Show a second angular speed given a reduced radius

[littlemayhem]

Homework Statement

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

Homework Equations

v = r\omega
a = \omega^{2}r
a = r\alpha
\omega^{2}_f = \omega^{2}_{i} + 2 \theta\alpha

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_{i} * \omega_{i} = 0.121
and then tried
r_{f} * \omega_{f} = 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^{2}, 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.

 

[Ikaros]

I just want a way to approach the problem.

Try the conservation of angular momentum.

 

[littlemayhem]

Thank you!

Okay:

L = r x p = r p sin \theta where sin 90 = 1 so

L = rp & p=mv so

L = rmv and v = \omegar so

L = r^{2}m\omega

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

r^{2}_i * \omega_i = r^{2}_f * \omega_f

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?

 

[Ikaros]

Nice work. Looks good to me.
\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}