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Taperecorder mechanics

  1. Sep 2, 2005 #1

    N2

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    hello, ive just started a mechanics course, and i dont quiet know how to start with this assignment:

    In a taperecorder the information to pass the recording head with constant velocity. Therefore is has to rotate with decreasing velosity as the empty wind is getting filled.

    What is the spools angular velocity, [tex] \omega [/tex], as function of
    [tex] b [/tex], the tape's width
    [tex] r_0[/tex], radius of the empty wind
    [tex] v_0[/tex], velocity at the recording head
    [tex] t[/tex], time

    My first thought was, that the angular acceleration must be constant, but how does this help? and how can i solve this problem?

    thanks
     
    Last edited: Sep 2, 2005
  2. jcsd
  3. Sep 2, 2005 #2

    Astronuc

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    Staff: Mentor

    For clarification, I believe that b is the tapes thickness. So in that case, the radius is increasing by b with each revolution, or dr = b.

    The linear velocity must be constant, so the angular velocity [itex]\omega[/itex]= v/r must be decreasing as r increases (proportional to r). If the angular velocity has to decrease, what does this say about the angular acceleration?
     
  4. Sep 2, 2005 #3

    N2

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    You are right; b is thickness.
    If the angular velocity decreases then the acceleration is negative. But is your point that a. acceleration, [tex]d\omega[/tex], is proportional dr and therefore to b ?
     
  5. Sep 2, 2005 #4

    LeonhardEuler

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    dr is only an infinitisimal quantity, while b is finite. I think you mean to say [itex]\frac{dr}{d \theta} = \frac{b}{2 \pi}[/itex]. This seems to be a good way to start the problem.
     
  6. Sep 2, 2005 #5

    Astronuc

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    Something like that, yes.

    So one has to work out [itex]\omega[/itex] = d[itex]\theta[/itex]/dt in terms of r and differentiate to get acceleration.

    Implies that the angular acceleration is negative, because angular velocity must be decreasing while the radius is increase in order to maintain a constant linear velocity.
     
  7. Sep 2, 2005 #6

    LeonhardEuler

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    I don't think finding the acceleration is important to this problem. It asks to find [itex]\omega[/itex] in terms of the other variables. The way I solved it was to write the equation as [itex]dr=\frac{b}{2\pi}d\theta[/itex] and then divide by dt to make [itex]\omega[/itex] appear. Then after seperation of variables, integration, etc. you get an answer. Maybe there is a simpler way, though.
     
  8. Sep 3, 2005 #7

    N2

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    integrating the equation containing [itex]\omega [/itex] - wouldn that just give you an expression for [itex]\theta[/itex] ?
     
  9. Sep 3, 2005 #8

    LeonhardEuler

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    Not the way I did it. I re-wrote [itex]\omega[/itex] as [itex]\frac{v}{r}[/itex]. Then, since v is constant, I just seperated variables and got an expression for r in terms of t.
     
  10. Sep 3, 2005 #9

    N2

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    ok thank you.
    the expression i end up with is now
    [tex]\omega (t) = \frac{v_0}{ \sqrt{b/ \pi \cdot v_0 t + r_0}}[/itex]
    but isent there i dimension problem with the units?

    [tex][s^{-1}] = \frac{[m/s]}{\sqrt{[m] [m/s] + [m]}}[/tex]
    dosent seem to add up? (the problem being [itex] r_0 [/itex])
     
  11. Sep 3, 2005 #10

    LeonhardEuler

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    Yes, re-check the work you did to find the value of the constant after you integrated. I get [itex]r_0^2[/itex]
     
  12. Sep 3, 2005 #11
    Another way might be to start with both reels with the same amount of tape, both moving the same velocity, say 1 inch per second at the recording head.


    Assume 1 ips, then whats the starting radius? It cant be the center because there is a hub. Imagine its exactly 1" radius from the center of the reel. So at 1 ips, with 1 inch from center, assume it can never change velocity because the reels are always equal, to get an initial calculation.

    If you start with a 1" radius you would add the tape thickness and recalculate for every revolution. The other reel isnt necessarily going the inverse speed, if that reel starts its overlap at some different part of rotation.

    The speed of the reel changes on the thickness of the tape. Constantly? Or only at the time the tape overlaps? It has to be when the tape overlaps, it accelerates up to the thickness of the tape then is constant until the next revolution.

    Imagine the reel is 5 feet in diameter and the tape is 1 inch thick, it winds on to the takeup side then overlaps, now the radius is changed by 1 inch and moves slower. It would probably increase, overshoot, then correct each rotation to the current radius +1 tape thickness
     
    Last edited: Sep 3, 2005
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