Answer
Each spout exerts a force of $0.883~N$
Work Step by Step
We can express the final angular velocity in units of rad/s:
$2.2~rev/s\times \frac{2\pi~rad}{1~rev} = 4.4\pi~rad/s$
We can find the angular acceleration:
$\omega_f = \omega_0+\alpha~t$
$\alpha = \frac{\omega_f - \omega_0}{t}$
$\alpha = \frac{4.4\pi~rad/s-0}{3.20~s}$
$\alpha = 4.32~rad/s^2$
Let $F$ be the force exerted by each spout:
$\tau = I~\alpha$
$3~r~F = I~\alpha$
$F = \frac{I~\alpha}{3r}$
$F = \frac{(9.20\times 10^{-2}~kg~m^2)(4.32~rad/s^2)}{(3)(0.15~m)}$
$F = 0.883~N$
Each spout exerts a force of $0.883~N$
