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Final Angular Velocity of a Top

  1. Mar 6, 2008 #1
    1. A top is a toy that is made to spin on its pointed end by pulling on the string wrapped around the body of the top. The string has a length of 64cm and is wound around the top at a spot where its radius is 2.0 cm. The thickness of the string is negligible. The top is initially at rest. Someone pulls the free end of thes stirng, theryb undwinding it and givig the top and an angular acceleration of +12 rad/s^2. What is the final angular velocity of the top when the string is completely unwound?



    2. Relevant equations
    [tex]\alpha[/tex] = [tex]\omega[/tex]/t
    [tex]\omega[/tex] = [tex]\theta[/tex] /t
    Circumfrence: [tex]\pi[/tex]r^2


    3. The attempt at a solution

    First I found that the circle's circumference is 4[tex]\pi[/tex].
    Then I divided 64cm/4[tex]\pi[/tex] to find that the rope wraps around 5.09 times.
    I know that 360 degrees is equal to 2[tex]\pi[/tex]radians, so 5.09 x 360 degrees = 1833.5 degrees.

    Then 1833.5 degrees x ([tex]\pi[/tex] radian / 180 degrees ) = 10.2 radians

    Therefore I know the angular displacement, which is from 0 to 10.2 radians
    Then I am stumped on how to find the time from that.

    I know that I can find the final angular velocity by using the time in the angular acceleration formula. Because Angular acceleration is equal to the final velocity / time. I know this because the initial velocity of the top was zero.

    In short I can't figure out how to find the time.
     
  2. jcsd
  3. Mar 6, 2008 #2

    dynamicsolo

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    Homework Helper



    There are two related ways to do this, because one is a method where the time remains concealed.

    In the approach you took, you know that the cord is 64 cm long and the circumference it is wrapped around is [tex]4\pi[/tex] cm. Therefore the top will turn 5.09 cycles (as you said) or, maybe a bit more helpfully, 64 cm/2 cm = 32 radians (since arclength around a circle is radius times angle).

    You can make a kinematic equation for constant angular acceleration analogous to the one for linear motion with constant linear acceleration. Instead of

    x_f = (1/2)·a·(t^2) , starting from rest at x=0,

    you have

    theta_f = (1/2)·(alpha)·(t^2) , starting from rest at theta = 0 .

    The top turns through 32 radians by the time the string unwinds and
    alpha = 12 rad/sec^2 , so

    (t^2) = 2·(32)/12 sec^2 . You can get the time and the final angular velocity from there.

    The other way combines the angle and angular velocity equations for constant acceleration to produce an "angular velocity-squared" equation analogous to the "velocity-squared" equation for linear kinematics. Instead of

    (v_f)^2 = (v_i)^2 + 2 · (a) · (delta_x) ,

    we have

    (omega_f)^2 = (omega_i)^2 + 2 · (alpha) · (delta_theta).

    You know that omega_i = 0, you know alpha, and you know the total angular displacement is delta_theta = 32 radians. Solve for omega_f .

    BTW, I think you dropped a factor here. The 5.09 cycles is correct, but since a cycle is [tex]2\pi[/tex] radians, which is slightly more than six, the total number of radians would have to be somewhat over 30 (in fact, exactly 32, as calculated above).
     
    Last edited: Mar 6, 2008
  4. Mar 7, 2008 #3
    oh thank you, I actually figured it out this morning, but i didn't think I could delete the thread. And I used the method where the time remained concealed! Thank you again!
     
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