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Centripetal Force, Gravity and Normal

  • Thread starter Biloon
  • Start date
1. The problem statement, all variables and given/known data

2. Relevant equations
[itex]F_{c} = \frac{mv^{2}}{r}[/itex]

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:

[itex]N = \sqrt{(\frac{mv^{2}}{r})^{2} + (mg)^{2}}[/itex]

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 [itex]\frac{mv^{2}}{r}[/itex]
(b) if calculated from the highest tip of the circle...

but where does [itex]2mg[/itex] comes from!?

Thank in advance.


Science Advisor
Homework Helper
Gold Member
The -2mg part originally comes from the KE converted to PE.

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


Science Advisor
Homework Helper
Gold Member
Yes that how I got the book answer.

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