Answer
a) $\alpha=9.70\frac{rad}{s^2}$
b) $a_{tan}=11.6\frac{m}{s^2}$
c) $a_R=585\frac{m}{s^2}$
d) $\sum F=4270N$
e) $\theta=1.14^o$
Work Step by Step
a) $\alpha=\frac{\omega^2-\omega_i^2}{2\theta}=\frac{(\frac{v}{r})^2-0}{2\times4\times360^o}=\frac{\Big(\frac{26.5\frac{m}{s}}{1.20m}\Big)^2-0}{2\times4\times2\pi}=9.70\frac{rad}{s^2}$
b) $a_{tan}=r\alpha=(1.20m)(9.70\frac{rad}{s^2})=11.6\frac{m}{s^2}$
c) $a_R=\omega^2r=(22.08\frac{rad}{s})^2(1.20m)=585\frac{m}{s^2}$
d) $\sum F=ma_{net}=m\sqrt{a_{tan}^2+a_R^2}=(7.30kg)\sqrt{(11.6\frac{m}{s^2})^2+(585\frac{m}{s^2})^2}=4270N$
e) $\theta=\arctan\Big(\frac{a_{tan}}{a_R}\Big)=1.14^o$
