A pulley one metre in a 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 is 

$$\begin{gathered} given final angular velocity \omega =0;time t=80sec;initial angular velocity{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ω}}_0\text{=600 revolution/sec} \\ convert{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ω}}_0\text{=} 600 to radian/sec \\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ω}}_0\text{=600 x }\frac{2\mathrm{\pi }}{60}\text{=20π radian/sec} \\ from equation of motion \\ \omega ={\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ω}}_0+\alpha t \\ substitute given values here, \\ to get \alpha angular acceleration \\ 0=20\mathrm{\pi }+\mathrm{\alpha }\left(80\right) \\ \mathrm{\alpha } =-\frac{\mathrm{\pi }}{4}\mathrm{radian}/{\mathrm{second}}^2 \\ \mathrm{to} \mathrm{find} \mathrm{Number} \mathrm{of} \mathrm{rotations} \mathrm{made} \mathrm{\theta }, \mathrm{consider} \mathrm{equation} \mathrm{of} \mathrm{motion}, substitute the obtained values \\ \mathrm{\theta } ={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2 \\ \mathrm{\theta }=20\mathrm{\pi }\left(80\right)+\frac{1}{2}(-\frac{\mathrm{\pi }}{4}){\left(80\right)}^2 \mathrm{\theta }=800\mathrm{\pi } \mathrm{radians} \\ \mathrm{convert} \mathrm{into} \mathrm{revolutions} \\ \mathrm{\theta }=\frac{800\mathrm{\pi }}{2\mathrm{\pi }} \\ \mathrm{Number} \mathrm{revolutions} \mathrm{made} \mathrm{by} \mathrm{pulley}=400 \end{gathered}$$