# Centripetal Force, Gravity and Normal

1. The problem statement, all variables and given/known data

2. Relevant equations
$F_{c} = \frac{mv^{2}}{r}$

3. The attempt at a solution

I think the normal force would be the magnitude of vector sum of centripetal force and gravitational force. So I did:

$N = \sqrt{(\frac{mv^{2}}{r})^{2} + (mg)^{2}}$

However, it isn’t one of the answer choice. The actual answer is (c), and I have got no clue why it is (c).

I could imagine why answer is:
(a) Since mg is perpendicular to point Q, normal force is only $\frac{mv^{2}}{r}$
(b) if calculated from the highest tip of the circle…

but where does $2mg$ comes from!?