Question

A turbine is spinning at $3800\mathrm{rpm}$. Friction in the bearings is so low that it takes $10\min$ to coast to a stop. How many revolutions does the turbine make while stopping?

Short answer

The turbine make$19012\mathrm{rev}$ while stopping.

Step-by-step solution

Given Information

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{f}}=3800\mathrm{rpm} \\ {\mathrm{\omega }}_{\mathrm{i}}=0 \end{gathered}$$

$\mathrm{t}=10\min$

Formula

$2∆\mathrm{\theta }=({\mathrm{\omega }}_{\mathrm{f}}+{\mathrm{\omega }}_{\mathrm{i}})\mathrm{t}$

Explanation

First convert angular velocity from revolution per minute to radian per second:

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{f}}=3800\frac{\mathrm{rev}}{\min }\times \frac{1\min }{60\mathrm{s}}\times \frac{2\mathrm{\pi }\mathrm{rad}}{1\mathrm{rev}} \\ =397.73\mathrm{rad}/\mathrm{s} \end{gathered}$$

Now, convert time in sec from min

$$\begin{gathered} \mathrm{t}=10\min \times \frac{60\mathrm{s}}{1\min } \\ =600\mathrm{s} \end{gathered}$$

By putting values in formula

$$\begin{gathered} \mathrm{\theta }=\frac{({\mathrm{\omega }}_{\mathrm{f}}+{\mathrm{\omega }}_{\mathrm{i}})\mathrm{t}}{2} \\ \mathrm{\theta }=\frac{(398\mathrm{rad}/\mathrm{s}+0\mathrm{rad}/\mathrm{s})(600\mathrm{s})}{2} \\ \mathrm{\theta }=\frac{(398\mathrm{rad}/\mathrm{s})(600\mathrm{s})}{2} \\ \mathrm{\theta }=119400\mathrm{rad} \end{gathered}$$

Converting radian into revolution

$$\begin{gathered} \mathrm{\theta }=2\mathrm{\pi N} \\ \mathrm{N}=\frac{\mathrm{\theta }}{2\mathrm{\pi }} \\ \mathrm{N}=\frac{119400\mathrm{rad}}{\left(2\right)(3.14\mathrm{rad})} \\ \mathrm{N}=19012.7\mathrm{rev} \end{gathered}$$