller coaster at the top of the loop. Answer: v = 12.83 m/s.

    I assume .4 mg is the torque, right? I have no clue what to do. Please help.
     
    Last edited: Nov 25, 2007
  2. jcsd
  3. Nov 25, 2007 #2

    rock.freak667

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    At the top of the loop...what forces provide the centripetal force? the weight and the force that the seat exerts....and so....you can find v....
     
  4. Nov 25, 2007 #3
    What do I do with the mg?
     
  5. Nov 25, 2007 #4

    rock.freak667

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    Well mg is just the formula to find weight...so that the force exerted by the seat is 0.4*(the weight of the seat)

    so that weight of seat+Force exerted by seat, provides the centripetal force
     
  6. Nov 25, 2007 #5
    So I would use the equation F = ma and a = v^2/r. I wouldn't use v = (2*pi*r)/(T) because T = period and that info is not provided. But I still don't understand what I'm suppose to do with the m and .4mg.
     
  7. Nov 25, 2007 #6

    rock.freak667

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    ok..seems you aren't getting this...

    Eq'n: Centripetal Force,[itex]F_c=\frac{mv^2}{r}[/itex]

    The weight of the seat(mg) + The force that the seat exerts on a passenger(0.4mg) provides the centripetal force.

    This means that

    [tex]F_c=mg+.4mg[/tex]

    since [itex]F_c=\frac{mv^2}{r}[/itex]

    then
    [tex]\frac{mv^2}{r}=mg+.04mg[/tex]


    What cancels out...and you have the value of 'r' and 'g'
     
  8. Nov 25, 2007 #7
    Solved! Thank you sooo much.
     
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