A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/seconds squared. How many revolutions does it make before stopping?
If only I could find the angular velocity first, in radians/second...
I'm assuming 0.700 rad/second squared is angular acceleration, and so far I have only been taught equations to find angular acceleration with time, and I don't know time.
How do I find angular velocity given this information? Once I get there, I can convert it to revolutions.
On
As per your question, the initial rate given is the angular velocity. And since it finally has to stop, the final angular velocity would be zero. You are already given the angular acceleration. Therefore, you can calculate the time taken by the gyroscope stops.
$$\omega = \omega_{0} + \alpha t $$
$$ 0 \text{ rad/s} = 32 \text{ rad/s} - 0.7 \text{ rad/s}^{2} t$$
$$ t = \frac{32 s^{-1}}{0.7 s^{-2}} = 45.7 \text{ s} $$
And for calculating revolutions
$$ Θ = \omega_{0} t + \frac{1}{2} \alpha t^{2} $$
$$ = 32 \text{ rad/s} \times 45.7 \text{ s} - \frac{1}{2} 0.7 s^{-2} \times \left( 45.7 \text{ s}\right)^{2} $$
$$ = 731 \text{ rad} $$
which is $731$ rad $ \times \frac{1 \text{ rev}}{2 \pi \text{ rad}} = 116$ revolutions.
Note how the rotation equations resemble the translation equations:
$$ v = V_{0} + a t $$
$$ s = V_{0} t + \frac{1}{2} a t^{2} $$
You do have the initial angular velocity; it is given as 32 rad/s.
The angular acceleration is $-0.7$ rad/$s^2$, it is negative because the gyro is slowing.
So to find the stopping time you have to solve $$ 32 - 0.7 t = 0 \\ t = 320/7 \approx 45.71 $$ Now you need to compute the number of revolutions, and here a trick is to note that the average angular velocity will be exactly half the initial angular velocity, since it is slowing with constant deceleration.
$$ \bar{\omega} = {16.0} \\ N_{\mbox{revolutions}}= \frac{t \bar{\omega} }{2\pi}= \frac{16\times 320}{14\pi} \approx 116.41 $$
So it will make $114$ complete revolutions, and a smidgen more.