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anonymous
 5 years ago
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?
anonymous
 5 years ago
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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It will do \[\frac{81}{2\pi}\]revolutions. I'll prove it in a second.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is kinematic rotational motion. We develop the equations as such:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...and since you've disappeared, I'm thinking you're not interested.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.013 rev total but now I'm not so sure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. 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 xy 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
 5 years ago
Best ResponseYou've already chosen the best response.0Nothing 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
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0Thx, didn't really know there was people that could help with physics

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0np. I'd appreciate another fan, that took forever to write out ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Really 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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