Question

Determine the final angular velocity of a particle that rotates 4500 ° in 3 seconds and an angular acceleration of 8 Rad / s ^ 2

Answer 1

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:

$4500^o\, \frac{\pi}{180^o}= 25\,\pi$

And now we can use the kinematic formulas for rotational motion:

$\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2$

Therefore we can find the initial angular velocity $\omega_0$  of the particle:

$\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2\\25\,\pi=\omega_0\,(3)+\frac{1}{2} (8)\,(3)^2\\25\,\pi-36=\omega_0\,(3)\\\omega_0=\frac{25\,\pi-36}{3} \\\omega_0\approx 14.18\,\,\,rad/s$

and now we can estimate the final angular velocity using the kinematic equation for angular velocity;

$\omega=\omega_0\,+\alpha\,t\\\omega=14.18+8\,(3)\\\omega=38.18\,\,\,rad/s$