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

$a_n=5.138m/s^2$ $a_t=0.754m/s^2$
We can determine the required normal and tangential components of acceleration as follows: We know that $\alpha d\theta=\omega d\omega$ $\implies 0.3\theta d\theta=\omega d\omega$ The angular velocity is given as $0.3\int_0^{2\pi} \theta d\theta=\int_1^{\omega} \omega d\omega$ This simplifies to: $0.3[\frac{(2\pi)^2}{2}-0]=[\frac{\omega^2}{2}-\frac{1}{2}]$ This simplifies to: $\omega=3.5844rad/s$ Now the normal component of the acceleration is $a_n=\omega^2 r$ We plug in the known values to obtain: $a_n=(3.584)^2(0.4)$ $a_n=5.138m/s^2$ and the tangential component of the acceleration is given as $a_t=\alpha r$ $\implies a_t=0.3\theta r$ $\implies a_t=0.3(2\pi)(0.4)$ This simplifies to: $a_t=0.754m/s^2$