## University Physics with Modern Physics (14th Edition)

(a) $a = 1.73~m/s^2$ (upward) (b) Please refer to the free-body diagram. (c) $F_N = 0.0115~N$ (upward) (d) $F_N = 0.00980~N$
(a) We can find the speed $v$ at the bottom of the swinging arm. $v = \frac{2\pi ~(0.700~m)}{(8)(0.5~s)} = (0.350~\pi)~m/s$ We can find the acceleration at the bottom of the swing. $a = \frac{v^2}{r} = \frac{(0.350~\pi~m/s)^2}{0.700~m}$ $a = 1.73~m/s^2$ (upward) (b) Please refer to the free-body diagram. (c) $\sum F = ma$ $F_N - mg = ma$ $F_N = ma+mg$ $F_N = (0.0010~kg)(1.73~m/s^2)+(0.0010~kg)(9.80~m/s^2)$ $F_N = 0.0115~N$ (upward) (d) If the arm were not swinging, then the normal force exerted by the blood vessel would be equal in magnitude to the weight of the drop. $F_N = mg = (0.0010~kg)(9.80~m/s^2)$ $F_N = 0.00980~N$