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

the final angular velocity of the particle is approximately 38.18 Rad/s

Step-by-step explanation:

To start with, let's make sure that units of angle measure are the same, converting everything into radians:

`4500^o\, (\pi)/(180^o)= 25\,\pi`

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

`\theta-\theta_0=\omega_0\,t+(1)/(2) \alpha\,t^2`

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

`\theta-\theta_0=\omega_0\,t+(1)/(2) \alpha\,t^2\\25\,\pi=\omega_0\,(3)+(1)/(2) (8)\,(3)^2\\25\,\pi-36=\omega_0\,(3)\\\omega_0=(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`