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Show a second angular speed given a reduced radius

  1. Aug 8, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle on a string at radius r=0.22m is moving in a (horizontal) circle with angular speed [itex]\omega[/itex] =0.55 rad/s. The string is shortened to 0.15m. Show that the new angular speed is 1.18rad/s

    2. Relevant equations

    v = r[itex]\omega[/itex]
    a = [itex]\omega^{2}r[/itex]
    a = r[itex]\alpha[/itex]
    [itex]\omega^{2}_f = \omega^{2}_{i} + 2 \theta\alpha[/itex]
    3. The attempt at a solution

    Okay. At first I thought this is really simple, which means I'm missing something. I assumed that there was no acceleration, and velocity was constant, but that can't be the case.

    I tried r[itex]_{i}[/itex] * [itex]\omega_{i}[/itex] = 0.121
    and then tried
    r[itex]_{f}[/itex] * [itex]\omega_{f}[/itex] = 0.177

    Which clearly isn't right for constant velocity - there must be acceleration, but I don't know over what period the acceleration occurs. I've looked at the kinematics equations for angular motion, but they either involve time or theta, variables not mentioned.

    I know it says speed, but I'm presuming it means acceleration - and vectors aren't really relevant in this case, anyway, since there's no change of direction.

    The only thing I can think is that we're supposed to work backwards, knowing that the acceleration is from 0.55 to 1.18 rad, but I don't know how to work that out in rad/s[itex]^{2}[/itex], because that involves time.

    I just want a way to approach the problem. We're studying moments of inertia mainly, I'm also thinking I might just be looking at this the wrong way, but I'm not sure. Any guidance would be appreciated - I really want to have a clear understanding from my own workings.
     
  2. jcsd
  3. Aug 8, 2011 #2
    Try the conservation of angular momentum.
     
  4. Aug 8, 2011 #3
    Thank you!

    Okay:

    L = r x p = r p sin [itex]\theta[/itex] where sin 90 = 1 so

    L = rp & p=mv so

    L = rmv and v = [itex]\omega[/itex]r so

    L = r[itex]^{2}[/itex]m[itex]\omega[/itex]

    if we equate {r[itex]^{2}[/itex]m[itex]\omega[/itex]}[itex]_{initial}[/itex] and {r[itex]^{2}[/itex]m[itex]\omega[/itex]}[itex]_{final}[/itex] and cancel the mass (which stays the same), we get

    [itex]r^{2}_i * \omega_i = r^{2}_f * \omega_f [/itex]

    with the values included, we have

    0.22 * 0.22 * 0.55 = 0.15 * 0.15 * 1.18

    which equal 0.02662 and 0.02655 respectively.

    I suppose that the small difference is attributable to the fact that the initial values were given only to two decimal places, and thus that my calculation is correct?
     
  5. Aug 8, 2011 #4
    Nice work. Looks good to me.
    [tex]\omega _f = \omega _i \frac{{r_i^2 }}{{r_f^2 }} = 0.55\frac{{0.22^2 }}{{0.15^2 }} = 1.18{\rm{rad}} \cdot {\rm{s}}^{ - 1}[/tex]
     
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