Calculate Angular Acceleration of Rotating Particle After Fixed Time

The angular position $\theta$ of the particle as a function of time $t$ is given by: $\theta(t) = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3}$

The angular velocity $\omega(t)$ is the first derivative of the angular position with respect to time: $\omega(t) = \frac{d\theta}{dt} = \frac{d}{dt}\left(\frac{t^4}{8} - \frac{t^3}{3}\right) = \frac{4t^3}{8} - \frac{3t^2}{3} = \frac{t^3}{2} - t^2$

The angular acceleration $\alpha(t)$ is the second derivative of the angular position with respect to time: $\alpha(t) = \frac{d\omega}{dt} = \frac{d}{dt}\left(\frac{t^3}{2} - t^2\right) = \frac{3t^2}{2} - 2t$

Substituting $t = 10\text{ s}$ into the expression for angular acceleration: $\alpha(10) = \frac{3(10)^2}{2} - 2(10)$ $\alpha(10) = \frac{300}{2} - 20 = 150 - 20 = 130\text{ rad/s}^2$

Thus, the angular acceleration of the particle after $10\text{ seconds}$ is $130\text{ rad/s}^2$.

Correct Option: C