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Period of uniform plank with spring not at Center of Mass

  1. Nov 15, 2008 #1
    Before begining, I would like to point out I did look at the following thread previous to this one, and thus have a few questions about that thread and it's application to this one: https://www.physicsforums.com/showthread.php?t=143536

    1. The problem statement, all variables and given/known data
    The problem: http://img122.imageshack.us/my.php?image=bphysicsry1.png

    2. Relevant equations
    Not quite sure, but I think this one comes into play: http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

    3. The attempt at a solution
    Alrighty, I looked through the thread mentioned above, and while the problems are similar, there is one notable exeption here.

    In the other thread's problem, I was given.

    In my problem, I is not given, and the spring is not acting through the CoM

    So, my question is, does the pendulum equation still work? And is I support merely 1/3mL2? And which length is l, the entire length of the bar, or the length of the spring to the pivot point? If not, how do I find I?

    I will say I'm slightly confused about the other thread, as I don't really get the "parallel axis theorem" or the more complicated matters of moment of inertia. Any help would be appreciated.

    One other thing, I'm also confused as to what should go on the bottom of the equation. I know it says "mgh", but since the spring is doing the pendullum motion, I'm not sure. Also, how the heck am I supposed to calculate a spring force without the displacement?
    Last edited: Nov 15, 2008
  2. jcsd
  3. Nov 15, 2008 #2
    If you've noticed from other equations, the period of a spring is influenced by the system's properties. Another point you might have encountered is that if a spring is vertical so that the weight of an attached mass pulls/pushes it down a bit, the gravitational force will not change the period of oscillation...it will simply shift the equilibrium point.

    Since the plank is pivoted, it will rotate, causing the spring to slant a bit; but the question specified for small angles of oscillations, so that does not have to be taken into account. Altogether, the problem is similar to if the spring simply had a block of mass attached to it instead of a pivoted plank. What you do want to take into account is that the force the spring is feeling is not simply the weight of the plank, thus, the angular frequency is not necessarily Sqr(k/m). If I'm thinking right, all you have to do is find the force the spring is feeling, relate it to a force of gravity by setting it equal to M*g (M being an arbitrary mass and g being acceleration due to gravity), find the mass, use it for the angular frequency, then substitute the M in terms of the mass of the plank. With that, you should be able to find the period...and if this is wrong...then please do so to correct me.
  4. Nov 15, 2008 #3
    Err, I'm sorry, can you put that in formulaic forms? Sorry for th etrouble, I just want to make sure I'm doing what you suggested right
  5. Nov 15, 2008 #4
    Actually...lets start anew:

    Our aim is to find the period of oscillation, so it would be helpful to find an equation so that
    d2x/dt2 = -ω2*x, in which the motion is simple harmonic and ω is angular frequency.

    We know that τ = I*α = r x F, in which τ is torque, I is moment of inertia, α is angular acceleration, and r is distance from point of rotation (distance to where the spring is). The moment of inertia for a rod pivoted at one end is (1/3)*M*L2, in which L is the entire length of the rod. As mentioned earlier, the weight of a mass attached to a spring does not influence the period, it simply shifts the equilibrium position, so we do not need take that into account.

    F = -k*x (spring force); the displacement x is also such that x = r*sinθ...since the problem specifies for small angles of oscilltions, sinθ may simply be taken as θ so that x = r*θ. τ = I*α = r x F = r*F*sinθ2, in which θ2 is the angle between r and F, but since θ is taken to be small, then θ2 could be taken as 90o throughout the oscillations, making τ = r*F = -r*k*x = -r2*k*θ = I*α. Thus,
    α = -(r2*k/I)*θ so that d2θ/dt2 = -ω2*θ, in which ω = Sqr(r2*k/I) for Sqr() being a square root function.

    T = 2π/ω = 2π*Sqr(I/(r2*k)) = 2π*Sqr(ML2/(3r2k))

    What I was saying in my previous post is that the period is simply [tex]T = 2 \pi \sqrt{\frac{m}{k}}[/tex], but the mass, m, is not the mass of the plank, but a relative mass that you'd have to give in terms of the mass of the plank...but I'm not sure if that was really a simpler approach or not.
    Last edited: Nov 15, 2008
  6. Nov 15, 2008 #5
    Uhhhh, quick question, your n stands for pi, correct? or does it stand for something else....

    Ok, so, let me see if I followed you here on how you derived that equation...

    So, the x/theta magically go away because of the small sign aproximation? See, that's what's bugging me, is where the small sign aproximation actually does, and how we can calculate torque without the x.

    To let me put things in more laymans terms, the following: [tex]T = 2 \pi \sqrt{\frac{I}{mgL}}[/tex]

    the bottom half of the fraction in that equation is merely the torque? And that the 2pi and the sqrt Get the a = I/T to a w quantity, correct?

    I know you worked it out, but I'm just trying to approach the total deriviation from a different angle.
  7. Nov 15, 2008 #6
    Also, when I plugged in that formula, I got the wrong answer, I got 1.555 seconds. :/
  8. Nov 15, 2008 #7
    Yes, that is pi.

    The small approximation is probably what they are expecting you to do. If the approximation isn't made, things would get much more complicated and we wouldn't be dealing with simple harmonic motion. In fact the equations for the SMH of a pendulum are also idealized; to be precise, pendulums do not follow simple harmonic motion, but their motion is close enough to SMH for small angles. The same concept applies to the equation you have there: [tex]T = 2 \pi \sqrt{\frac{I}{mgL}}[/tex] The torque is -mgLsinθ, in which -mgLsinθ = I*α so that α = -mgLsinθ/I...they simply approximated sinθ as θ for small angles of θ; this gives d2θ/dt2 = -(mgL/I)*θ.

    I didn't get 1.555s as the answer...I got .822s...what is the actual answer?
    Last edited: Nov 15, 2008
  9. Nov 15, 2008 #8
    I don't know, I use an "online homework" system, which means we submit homework a set number of times. The system tells if we're right or wrong within 1% of the right answer. Each time we get a wrong answer, the total points goes down on what the problem is worth.

    Ok, your answer is right, not sure what I did wrong there, thank you though. :D

    Might have plugged it in on my calc wrong, or something odd happened. >.>
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