Before braking angular velocity was 300 rad per second. It took 10 seconds to stop. Braking was steady. How many rotations did wheel ?
$\displaystyle \omega_0 = 300 \, rad/s$ and the wheel stopped in $\displaystyle t = 10 \, s$, so $\displaystyle \omega = 0 \, rad/s$. The braking was "steady" which I will take to mean that the angular acceleration $\displaystyle \alpha$ was constant. We wish to know what angle the wheel rotated through during this time period, so I'm going to set $\displaystyle \theta _0 = 0 \, rad$.
One way to do this is to use:
$\displaystyle \omega = \omega _0 + \alpha t$
to find the $\displaystyle \alpha$ then use:
$\displaystyle \theta = \theta _0 + \omega _0 t + \frac{1}{2} \alpha t^2$
to find $\displaystyle \theta$.
The more direct way is to use:
$\displaystyle \theta = \theta _0 + \frac{1}{2}( \omega _0 + \omega )t$
$\displaystyle \theta = \frac{1}{2} \omega _0 t$
$\displaystyle \theta = \frac{1}{2} (300 \, rad/s) (10 \,s ) = 1500 \, rad$
Now we need to find out how many revolutions this is. This is a simple unit conversion: there are $\displaystyle 2 \pi \, rad$ for every revolution. Thus:
$\displaystyle \frac{1500 \, rad}{1} \cdot \frac{1 rev}{2 \pi \, rad} \approx 238.732 rev$
So the answer is about 240 revolutions.
-Dan