## College Physics (4th Edition)

Each spout exerts a force of $0.883~N$
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 $0.883~N$