- #1
littlemayhem
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Homework Statement
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
Homework 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]
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.