Answer
a) $816 kg\frac{m^2}{s^2}$
b) $816Nm$
c) $932W$
Work Step by Step
a) The angular momentum delivered to the waterwheel is that lost by the water.
$$\Delta L_{wheel}=-\Delta L_{water}=L_{initial water}-L_{final water}=mv_1R-mv_2R$$
$$\frac{\Delta L_{wheel}}{\Delta t}=\frac{mv_1R-mv_2R}{\Delta t}=\frac{mR}{\Delta t}(v_1-v_2)=(85kg/s)(3m)(3.2m/s)$$
$$=816kg.m^2/s^2$$
b) The torque is the rate of change of angular momentum, from Eq. 8-16.
$$\tau_{on wheel}=\frac{\Delta L_{wheel}}{\Delta t}=816kg.m^2/s^2=816Nm$$
c) Power is given by $P=\tau \omega$.
$$P=\tau \omega=(816Nm)(\frac{2\pi rev}{5.5s})=932W$$
