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Problem with variables

  1. Nov 7, 2004 #1
    ok, heres a problem that you have to work out with all variables, and i frankly have trouble doing that.... i got part a, and i did a lot of work on part b, but cant seem to come up with the right answer.... any help appreciated....

    In Figure 11-32, a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section. For the following answers use g for the acceleration due to gravity, and m, r, and R, as appropriate, where all quantities are in SI units.

    Fig. 11-32
    [​IMG]

    (a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R. Assume R r.)

    (b) If the marble is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?

    ok, for part a, i got 2.7R whichi is correct, so then i moved on to part b, and i was told by my TA to start with the formula mg6R = .5mv^2 + .5Iomega^2 + mgR.... so i did, and tried to solve for v. i used 2/5mR^2 for I and for omega, used v/R...... so if i do that, and solve for v, then i use a=v^2/r and then when i have that, F=ma so i have the idea, i just cant get it right. anybody have any ideas? thanks in advance
    John
     
  2. jcsd
  3. Nov 7, 2004 #2
    anybody have any ideas?
     
  4. Nov 7, 2004 #3
    maybe i can find some help on another forum or something
     
  5. Nov 7, 2004 #4
    Ok you work with conservation of energy and do, mgh=mgR+7mv^2/10. So at point Q you have F=mv^2/R=(50/7)mg
     
  6. Nov 7, 2004 #5
    awesome, cant believe i didnt think of that, lol
     
  7. Nov 7, 2004 #6
    btw u can find this one in almost every physics book it a classic


    mg6R = 1/2mv^2+1/2Iw^2 + mgR

    mg6R = 1/2mv^2 + 1/2 (2/5mr^2)w^2 + mgR

    6gR = 1/2v^2 + 1/5r^2(v/r)^2+gR

    5gR = 7/10 v^2

    50/7gR = v^2

    F = 50/7mg
     
    Last edited: Nov 7, 2004
  8. Nov 7, 2004 #7
    ah, really? its not in ours (since its asked as a question in ours... lol)
     
  9. Nov 7, 2004 #8
    Help...

    Hey HobieDude16,

    Can you show me how you got part A? I'm learning angular momentum too and I have a similar problem except the ball is not on a ramp. Thanks.
     
  10. Nov 7, 2004 #9
    Hobie sent you a message hope you can help :)
     
  11. Nov 7, 2004 #10
    sqrt(9.8*r)=sqrt(10/7*9.8*(h-2R))
    basically, i looked through the book, found examples of a guy riding a bike around a loop, and then found an example of a ball rolling down a hill, and combined and adjusted to work... maybe that might help
     
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