## Engineering Mechanics: Statics & Dynamics (14th Edition)

$\alpha=60rad/s^2$ $\omega=90rad/s$ $\theta=90 rad$
We can determine the required angular acceleration, angular velocity and the angular displacement as follows: As $a_t=\alpha r$ We plug in the known values to obtain: $10t=\alpha \times 0.5$ This simplifies to: $\alpha=20t$ At $t=3s$ $\alpha=20(3)=60rad/s^2$ The angular velocity can be calculated as $d\omega=\alpha dt$ $\implies \int_0^\omega d\omega=\int_0^ t 20t dt$ This simplifies to: $\omega=10t^2$ We plug in the known values to obtain: $\omega=10(3)^2=90rad/s$ Now we calculate the angular displacement as $d\theta=\omega dt$ $\int_0 ^\theta d\theta=\int_0^t 10t^2 dt$ $\theta=(\frac{10}{3})t^3$ At $t=3s$ $\implies \theta=(\frac{10}{3})(3)^3$ $\implies \theta=90 rad$