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Homework Help: Two difficult problems any hints?

  1. Apr 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A 1000kg car carrying two 100 kg football players travels over a bumpy road with the bumps spaced 3 m apart. The driver finds that the car bounces up and down with maximum amplitude when he drives at a speed of 5 m/s. The car then stops and picks up three more 100 kg passengers. By how much does the car body sag on its suspension when these three aditional passengers get in?

    A 200g uniform rod is pivoted at one end. The other end is attached to a horizontal spring. The spring is neither stretched nor compressed when the rod hangs straight down. What is the rod's oscillation period? You can assume that the rod's angle from the vertical is always small.

    2. Relevant equations

    3. The attempt at a solution
    #1 I have stared at this problem for awhile, trying to figure out what "maximum amplitude should tell me. vmax= omega* A But that doesn't really lead me anywhere.
    Hmmm... w= sqrt k/m and I don't know the k. Any hints would be greatly appreciated.

    #2 If the rod weren't attached to the spring
    I=(1/3)*(.200kg)(.20m)^2 I= .0027 kgm^2

    so omega= sqrt [(.200kg*9.80*.10m)/.0027kgm^2] and from there I could find the period. However, I cannot visualize how the spring would affect this physical pendulum other than decreasing its amplitude possibly.
  2. jcsd
  3. Apr 20, 2008 #2
    any ideas for these problems??? they are driving me crazy
  4. Apr 20, 2008 #3
  5. Apr 21, 2008 #4


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    Hi bcjochim07;! :smile:

    Hint: You're not being asked for anything involving maximum amplitude.

    You're being asked for the resonant frequency, which you can work out from the speed and from the humpitude. :smile:
  6. Apr 21, 2008 #5
    so f0=fext

    this frequency is 1 bump/3m* 5m/s = 1.67 bumps/sec

    so to solve for k

    1.67= (1/2pi)*sqrt[k/1200]
    k= 132120 N/m

    then I can use the same formula with the new mass to calculate the new frequency
  7. Apr 21, 2008 #6


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    Hi bcjochim07! :smile:

    Looks good!

    (except … I can never remember … is frequency = bumps/second, or half of bumps/second? :smile:)

    ok … #2
    This question is designed to test whther you can only apply the formula for frequency of a pendulum, or whether you actually understand where the formula comes from.

    A pendulum has a harmonic movement because, loosely speaking, the horizontal "restoring force" is a (negative) constant times the displacement.

    So … what is that constant? :smile:

    (Then add it to the spring constant.)
  8. Apr 21, 2008 #7
    frequency is oscillations per sec

    This acts as a pendulum also??
  9. Apr 21, 2008 #8


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    Sorry … not following you. :confused:

    Are you still on #1? I was on #2 (the uniform rod behaving like a pendulum).
  10. Apr 21, 2008 #9
    oohh... ok sorry I didn't see your "ok"#2"
  11. Apr 21, 2008 #10
    so this constant is -mg/L
    k= 9.8 so I add this to the spring constant to get a new constant

    Then do I use the formula for frequency of spring with this new constant plugged in?

    What about the mass--since it is an extented object how will I take that into account in the formula?
  12. Apr 21, 2008 #11


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    Hi bcjochim07! :smile:

    Do you remember someone saying this … :rolleyes:
    ok … that gives you an restoring acceleration … combine it with the spring. :smile:
  13. Apr 21, 2008 #12
    I don't understand what you mean by combining it with the spring. Do you mean I'm supposed to combined the forces?

    angular acceleration= omega^2*r
  14. Apr 21, 2008 #13


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    Nooo … radial acceleration = omega^2*r.

    You want the tangential acceleration (of the pendulum). :smile:
    "combining the accelerations" makes more sense …

    this is because the force of gravity acts all over the rod, but you only want to consider the movement of the tip of the rod.
  15. Apr 21, 2008 #14
    Frestoring=-kx=ma a=-kx/m

    omega= sqrt[Mgl/I] for a physical pendulum so from getting omega, could I find the angular acceleration and then go to tangential of the tip of the rod?

    Then I would combine these accelerations, but then how do I get back frequency?

    I don't think I can say -(mg/L)s=ma since this isn't a simple pendulum. Right?

  16. Apr 21, 2008 #15


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    Yes … multiply the angular acceleration by length to get the tangential acceleration.
    You tell me … if acceleration is -k (a constant) times displacement, what is the frequency? :smile:
  17. Apr 21, 2008 #16
    k=ma/x??? I am really getting frustrated because I don't where I'm going with this
  18. Apr 22, 2008 #17


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    I'm not going to do it for you!

    Show us what you've done so far …

    the period without the spring,
    the tangential acceleration without the spring,
    the tangential acceleration only from the spring …

    and then how you think you might combine them (make a guess if necessary!) :smile:
  19. Apr 22, 2008 #18
    Ok, so from the I can find the angular frequency of the physical pendulum, then I will go to angular acceleration and then to tangential acceleration. The acceleration caused by the spring is a=-kx/m. So if I add those accelerations together, then I have a new tangential acceleration. There is a formula at=alpha*r and so maybe I could do something with that but I don't know what the r would be. I am I even remotely on the right track?
  20. Apr 22, 2008 #19
    How do I relate the angular acceleration to the angular frequency?
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