Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 352: 72

$ 10.83\;\rm N\cdot m$

When we cut the power, the only force acting on the turbine is the friction force. Thus, the only torque exerted on the turbine is the one due to the friction of the bearings. $$\sum\tau=-\tau_f= I\alpha\tag 1$$ The negative sign is due to the direction of the torque that is opposing its rotation direction. We know that the angular acceleration is given by $\alpha=\dfrac{\omega_f-\omega_i}{t}$ and we know that the final angular speed is just half the initial one $\omega_f=\frac{1}{2}\omega_i$. Thus, $\alpha=\dfrac{\frac{1}{2}\omega_i-\omega_i}{t}$ Plugging into (1); $$- \tau_f= I\dfrac{-\frac{1}{2} \omega_i}{t}$$ Thus, $$t = \dfrac{ \frac{1}{2}I }{\tau_f} \omega_i $$ Plugging the known; $$t = \dfrac{ \frac{1}{2}\times 2.6 }{\tau_f} \omega_i $$ $$t = \dfrac{ 1.3 }{\tau_f} \omega_i \tag 2$$ Now we have $t$ as a function of $\omega_i$ and its slope is $${\rm Slope}= \dfrac{ 1.3 }{\tau_f}$$ Solving for the friction torque; $$\boxed{\tau_f = \dfrac{ 1.3 }{ {\rm Slope}}}$$ Now we need to draw the function in equation (3) and then find its slope then plug the result into the boxed formula to find the friction torque. So we have to convert the angular speeds to rad/s rather than rpm. See the table below. \begin{array}{|c|c|} \hline \omega_i{\rm\;(rad/s)}& t\;{\rm (s)}\\ \hline 50\pi & 19 \\ \hline 60 \pi& 22 \\ \hline 70\pi & 25 \\ \hline 80\pi & 30 \\ \hline 90\pi& 34 \\ \hline \end{array} From the figure below, we can see that the slope is 0.12. Plugging into the boxed formula above; $$ \tau_f = \dfrac{ 1.3 }{ 0.12}=\color{red}{\bf 10.83}\;\rm N\cdot m$$