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Record and turntable

  1. Oct 27, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider a turntable to be a circular disk of moment of inertia I_t rotating at a constant angular velocity w_i around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis.
    Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is I_r. The initial angular velocity of the second disk is zero.

    There is friction between the two disks.

    After this "rotational collision," the disks will eventually rotate with the same angular velocity.

    [​IMG]

    (a)Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, K_f, of the two spinning disks?
    Express the final kinetic energy in terms of I_t, I_r , and K_i the initial kinetic energy of the two-disk system. No angular velocities should appear in your answer.

    (b)Assume that the turntable deccelerated during time [tex]\Delta[/tex]t before reaching the final angular velocity ([tex]\Delta[/tex]t is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque,[tex]\tau[/tex] , acting on the bottom disk due to friction with the record?
    Express the torque in terms of I_t, w_i, w_f, and [tex]\Delta[/tex]t

    2. Relevant equations

    K_i - K_f = [tex]\Delta[/tex]K lost to friction --->

    (1/2)I_1*w_1i^2 + (1/2)I_2 * w_2i^2 = (1/2)I_1 * w_1f^2 + (1/2)I_2 * w_2f^2



    3. The attempt at a solution

    Ok, i've only attempted part (a) at the moment so i need to concentrate on that first. then i'll move onto part (b)

    What is confusing is what the question is asking for. delta K is just the initial - final energies, but where does that stand in the solution they are looking for.

    This is what i know --->
    K_i = (1/2)I_t * w_i^2

    K_f = (1/2)w_f^2(I_t + I_r)

    How and where do I get rid of the w's and where is delta K in the solution?
     
  2. jcsd
  3. Oct 27, 2007 #2

    Doc Al

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    Staff: Mentor

    You don't need delta K, you just need to express K_f in terms of K_i and the other variables. First find w_f. You'd do that just like you did in that similar problem the other day. Once you find w_f, then worry about finding K_f and expressing it the way they want it. (It will take a little algebraic rearranging, that's all.)
     
  4. Oct 27, 2007 #3
    Rotational kinetic energy is not conserved?
    But there is a rotational quantity that is conserved. What is it?
     
  5. Oct 27, 2007 #4
    I've already figured w_f from the first part of this problem that i didn't post--->

    w_f = (I_t * w_i)/(I_t + I_r) ---> i plug this into K_f --->

    K_f = (1/2)w_f^2(I_t + I_r) ---> so --->

    (1/2)[(I_t * w_i)/(I_t + I_r)]^2 * (I_t + I_r) ---> reduces to --->

    (1/2) * (I_t * w_i)^2 / (I_t + I_r) ---> a little stuck here. i'm trying to figure out how i can get a K_i out of this expression...
     
  6. Oct 27, 2007 #5


    First thing, why do you need delta K? Neither it has been asked, nor it appears in the two equations you have quoted!
    You need Kf in terms of It, Ir and Ki. In the above two equations, you have two unknowns, wf and wi. To get a required condition, you need just one more equation. So, somehow try to relate wf and wi.
    Hint: You know some conservation laws, right?
     
  7. Oct 27, 2007 #6
    Substitute unwanted wi in terms of Ki.
     
  8. Oct 27, 2007 #7
    (K_i * I_t) / (I_t + I_r) ---> i believe this is correct....Now onto the second part...
     
  9. Oct 27, 2007 #8
    ya.. show ur attempt on the second part.
    i would just like to point out one thing: note the striking similarity between

    a particle, mass m1, with speed vi collides and sticks to a second particle, mass m1, at rest.
    conservation of momentum: (m1+ m2).vf = m1.vi

    a disk, mass moment of inertia I1, with rotational kinetic energy Ki collides and sticks to a second disk, mass moment of inertia I2, at rest.
    and in this problem: (I1+ I2).Kf = I1.Ki

    Have fun.
     
  10. Oct 27, 2007 #9
    part (b) solved:

    w_f = w_i - [tex]\alpha[/tex]*[tex]\Delta[/tex]t --->

    -[tex]\alpha[/tex] = (w_f - w_i) / [tex]\Delta[/tex]t --->

    [tex]\tau[/tex] = I*[tex]\alpha[/tex] --->

    [tex]\tau[/tex] = I[tex]^{t}[/tex] * (w_f - w_i) / [tex]\Delta[/tex]t
     
  11. Oct 28, 2007 #10

    Doc Al

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    Staff: Mentor

    Good! Realize that this is just the angular form of impulse. Just like linear impulse equals change in linear momentum, angular impulse equals change in angular momentum:
    [tex]\tau \Delta t = \Delta L = I \Delta \omega[/tex]
     
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