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Rotational Motion of compact disc

  1. Jan 5, 2007 #1
    1. The problem statement, all variables and given/known data
    A digital audio compact disc carries data, with each bit occupying 0.6 (mu)m, along a continuous spiral track from the inner circumference of the disc to the outside edge. A CD player turns the disc to carry the track counterclockwise above a lens at a constant speed of 1.3 m/s. Find the required angular speed a) at the beginning of the recording, where the spiral has a radius of 2.3 cm and b) at the end of the recording, where the spiral has a radius of 5.8 cm c) A full-length recording lasts for 74 min 33s. Find the average angular acceleration of the disc. d) Assuming that the accleration is constant, find the total angular displacement of the disc as it plays. e) Find the total length of the track.

    2. Relevant equations

    3. The attempt at a solution

    For a) I converted the radius of 2.3 cm to meters: 0.023 meters
    I used the equation: v=rw to solve for the angular speed (w):
    1.3m/s=(0.023m)W --> W=56.52 rad/s

    b) same method as a): v=rW --> 1.3m/s=(0.058m)W --> W=22.4 rad/s

    c) I used the equation: W=W_o+(alpha)t
    t=74 min 33s or 4473s

    I used the answers for a) and b) for the initial/final angular speeds. Is this correctly done? 22.4 rad/s=56.52 rad/s+(alpha)(4473s)
    Solving for (alpha) I got (alpha)= -7.628x10^-3

    I'm a little ambiguous about the answers I've obtained for a) and b). I thought that the angular speed for a larger radius should be larger than the circumference with the smaller radius. Does this only apply to the tangential speed/velocity? I remember reading somewhere that the outside of the circular rotational motion moves faster than the inside if they were both on the same reference line. Are my attempts correct so far? How can the angular accleration(alpha) equal such a small negative number? I don't really understand how the angular velocities and accelerations work because they are related to theta =/ Can someone please explain it to me? Thanks! :smile:
  2. jcsd
  3. Jan 5, 2007 #2
    You did your calculations right. The reason the angular velocity decreased as you went farther out on the CD is because the linear velocity stayed constant. The laser can only read a certain number of bits off the CD every second, so if it kept rotating at the same angular speed as the laser moved out, it would actually be trying to read more bits than it should. As a result of this, the CD player is programmed to slow its angular velocity as it moves farther out the CD to keep the linear speed of the bits above the laser constant. Mathematically, this makes sense too; if v remains constant in [itex]v = r \omega[/itex] and the radius increases, [itex]\omega[/itex] must decrease to compensate.

    The angular acceleration you got is correct too--74 minutes is a pretty long time, so the CD decelerates only very slightly every second to go from 56.5 rad/s in the beginning to 22.4 rad/s at the end. So it makes sense that you got a small negative number for the acceleration.

    Your book should probably mention the rotational analogs of kinematic equations--this is how you'll obtain the total rotation of the CD. Recall from kinematics that

    x = x_0 + v_0 t + \frac{1}{2}at^2

    This relationship works the same way when dealing with angular calculations. Since x is like [itex]\theta[/itex], v is like [itex]\omega[/itex] and a is like [itex]\alpha[/itex]:

    \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2

    Edit: whoops, added 1/2 to equation.
    Last edited: Jan 5, 2007
  4. Jan 5, 2007 #3
    Also excellent.

    That's the one to use.

    All of these are correct, although I haven't checked your arithmetic.

    The angular speed would ordinarily be the same at all points along the radius, unless something is done to change it, angular speed is the same everywhere, it is linear speed that changes.

    Given a certain angular velocity, the linear speed has to be different (faster farther away from the center). However, a CD player changes the angular velocity so that the linear speed is always the same, so the angular velocity is slower on the outer tracks.

    The angular acceleration is so small simply because the playing time of the disc is so large. If it takes almost 5,000 seconds to move from the outer track to the inner track, the change in angular acceleration is going to be quite small, since that is alpha/t.

  5. Jan 5, 2007 #4
    Oops. I see gabee was writing at the same time as me and much faster :approve:

    Since you know the beginning and ending w's, you could also use 1/2(w0 + wf)t.

    Just like to suggest that, given a constant linear speed and a time, you don't need these particular equations to answer (e).

    Last edited: Jan 5, 2007
  6. Jan 5, 2007 #5
    Ohhh, this is making sense to me now. Also, would the angular acceleration be negative because the disc is rotating counter-clockwise?
  7. Jan 5, 2007 #6
    Editing, Trying out Latex so it may take a while =]

    Well, I've just finished d) and e) and I would like someone to check my work for me just to make sure I did everything right. Thanks! =D

    d) I used the equation: [tex]\theta=\omega_{i}t+\frac{1}{2}\alpha t^2[/tex]

    [tex]\theta=(56.52)(4473s)+ \frac{1}{2} (-7.628 \times 10^{-3})(4473)^2[/tex]

    [tex]\theta=176,504.48 rad[/tex]

    e) I used the equation: [tex]x=x_o+v_{ox}t+\frac{1}{2}a_{x}t^2[/tex]

    [tex] x=(1.3m/s)(4473s)[/tex]


    Edit: There is no acceleration right? The tangential velocity is constant=no accel? Thats why I canceled it out.

    Okay, are my procedures correct? Thanks =D
    Last edited: Jan 5, 2007
  8. Jan 5, 2007 #7
    It depends on whether or not it is speeding up or slowing down.

    If the disc is speeding up, then the acceleration would have the same sign as the direction of rotation. If it is slowing down, then it would have the opposite sign from the rotation. It's the same as in linear motion.

    Usually, counter-clockwise rotation is taken as positive, but that's just a convention.

  9. Jan 5, 2007 #8
    That's right, thanks for catching the error with the [itex]\frac{1}{2}[/itex] :)
  10. Jan 5, 2007 #9
    Oh, lol I didn't even notice. =P No big deal though. Thanks for your beautiful presentation with the Latex that inspired me to do the same. I took about 15 min trying to do the fraction lol :tongue: I'm a slow learner. hehe
  11. Jan 5, 2007 #10
    Hmm.. Wait, is the tangential velocity the same as linear velocity? Could those two terms be used... what's that word that I'm looking for? :rolleyes: inter.. lol
  12. Jan 5, 2007 #11
    Yes, tangential velocity and linear velocity mean the same thing in this case and can be used interchangeably. :)
  13. Jan 5, 2007 #12
    Yeah! That was the word I was looking for lol. Is there a situation where the tangential velocity and linear velocity aren't the same?
  14. Jan 5, 2007 #13
    Well, they are always the same as far as I know, you just wouldn't use the term tangential velocity unless you are talking about rotational dynamics. (You wouldn't use the term "tangential" to describe a particle traveling in a straight line.)
  15. Jan 5, 2007 #14
    Oh ok. Also, did I manage to solve part e) correctly?

    I used the equation: [tex]x=x_o+v_{ox}t+\frac{1}{2}a_{x}t^2[/tex]

    [tex] x=(1.3m/s)(4473s)[/tex]


    There is no acceleration right? The tangential velocity is constant=no accel? Thats why I canceled it out.

  16. Jan 5, 2007 #15
    That's right, it sounds like you have a good grasp of the concepts so far.

    No problem! Good luck in the rest of your physics studies.
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