# Centripetal Force, Gravity and Normal

1. Jun 20, 2014

### Biloon

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!?

2. Jun 20, 2014

### CWatters

The -2mg part originally comes from the KE converted to PE.

Try writing an equation for the energy of the ball at Q.

3. Jun 20, 2014

### Biloon

Oh thanks, so we approach this problem using conservation of energy.

let v = initial velocity

We have this at Point Q:
KE = KE' + PE..

1/2mv^2 = 1/2mv'^2 + mgr.

mv^2/r - 2mg = mv'^2/r

We also have Force at point Q:
Fn = Fc = mv'^2/r = mv^2/r - 2mg

4. Jun 20, 2014

### CWatters

Yes that how I got the book answer.