Answer
(a) $3.81N$
(b) $195N$
Work Step by Step
(a) The tension at the top of the circle can be determined as
$\Sigma F_y=-mg-T=-ma_c$
$\implies T=ma_c-mg$
$\implies T=m(\frac{v^2}{r}-g)$
We plug in the known values to obtain:
$T=(3.25)[\frac{(3.23)^2}{0.950}-9.81]$
$T=3.81N$
(b) We can find the tension at the bottom of the circle as follows:
$\Sigma F_y=-mg+T=ma_c$
This can be rearranged as:
$T=ma_c+mg$
$\implies T=m(\frac{v^2}{r}+g)$
We plug in the known values to obtain:
$T=(3.25)[\frac{(6.91)^2}{0.950}+9.81]$
$T=195N$
