Answer
(a) $a = 4.64~m/s^2$ (upward)
(b) $T = 105~N$
Work Step by Step
(a) $a = \frac{v^2}{r} = \frac{(4.20~m/s)^2}{3.80~m}$
$a = 4.64~m/s^2$
The magnitude of the acceleration is $a = 4.64~m/s^2$. Since the ball is moving in a circle and there is no tangential acceleration at the bottom point, the acceleration is directed straight up toward the middle of the circle.
(b) We can find the tension $T$ in the rope.
$\sum F = \frac{mv^2}{r}$
$T - mg = \frac{mv^2}{r}$
$T = \frac{mv^2}{r} + mg$
$T = \frac{(71.2~N)(4.20~m/s)^2}{(9.80~m/s^2)(3.80~m)} + 71.2~N$
$T = 105~N$
