#### Answer

(a) $a = 1.73~m/s^2$ (upward)
(b) Please refer to the free-body diagram.
(c) $F_N = 0.0115~N$ (upward)
(d) $F_N = 0.00980~N$

#### Work Step by Step

(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.
$F_N = mg = (0.0010~kg)(9.80~m/s^2)$
$F_N = 0.00980~N$