A pulley one metre in diameter rotating at 600 revolutions a minute is brought to rest in 80 sec by a constant force of friction on its shaft.  How many revolutions does it makes before coming to rest ?

$$\begin{gathered} \mathrm{given} \mathrm{final} \mathrm{angular} \mathrm{velocity}: \mathrm{\omega }=0; \\ \mathrm{time} \mathrm{t}=80\sec ; \end{gathered}$$

$$\begin{gathered} \mathrm{initial} \mathrm{angular} \mathrm{velocity}:{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ω}}_0\text{=600 revolution/min} \\ \text{=600 × }\frac{2\mathrm{\pi }}{60}\text{=20π radian/sec} \end{gathered}$$

$$\begin{gathered} \mathrm{from} \mathrm{equation} \mathrm{of} \mathrm{motion}: \\ \mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t} \\ \Rightarrow 0=20\mathrm{\pi }+\mathrm{\alpha }\left(80\right) \\ \Rightarrow \mathrm{\alpha } =-\frac{\mathrm{\pi }}{4} \mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{\theta } ={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2 \\ \Rightarrow \mathrm{\theta }=20\mathrm{\pi }\left(80\right)+\frac{1}{2}(-\frac{\mathrm{\pi }}{4}){\left(80\right)}^2 \end{gathered}$$

$$\begin{gathered} \Rightarrow \mathrm{\theta }=800\mathrm{\pi } \mathrm{rad} \\ \Rightarrow \mathrm{\theta }=\frac{800\mathrm{\pi }}{2\mathrm{\pi }}=400 \mathrm{rev} \end{gathered}$$

$\mathrm{Number} \mathrm{revolutions} \mathrm{made} \mathrm{by} \mathrm{pulley}=400$