anonymous
  • anonymous
Ball moving at 18 rads/s slows down at a rate of 2 rads/s^2 how many rev's does it do before it stops?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
It will do \[\frac{81}{2\pi}\]revolutions. I'll prove it in a second.
anonymous
  • anonymous
This is kinematic rotational motion. We develop the equations as such:
anonymous
  • anonymous
...and since you've disappeared, I'm thinking you're not interested.

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anonymous
  • anonymous
I'm here sry
anonymous
  • anonymous
13 rev total but now I'm not so sure
anonymous
  • anonymous
Okay. Angular acceleration is the second derivative of \[\theta (t)\] that is,\[\frac{d^2 \theta}{dt^2}=\alpha\]If you integrate with respect to time, you get\[\frac{d \theta }{dt }=\alpha t + \omega _0\]where omega_0 is the initial rotational speed. Integrate again to get the angle swept out as a function of time\[\theta =\frac{1}{2}\alpha t^2+\omega_0t+\theta_0\]Theta_0 is the initial angle the system starts at (with respect to the standard x-y axis). Since we're only interested in the change in theta (so we can work out the number of revolutions), we can write\[\Delta \theta = \frac{1}{2}\alpha t^2+\omega_0t\](more's coming).
anonymous
  • anonymous
Nothing is mentioned about time in the question, so we need to eliminate it from the equation above. We know, from out calculations that,\[\omega = \alpha t + \omega _0\]so that\[t=\frac{\omega - \omega_0}{\alpha}\]Substitute this into the equation for the total angle swept out to get,\[\Delta \theta = \frac{1}{2}\alpha \left( \frac{\omega - \omega_0}{\alpha} \right)^2+\omega_0\left( \frac{\omega-\omega_0}{\alpha} \right)\]You can simplify this, but since you're only after a number, there's not much point. So, to use it, we know that it has\[\omega_0 = 18 ^cs^{-1}\]\[\alpha = -2^cs^{-1}\]and that, since in the end, it's not moving,\[\omega = 0\]Substituting these values into Delta theta gives you\[\Delta \theta = 81^c\]
anonymous
  • anonymous
Now, this is the total amount of radians swept out, not the revolutions, but we know that there are \[2\pi \frac{rad}{rev}\](i.e. 2pi radians per revolution) and so the number of revolutions is\[n=\frac{81 rad}{2\pi rad/rev}=\frac{81}{2\pi}rev \approx 12.89rev\]
anonymous
  • anonymous
You could have also solved for the time in omega = alpha t + omega_0, and then sub'd that value into the Delta theta expression. The time you get is 9 seconds.
anonymous
  • anonymous
Thx, didn't really know there was people that could help with physics
anonymous
  • anonymous
np. I'd appreciate another fan, that took forever to write out ;p
anonymous
  • anonymous
Really thx a lot though I might be asking so more here soon, that's the best I've seen explained yet

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