Answer
$a_{cp}=5.20m/s^2;a_t=4.46m/s^2;a=6.85m/s^2;\phi=49.4^{\circ}$
Work Step by Step
We can find the centripetal acceleration as
$a_{cp}=r\omega^2$
We plug in the known values to obtain:
$a_{cp}=(7.20m)(0.850rad/s)^2$
$a_{cp}=5.20m/s^2$
The tangential acceleration is given as
$a_t=r\alpha$
We plug in the known values to obtain:
$a_t=(7.20m)(0.620rad/s^2)$
$a_t=4.46m/s^2$
The total acceleration is given as
$a=\sqrt{a_{cp}^2+a_t^2}$
We plug in the known values to obtain:
$a=\sqrt{(5.20m/s^2)^2+(4.46m/s^2)^2}$
$a=6.85m/s^2$
We can find the required angle as follows:
$\phi=tan^{-1}(\frac{a_{cp}}{a_t})$
We plug in the known values to obtain:
$\phi=tan^{-1}(\frac{5.20m/s^2}{4.46m/s^2})$
$\phi=49.4^{\circ}$
