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Homework Help: Constant Angular Acceleration Question

  1. Apr 6, 2010 #1
    Dario, a prep cook at an Italian restaurant, spins a salad spinner 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration.

    a) Find the angular distance the salad spinner travels as it comes to rest

    b) What is the angular acceleration of the salad spinner as it slows down?

    2. Relevant equations
    constant rotational equations

    3. The attempt at a solution
    a) ok, i found initially when the spinner is spinning 20 times in 5seconds to have a angular velocity of 1440 degrees. For the second part when it spins 6 more times, i calculated: [tex]\frac{6 * 360}{5}[/tex]

    so angular distance is 1440-432 = 1008 degrees

    b) To find angular acceleration i used the formula : [tex]\omega final^2 = \omega initial^2 + 2\alpha\theta [/tex]

    [tex]432^2 = 1440^2 + 2\alpha * 1008[/tex]

    [tex]\alpha = \frac{186624 - 2073600}{2016}[/tex]

    [tex]\alpha = 936 rad/s^2[/tex]

    Can someone please check if this is correct.
    Last edited: Apr 6, 2010
  2. jcsd
  3. Apr 6, 2010 #2
    The angular velocity is equal to the change in theta divided by the change in time. You know that there is 20 revolutions on 5 seconds. Angular velocity is measured in radians per second so you must multiply 20(2pi) then divide by the time interval of 5 seconds. This will give you your initial angular velocity.
  4. Apr 6, 2010 #3
    how would i find my final angular velocity?
  5. Apr 6, 2010 #4
    You can use the equation of change in theta is equal to (1/2)(wfinal+winitial) times the change in time. Since we know the data at the initial point we can plug in the numbers which would be 20(2pi)=(1/2)(wfinal+[(20*2pi)/5])(5) then solve for wfinal
  6. Apr 6, 2010 #5
    thanks i got the answer