Physics: Principles with Applications (7th Edition)

(a) $\omega = 230~rad/s$ (b) $v = 40~m/s$ $a_R = 9300~m/s^2$ The tangential acceleration is zero.
(a) $\omega = (2200~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = 230~rad/s$ (b) Note that $r = diameter/2$ which is (0.35/2) m. Therefore, $v = \omega r = (230~rad/s)(0.35/2~m) = 40.25~m/s \approx 40~m/s$ We can calculate the centripetal acceleration $a_R$ as: $a_R = \frac{v^2}{r} = \frac{(40.25~m/s)^2}{(0.35/2)~m} = 9300~m/s^2$ Since we can assume that the angular velocity is constant, the tangential acceleration is zero.