Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Turntable physics

  1. Mar 21, 2004 #1
    Hi all,

    I haven't studied rotational physics in a long time so this may be a stupid question, but here goes. I'm trying to model the properties of a turntable (record player). Using the theories of rotational motion, is it possible to calculate the time it would take for the spinning turntable platter to come to a complete stop when the power to the motor is cut?

    Given that I know the starting velocity is 33 rpm, the final velocity is 0rpm and the weight of the platter is 1.996kg is it possible to calculate the time it takes to come to a complete stop?

    Thanks for you help!
     
    Last edited: Mar 28, 2004
  2. jcsd
  3. Mar 21, 2004 #2
    Yes, but you would need to know the coefficient of friction between the turntable and the platter. If there were no friction, the platter would never stop. Once you have the magnitude of the friction force you can find the angular deceleration of the platter and how long it would take for its angular velocity to become 0.
     
  4. Mar 21, 2004 #3
    Ah yes, I thought that would be the missing factor. I assume this (the friction between the turntable and the platter) is something that is not so easy to measure/calculate?
     
  5. Mar 22, 2004 #4
    If you can find such a turntable, it is very easy to calculate. Just measure how much force it takes to turn it. All you need is a small scale.

    The other thing you could do is to time how long it takes to come to a stop. With a little algebraic manipulation you can solve the equation for the coefficient of friction.

    Or you could guess. Maybe around .1 ? Does it have bearings?
    -Mike
     
    Last edited by a moderator: Mar 22, 2004
  6. Mar 22, 2004 #5
    [tex]V(t) = V_0 + at[/tex]
    In our case the platter stops so V(t) is zero:
    [tex]t = -\frac{V_0}{a}[/tex]
    (a will also be negative so the time will be positive.)

    Finding a is just a matter of: (fk being kinetic friction)
    [tex]\Sigma F = ma = -f_k = -N\mu = -mg\mu[/tex]
    [tex]a = -g\mu[/tex]

    Plug that into the formula above:
    [tex]t = \frac{V_0}{g\mu}[/tex]
    And you're done. :smile:
    Or if you have the time and want to find the coefficient of friction:
    [tex]\mu = \frac{V_0}{gt}[/tex]
     
  7. Mar 25, 2004 #6
    OK,

    Thanks Chen. I've measured the time it takes for the turntable to stop, and get an average of about 15 seconds. I'm trying to put that into your last formula, to find the coefficient of friction but am unsure of a couple of variables.

    V(0) Velocity is 33.33 rpm so = 3.49 rad/s
    t Time is = 15s

    So that would make: Coefficient = 3.49 / (g * 15)

    Stupid question, what is g?
    Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Turntable physics
  1. Trampoline Physics (Replies: 3)

  2. Statistical Physics (Replies: 2)

  3. Plasma physics (Replies: 2)

Loading...