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Homework Help: Calculate the speed of the block

  1. Dec 8, 2004 #1
    Hi. I have a problem.
    Two metal disks with different radius and different weight are welded together. The smaller disk has a string wrapped around it with a block on the other end. I have to calculate the speed of the block after 2 meters. I'm trying to use this equation: v=sqrt(2gh/1+M/2m) to solve this. Where M is the total mass of the disks and block while m is the mass of the block. This is not working. Do you guys know a way around this, I guess I need some kind of a relation between the two disks.
     
  2. jcsd
  3. Dec 8, 2004 #2

    Galileo

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    I don't understand the entire problem. How are the disks set in motion?
    Are they allowed to roll on a table with the block hanging over a pulley at the end of the table? Or something else?
     
  4. Dec 8, 2004 #3
    If i've understood correctly the two disks act as a pulley with the mass at the end of the string. If it is the case you can work it as a typical solid rigid problem calculating the angular acceleration and relating it with the linear acceleration of the mass. Then it becomes a "semi-free" fall problem.

    Or you can take the simpler approach of conservation of energy:

    Supposing the pulley is fixed so its potential energy remains constant, the initial energy of the system is

    [tex]mgh_1[/tex]

    and the final would be

    [tex]\frac{1}{2}(I_1+I_2)\omega^2+\frac{1}{2}mv^2+mgh_2[/tex]

    [tex]I_1[/tex] and [tex]I_2[/tex] are the moments of inertia of the disks.

    you can relate [tex]v[/tex] and [tex]\omega[/tex] by [tex]v=\omega r[/tex]

    where [tex]r[/tex] is the radius of the disk where the string is wrapped.

    I think you can work the rest. I think there's no major mistake, feel free to correct me.
     
    Last edited: Dec 8, 2004
  5. Dec 8, 2004 #4
    You understod the probmlem I think but I'm not quite following you Evil_kyo. I only need one kind of h since it's falling from 0 to 2 m. I have been trying to solve this with conservation of energy. I can't see from you equation what the tangential speed v should be, is the equation equal to zero or mgh?

    With thanks,

    Gunnar
     
  6. Dec 8, 2004 #5
    The only kind of h that you need is nothing but Evil_Kyo's [tex]h_1 - h_2[/tex](taking origin at the pulley)
    Since your [tex]h_1 = 0[/tex] and [tex]h_2 = -2[/tex],

    Your only kind of h = [tex]\0-(-2) = 2[/tex]

    Tangential speed cannot be mgh!!!
     
    Last edited: Dec 9, 2004
  7. Dec 8, 2004 #6
    As prasanna has noted i haven't selected a origin for the potential energy.
    [tex]h_1[/tex] and [tex]h_2[/tex] mean the initial and final position of the falling mass attached to the pulley. You don't know it but you know its difference [tex]h_1-h_2[/tex] that equals the length traveled by the falling mass 2 metres.

    Conservation of energy states that the initial energy of the system equals the final energy, then

    [tex]mgh_1=\frac{1}{2}(I_1+I_2)\omega^2+\frac{1}{2}mv^2+mgh_2[/tex]

    rearranging

    [tex]mg(h_1-h_2)=\frac{1}{2}(I_1+I_2)\omega^2+\frac{1}{2}mv^2[/tex]

    now we have the equation relating the angular speed [tex]\omega[/tex] of the pulley and the linear speed [tex]v[/tex] of the falling mass.

    [tex]v=\omega r[/tex]

    Imagine the pulley rotates an angle [tex]\omega[/tex] in a unit of time, then the lenght of string unrolled is equal to the lenght of arc subtended by this angle [tex]s=\omega r[/tex]. If the pulley rolls [tex]\omega[/tex] in a unit of time the mass falls [tex]\omega r[/tex] in a unit of time, then [tex]\omega r[/tex] is the speed [tex]v[/tex] of the mass.

    Using this equation we can simplify

    [tex]mg(h_1-h_2)=\frac{1}{2}(I_1+I_2)(\frac{v}{r})^2+\frac{1}{2}mv^2[/tex]


    [tex]mg(h_1-h_2)=\frac{1}{2}(\frac{I_1+I_2}{r^2}+m)v^2[/tex]

    [tex]v^2=\frac{2mgr^2(h_1-h_2)}{I_1+I_2+mr^2}[/tex]

    It's finished if there is no major mistake, i hope this helps you.
     
  8. Dec 8, 2004 #7
    It works, of course. Thanks a lot.
    I love this place.
     
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