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I am having difficulty understanding circular motion problems in an intuitive way - in particular, when an airplane is performing a loop-the-loop:

Suppose an airplane performs a loop-the-loop, so it draws a vertical circle. The speed at the top of the loop is 134 m/s, whereas at the bottom it's 201 m/s. The radius is 366m. What is the pilot's apparent weight at the bottom of the loop, if his true weight is 73kg?

Well then, my first step is to find what forces are being applied on the pilot when he is at the bottom.

  • Clearly there is $mg$, which goes downwards. We are told it is $73$kg.
  • But, there must also be a centripetal force $F_c$ acting against this weight (to keep the plane in the circular motion).

So, assuming that upwards is positive, the forces the pilot is experiencing are:

$$\Sigma F = F_c - mg$$

$F_c$ is positive since it pulls the airplaine towards the center, and $mg$ is negative because its pushing him away.

Then it's just a matter of plugging the values:

$$\Sigma F = m\frac{v^2}{r} - mg$$

$$\implies F = (73/9.8) \cdot\frac{201^2}{366}-73$$

$$\implies F = 749.25$$

So the apparent weight of the pilot is about $750$kg.

In a similar way, the apparent weight when the pilot is at the top would be (assuming that downwards is positive):

$$\Sigma F = F_c + mg$$

Because both the centripetal force and $mg$ are pulling the pilot to the center.


This is wrong - the book proceeds to say that at the bottom it should be

$$\Sigma F = mg + F_c$$

But that means that both the centripetal force and the weight are on the same direction - but that shouldn't be possible because the weight is moving away from the center (downwards) and the centripetal force should be moving it closer to the center (upwards). Why is it this way?

And then at the top:

$$\Sigma F = mg - F_c$$

Which just rises a similar question: at the top, the weight moves to the center, and the centripetal force should also move to the center - so why are they on opposite directions?

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  • $\begingroup$ The question asks about the apparent weight of the pilot, which is the weight a balance would show if he was standing on one. Therefore, what you need to calculate isn't the net force, but the normal force acting on the pilot $\endgroup$ – Nickolas Alves Nov 6 '17 at 4:46
  • $\begingroup$ @NickolasAlves I see! Well, I understand it for the bottom one (because the airplane seat is pushing a normal force of $mg$ towards the center, so it's the same direction as $F_c$). But I don't quite get it for the top: if the pilot is at the top, is the airplane seat still producing a normal force? I imagine that $mg$ and $F_c$ both go to the center of the loop so they should have the same direction. $\endgroup$ – Omega Nov 6 '17 at 4:57
  • $\begingroup$ The reference frame for the pilot is not an inertial frame of reference. There is a centrifugal force acting on him making him "escape" from the plane, so that there is something compressing him against the floor $\endgroup$ – Nickolas Alves Nov 6 '17 at 5:01
  • $\begingroup$ Possible duplicate of Normal force on vertical circular motion $\endgroup$ – sammy gerbil Nov 6 '17 at 5:16
  • $\begingroup$ The top of the loop should be $F_c = N + mg$. The centripetal force should be a net force in these types of problems, especially at points you've described where there is just a radial force present (gravity causes angular acceleration at every other instant except for the top and bottom) $\endgroup$ – sangstar Nov 6 '17 at 11:55
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Centripetal force is not a force itself. It is that part of the resultant force $F$ which causes centripetal acceleration $a_c=\frac{v^2}{r}$ - ie motion in a circle. If the motion is purely circular then the resultant force $F$ is the centripetal force.

The pilot's apparent weight is the normal force $N$ on him from his seat. So the resultant of $N$ and the gravitational force $mg$ equals the centripetal force $ma_c$.

At the top of the loop $N$ will be downwards, assuming the plane is upside down at this point.

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