Answer:
3.34
Explanation:
The relationship between final angular speed, initial angular speed, angular acceleration, and change in radians is given by
[tex]\omega^2_f=\omega^2_i+2\alpha(\theta_f-\theta_i)[/tex]Now in our case
[tex]\begin{gathered} \omega_i=59rpm \\ \omega_i=59\times\frac{2\pi}{60}\text{rad}/s \\ \end{gathered}[/tex][tex]\omega_f=17\times\frac{2\pi}{60}\text{rad}/s[/tex][tex]\theta_f-\theta_i=300^o=\frac{5}{3}\pi(radians)[/tex]Therefore, the above equation gives
[tex](17\times\frac{2\pi}{60})^2=(59\times\frac{2\pi}{60})^2+2(\frac{5}{3}\pi)\alpha[/tex]subtracting (17 2pi/ 60 )^2 from both sides gives
[tex](17\times\frac{2\pi}{60})^2-(59\times\frac{2\pi}{60})^2=2(\frac{5}{3}\pi)\alpha[/tex]dividing both sides by 2 * 5 / 3 pi gives
[tex]\frac{(17\times\frac{2\pi}{60})^2-(59\times\frac{2\pi}{60})^2}{2(\frac{5}{3}\pi)}=\alpha[/tex]simplifying the above gives
[tex]\alpha=3.34[/tex]which is our answer!
