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Vibrations physics homework

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    The disk, hvng a weight of 15kN, is pinned at it center O and supports the blck A that has a weight of 3 kN. If the belt which passes over the disk is not allowed to slip at its contacting surface, determine the natural period of vibration of the system.

    2. Relevant equations
    [tex]\Sigma[/tex]Mo=Mo
    T=1/f
    Mo=I [tex]\alpha[/tex]
    I=(1/2)*M*r2

    3. The attempt at a solution
    M=I [tex]\alpha[/tex]
    I=1/2*M*r2
    I=(1/2)*(15000/9.81)*(0.75)2
    I=430 kg*m2
    therefore Mo=430*[tex]\alpha[/tex]

    [tex]\Sigma[/tex]Mo=Mo
    430[tex]\alpha[/tex]=(0.75)(K*x)-(0.75)*(3000)
    x=0.75sin([tex]\vartheta[/tex])
    sin([tex]\vartheta[/tex])=1 because [tex]\vartheta[/tex] is very small
    therefore [tex]\vartheta[/tex]=0.75

    430[tex]\alpha[/tex]=(0.75)(80000)(0.75)[tex]\vartheta[/tex]-(0.75)*(3000)

    At this point I'm not sure what to do. I have 2 terms in angular components and one in rectangular components. I'm not sure if I can just convert the moment provided by the weight into an angular compononent or if I need to somehow include its moment with the momnt provided by the spring. Any advice would be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Nov 9, 2009 #2

    ideasrule

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    Homework Helper

    Re: Vibrations

    Where did you get this?

    I once made the same mistake on a major exam and it cost me dearly. The weight does not exert 0.75*3000 N because it's accelerating. Draw a free-body diagram on the weight to see what it exerts.

    You don't need to assume that theta is very small; no approximations are required here. Even if you did, sin(theta) is equal to 0 when theta is small, not 1.

    How did you get this? The only equation you need is this:

    430[tex]\alpha[/tex]=(0.75)(K*x)-(0.75)*(3000)

    when you correct the mistake, you'll see that it takes the form d^2x/dt^2 + m^2x=0. The period is just m. (Huge hint: x=r*theta).
     
  4. Nov 9, 2009 #3
    Re: Vibrations

    T=1/f
    period=1/frequence



    Isn't the spring holding the system in equilibrium? We are just slightly disrupting the system to provide a small vibration so we can equate for the natural vibration arent we?






    430[tex]\alpha[/tex]=(0.75)(80000*0.75*theta)-(0.75)*(3000)

    That is the equation I used, I just plugged in the 80000 in for the value of k and 0.75*theta in for the value of x.
     
  5. Nov 9, 2009 #4

    ideasrule

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    Re: Vibrations

    I didn't understand the second sentence, but the system isn't in equilibrium because it's vibrating. It's always good to do things rigorously and only approximate if absolutely necessary.
     
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