Answer:
the final angular velocity of the particle is approximately 38.18 Rad/s
Explanation:
To start with, let's make sure that units of angle measure are the same, converting everything into radians:
[tex]4500^o\, \frac{\pi}{180^o}= 25\,\pi[/tex]
And now we can use the kinematic formulas for rotational motion:
[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2[/tex]
Therefore we can find the initial angular velocity [tex]\omega_0[/tex] of the particle:
[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2\\25\,\pi=\omega_0\,(3)+\frac{1}{2} (8)\,(3)^2\\25\,\pi-36=\omega_0\,(3)\\\omega_0=\frac{25\,\pi-36}{3} \\\omega_0\approx 14.18\,\,\,rad/s[/tex]
and now we can estimate the final angular velocity using the kinematic equation for angular velocity;
[tex]\omega=\omega_0\,+\alpha\,t\\\omega=14.18+8\,(3)\\\omega=38.18\,\,\,rad/s[/tex]
