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

Homework Help: Rotation/angular acceleration

  1. Apr 1, 2010 #1
    1. The problem statement, all variables and given/known data

    A mass attached to a 48.9 cm long string starts from rest and is rotated 31 times in 78.0 s before reaching a final angular speed. Determine the angular acceleration of the mass, assuming that it is constant.

    2. Relevant equations

    honestly, this is my first problem. I figure that at least angular kinematics will come into play, but I'm not sure where to start...
    I'm thinkin wf=0+a(lpha)t is where I'll want to go, since position is useless here.

    3. The attempt at a solution
    well there's .397 rotations/second, for what it's worth.
    .397 rot/s*2pi radians/rot=15.8rad/s.
    so then I put that into the equation with the known value of time, but hey, that doesn't actually work. so I guess I did my rotation velocity incorrectly, but in all honesty, I don't know how to find that, because I don't really know how to approach this problem. it seems that the length of the string should come into play at some point, but I don't really know how.
     
    Last edited: Apr 1, 2010
  2. jcsd
  3. Apr 1, 2010 #2

    Filip Larsen

    User Avatar
    Gold Member

    If something has a constant angular acceleration, you can easily get the total angle by integrating twice (in same way as you can find linear distance from constant linear acceleration). Once you have total angle as a function of time you are ready to insert the given values.

    And you are right, average angular velocity is not very useful here and the length of the string is not really needed either (if there are other questions for this problem, the length may perhaps be need there).
     
  4. Apr 1, 2010 #3
    well, we have to find the angular speed, and I think that's where we will use the length of the string(maybe?).

    uh, I guess I don't really understand "get the total angle by integrating twice". I cannot say we've ever actually mentioned integrating in my physics class[it's a non calculus based class, so maybe that's why..?]
     
  5. Apr 1, 2010 #4
    31 times is 2pi*31 rads. omega is rads per second, so divide that by the total time to get the average angular velocity. You then know it starts from rest, and that acceleration is constant, so simply double this average to find the final angular velocity.
     
  6. Apr 1, 2010 #5
    although I have no idea as to what you meant by this, it worked. haha.
    thank you both. =]
     
  7. Apr 1, 2010 #6
    Sorry, let me elaborate.

    So if the angular velocity is increasing at a constant rate, then the average angular velocity is the velocity at half the total time.

    You know the equation
    [tex]
    \omega_{avg}=\frac {\omega_{f} - \omega_{i}}{2}
    [/tex]

    And you know the average omega by the method I explained, which is angular kinematics
    [tex]
    \Delta \theta = \omega_{avg}t
    [/tex]
     
  8. Apr 1, 2010 #7

    Filip Larsen

    User Avatar
    Gold Member

    I assumed you had started on calculus, but if you haven't it should still be ok.

    Just like you from a constant linear acceleration, a, and initial speed and distance of zero can get the speed as v = a*t and the distance as s = 1/2 * a * t2 (which are equations I hope you have seen before) you can do exactly the same with angular acceleration and get total angle = 1/2 * angular acceleration * t2. Since you know the total angle (31 * 2 * pi) and you know the time, you can solve that last equation for angular acceleration.
     
  9. Apr 1, 2010 #8
    ah, okay, I think I get what you are both saying. we've never talked about the movement along an angle[our teacher likes to assign problems that cover more than we have currently learned; consequently, I always have trouble on the homework, but not the quizzes/exams...], or anything like the total angle.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook