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Rotational motion conservation of energy; time

  1. Jul 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A hollow cylinder (hoop) is rolling on a horizontal surface at speed v=3.3m/s when it reaches a 15 degree incline. How far up the incline will it go? How long will it be on the incline before it arrives back at the bottom.


    2. Relevant equations
    SOH CAH TOA
    MGH=KEtrans +KErot
    w=v/r
    x=theta*r
    [tex]\omega[/tex]^2=wo^2 +2[tex]\alpha[/tex][tex]\vartheta[/tex]

    3. The attempt at a solution
    I have finished the first part of the problem and solved for 4.3 m using conservation of energy and a little trig. The second part is giving me trouble because I'm not sure if I calculated the radius of the hoop correctly (.525m) which could be throwing off my substitutions and calculations. I used the last equation listed with 2[tex]\pi[/tex] for theta. I know the time should be 5.2 seconds but I'm doing something wrong. I know i have to relate speed and time where vf=0 and take that time and multiply is by 2 to get the total time on ramp. Please help?
     
  2. jcsd
  3. Jul 2, 2008 #2
    What's the mass of the hoop? (this is only relevant to check your work)

    (and with your work, I arrived at 5.21 seconds on the ramp)


    You can simplify this problem immensely for the second part. Just treat it like an object decelerating from 3.3 m/s to 0 m/s over 4.3 meters. Once you've solved for that acceleration, you have a nifty formula relating velocity, acceleration, and time.
     
    Last edited: Jul 2, 2008
  4. Jul 2, 2008 #3
    The mass of the hoop is not given, i was under the impression mass does not matter because it all cancels out? Which equation can I use to determine the mass of the hoop? I'll do the work but I just dont know which one to use. I don't see an equation relating rotational motion and mass..
     
  5. Jul 2, 2008 #4
    Yeah, ignore that entirely. I misread what you wrote. You don't need to solve for the mass.
     
  6. Jul 2, 2008 #5

    Doc Al

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    Staff: Mentor

    To find the time on the ramp you do not need to know (nor do you have sufficient information to determine) the radius of the hoop. (Nor do you need the mass of the hoop.)

    Hint: What's the average translational speed of the hoop as it goes up the ramp?
     
  7. Jul 2, 2008 #6
    hmm. I'm sort of confused by what you mean by average translational speed of the hoop, do you mean to plug my numbers back into PE=KEtrans +KErot? would the average translational speed be Vf-Vo/2?
     
  8. Jul 2, 2008 #7

    Doc Al

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    Staff: Mentor

    No. All you need is simple kinematics.
    Almost: Vave = (Vf + Vo)/2
     
  9. Jul 2, 2008 #8
    I plugged my numbers back into x=1/2at^2 +Vo*T +X0 substituting (Vf-Vo)/T But I still get the wrong answer. What am i missing?
     
  10. Jul 2, 2008 #9

    Doc Al

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    Staff: Mentor

    The only formula you need is distance = ave speed X time. You don't need the acceleration.
     
  11. Jul 2, 2008 #10
    when I use that eq I get (0-3.3m/s)/2=-1.65m/s.
    4.3m=-1.65*T I get -2.6 for T which isn;t right. Sorry if I'm not getting something this simple but im just not seeing something.
     
  12. Jul 2, 2008 #11

    Doc Al

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    Staff: Mentor

    (1) The average speed is 1.65 m/s, not -1.65 m/s. (Why are you subtracting?)
    (2) Going up the incline is just half the time.
     
  13. Jul 2, 2008 #12
    yes! thank you! I forgot to multiply it by two! It was negative because I thought it was Vf-V0 leaving -3.3m/s. Thanks a bunch! :biggrin:
     
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