# Loop-the-loop normal force problem. Possible textbook error?

1. May 11, 2014

### zakh508

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

A pilot of mass m flies a jet plane on a "loop-the-loop" course by flying in a vertical circle of radius R at a constant speed v. The force that the cockpit seat exerts on the pilot at the top of the loop is

(A) $mg$.
(B) $mg\left(\frac{v^{2}}{R+1}\right)$.
(C) $\frac{mv^{2}}{R(g+1)}$.
(D) $m\left(\frac{v^{2}}{R-g}\right)$.
(E) $\frac{mv^{2}}{R}$.

2. Relevant equations
$a_{c}=\frac{v^{2}}{R}$

3. The attempt at a solution
I determined that because the centripetal acceleration is pressing the pilot into the seat while gravity is pulling him away from the seat that $F_{N}=\frac{v^{2}}{R}-mg$. I simplified this to $F_{N}=m\left(\frac{v^{2}-g}{R}\right)$. (Just realized this is wrong, should be $F_{N}=m\left(\frac{v^{2}}{R}-g\right)$) The only answer in the book close to that is D but I have no idea how g ended up in the denominator. Could the textbook be wrong or am I missing something?

EDIT: The book's explanation is that the normal force plus force of gravity have to equal mass times centripetal acceleration, therefore $F_{N}+mg=\frac{mv^{2}}{R}$ which they simplified to D. So their explanation makes more sense now but I still think it's wrong.

Last edited: May 11, 2014
2. May 11, 2014

### haruspex

I disagree with all the offered answers and yours.
Centripetal acceleration is not a force, because an acceleration is not a force. Centripetal force is not an applied force - it is a required resultant force (to produce an observed acceleration). So it doesn't push the pilot anywhere.
Also, you seem to be describing the bottom of the loop, whereas the question says the top of the loop.
Dimensional analysis shows that your answer and all the offered answers except A and E must be wrong. But they cannot be right for other reasons.

What are the forces acting on the pilot?
What is the pilot's acceleration?
What equation does that give you?

3. May 11, 2014

### jackarms

Well, the answer should be zero anyway. At the top of the loop, only gravity would be supplying the centripetal acceleration. Answers B, C, and D are all wrong because of dimensions -- you can't add a constant to $R$ for instance, or subtract $g$ (which is an acceleration) from $R$ (which is a distance). Assuming the force isn't zero though, you have the correct answer of $F_{N} = m\frac{v^{2}}{R} - mg$, which (check your simplification), would be $N = m(\frac{v^{2}}{R} - g)$

The book could be wrong.

4. May 11, 2014

### zakh508

I interpreted it as the seat needs to press into the pilot in order to keep him in the loop. I wrote in an edit that the book does give D as the answer and gives the explanation that gravity and the normal force from the seat provide the centripetal force. I'm pretty sure the book is wrong, but I think I have a solid handle on the theory now anyways.

5. May 11, 2014

### haruspex

There's a force from the seat and a force from gravity. The resultant is the centripetal force. If the required centripetal force is less than that from gravity then the normal force from the seat will be positive. There's no reason that the plane should be in free fall at this point.
Except, what about the sign? Assuming up is positive, then you'd have to take R as negative above to get the right relationship between v2/r and FN.
g is always a bit ambiguous because you can either let the gravitational acceleration be g, regardless of sign convention (in which case g will have a negative value when up is positive), or take g to be the absolute value and apply the appropriate sign in the equation. I prefer the former, so if R is positive and up is positive we have
$F_{N} = -m\frac{v^{2}}{R} - mg$
where g is negative.

6. May 11, 2014

### jackarms

Yes, I'm sorry. I was thinking of something of problems such as a bike going around a loop and finding the minimum velocity such that the bike would stay on the loop, in which case the normal force would be zero and it would be in free fall. But here it's different.

Okay, that makes sense. I was thinking more about finding the magnitude of the normal force since its direction was already known.