Is the AAPT 2023 Physics Bowl Answer Key Incorrect?

[kuruman]

I was looking at the AAPT 2023 Physics Bowl questions (and answers) found here
https://aapt.org/programs/physicsbowl/currentexam.cfm
and multiple choice question 6 caught my eye:

Consider one horizontal circular path that a ball makes on the inside surface of a cone. In this situation, the normal force on the ball
a. is mg.
b. is always greater than mg.
c. may be greater or less than mg.
d. is always less than mg.
e. depends upon the speed of the ball.

The posted correct answer is B
The normal vector has a greater magnitude than its vertical component (magnitude of mg). The magnitude of the normal vector is also greater than the horizontal component, the net force directed to the center of circular motion.

Only one answer is allowed which implies that the rest are incorrect. Not so. Since the ball moves in a horizontal circle, the vertical component of the normal force must be equal to the weight and the horizontal component provides the centripetal acceleration.
Let = the half-angle of the cone. Then $$\begin{align} & N\sin\!\alpha=mg \nonumber \\ & N\cos\!\alpha =\frac{mv^2}{r} \nonumber \\ & \sqrt{N^2\sin^2\!\alpha+N^2\cos^2\!\alpha}=N=\sqrt{(mg)^2+\left(\frac{mv^2}{r}\right)^2}.
\nonumber \end{align}$$It looks like both B and E are correct answers. I expected better from the AAPT.

(Edited to fix the equations' format.)
(Re-edited more than a year later to fix misconception. See post #9 below)

 

[Vanadium 50]

Only one answer is allowed which implies that the rest are incorrect.

I believe the instructions say "Choose the best answer."

I also don't think the distractors are good. If you just gave me the answers and not the question, I could exclude C and E as likely possibilities.

 

[kuruman]

I believe the instructions say "Choose the best answer."

I didn't see this instruction anywhere here. Maybe I have the wrong document. In any case, I really don't see why the normal force "depends upon the speed of the ball" is not as good as "is always greater than the weight". The mathematical expression for the normal force says it all.

I agree that the distractors could be better. For example, replace E with

"The normal force is equal to $mg\cos\!\alpha$ where $\alpha$ is the cone half-angle."

Then the centripetal is component is cut out and the question is entirely focused on the comparison between the normal force and the weight.

 

[PhDeezNutz]

I know this question is more than a year old but it intrigued me so I decided to practice some Lagrangian mechanics. I'm getting Answer B let me know what you think if you get the chance.

EL - equation of course is

$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = \lambda \frac{\partial f}{\partial q_i}$

where $q_i$ is the generalized coordinate, $\lambda$ is the Lagrange multiplier and $f$ is a constraint of the form $f \left(q_1,....q_n \right) = 0$ for the generalized coordinates $i = 1,2,....n$. And the expression $\lambda \frac{\partial f}{\partial q_i}$ is generalized force of constraint component relating to that coordinate. I believe the force of constraint is always Normal to the surface of constraint.

With the way I am treating the problem I have only one equation of constraint so I didn't bother summing/indexing $\lambda$ and $f$. I'm doing it in cylindrical coordinates

$x = r \cos \theta$
$y = r \sin \theta$
$z = z$

the equation of the constraint surface is $z = C \sqrt{ x^2 + y^2} = Cr$ where

$C$ is a constant relating to how wide the cone opens and it can be greater or less than $1$.

So $f\left( r, \theta, z \right) = Cr - z = 0$

The unconstrained Lagrangian in Cartesian is

$L = \frac{1}{2} m \left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) - mgz$

In cylindrical that is

$L = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2 \right) - mgz$

again $f\left( r, \theta, z \right) = Cr - z = 0$

The EL equations become

$m \ddot{r} - mr \dot{\theta}^2 = \lambda C$

$mr^2 \ddot{\theta} = 0$

$m \ddot{z} - mg = - \lambda$

if we further require for the path to be at a constant height $\ddot{z} = 0$ and we have

$m \ddot{r} - mr \dot{\theta}^2 = \lambda C$

$mr^2 \ddot{\theta} = 0$

$- mg = - \lambda$

which then implies $- \lambda = - mg$ and then we can plug that in above

$m \ddot{r} - mr \dot{\theta}^2 = mg C$

$mr^2 \ddot{\theta} = 0$

$- mg = - mg$

The force of constraint magnitude is $N = \sqrt{ m^2 g^2 + C^2 m^2 g^2}$ which is always greater than $mg$ regardless of how small $C$ is.

 

[PhDeezNutz]

Then again $m \ddot{r} = mgC + mr \dot{\theta}^2$ and maybe that's the $r$- component of force which would make sense but I swear I remember $\lambda \frac{\partial f}{\partial q_i}$ being the components of force by themselves. I'm gonna find my copy of Marion and Thornton and look really quick.

Sorry if I have wasted anyones time.

 

[PhDeezNutz]

@kuruman I realized after looking over my work again that we believe the same thing.

the equation
$m \ddot{r} - mr \dot{\theta}^2 = mgC$

literally indicates a dependence on angular speed. so yes I agree that B and E are both valid answers.

Sorry for the brain flatulence.

 

[DaveE]

Since the question states that the ball travels a circular path inside a cone, doesn't that determine the ball's velocity vs. radius? Fast balls spiral up to some equilibrium position, etc. I think that needs to be taken into account, although I'm both too lazy and too ignorant to actually do it (easily).

 

[A.T.]

Consider one horizontal circular path that a ball makes on the inside surface of a cone. In this situation, the normal force on the ball
a. is mg.
b. is always greater than mg.
c. may be greater or less than mg.
d. is always less than mg.
e. depends upon the speed of the ball.

The posted correct answer is B

B and E are correct answers.

I suspect E was meant to be: 'It depends on the speed of the ball, whether N is greater, equal or less than mg', which would be wrong. But the way they combined the question with the choices changes that meaning. They should have written each choice as a full sentence, if they wanted to include E.

 

[kuruman]

I know this question is more than a year old but it intrigued me so I decided to practice some Lagrangian mechanics. I'm getting Answer B let me know what you think if you get the chance.

I think that you exposed a blind spot that I had. Answer B is the correct answer and I have already indicated that. However, I deluded myself when I looked at the expression for the magnitude of the normal. I will reproduce the equations below and then comment on them. $$\begin{align} & N\sin\!\alpha=mg \\ & N\cos\!\alpha =\frac{mv^2}{r} \\ & \sqrt{N^2\sin^2\!\alpha+N^2\cos^2\!\alpha}=N=\sqrt{(mg)^2+\left(\frac{mv^2}{r}\right)^2}.
\end{align}$$Comments

1. Equation (1) can be rewritten as $N=mg/\sin\!\alpha$ which clearly shows that, for cone half-angle $0<\alpha<\frac{\pi}{2}$, the normal force is greater than the weight. This justifies the choice of (B) as correct.
2. Could (E) also be correct? Arguably, we can use the Pythagorean theorem to combine the vertical and horizontal components in equations (1) and (2) to get the magnitude as seen in equation (3). Here is where I pulled the wool over my own eyes. I know mantra that "the normal force adjusts itself to provide the observed acceleration", I know that the speed $v$ depends on the distance $z$ from the cone's apex. Therefore, answer (E) could also be correct, right? Wrong.
3. Why wrong? We have already established in Comment 1 that the magnitude of the normal force as a ratio of constant quantities must be constant. The only way to have $N$ constant given equation (3) is to have $v^2/r =\text{const.}$.
4. Is it obvious that $v^2/r =\text{const.}$? It wasn't to me but it should have been as explained below. There are two physical considerations that need to be taken into account: (a) The normal force must be perpendicular to the cone surface and form angle $\varphi=90^{\circ}-\alpha$ with respect to the vertical; (b) the vertical component of the normal force is fixed and equal to the weight $mg$. We have a right triangle having right sides $mg$ and $mv^2/r$ and hypotenuse $N$. We know one right side (the weight), and the angles adjacent to it ($\varphi$ and $90^{\circ}$). This makes the triangle unique no matter where one draws it on the surface of the cone. In other words, since the vertical component of the normal force is fixed and the normal force has to be perpendicular to the cone's surface, the centripetal force $F_c=mv^2/r$ has a fixed value and does not depend on the speed.

I apologize to the AAPT author(s) of this question for doubting their work.

 

[PhDeezNutz]

the equation
$m \ddot{r} - mr \dot{\theta}^2 = mgC$

@kuruman in regards to post #9. Let me see if I am tracking. (Still using the lagrangian approach)

I tried to backtrack in the quoted post above that since $m \ddot{r} - mr \dot{\theta}^2 = mgC$ that the component of $N$ related to $r$ depended on $\dot{\theta}$ (squared) and therefore the Normal DID depend on $\dot{\theta}$.

BUT according to constraint $Cr = z = constant$ requires that $\ddot{r}$ is constant (because $r = \text{constant}$ and therefore

$mr \dot{\theta}^2 = mgC$ and therefore $\dot{\theta} = \text{constant}$

If $\dot{\theta} = \text{constant}$ then $v = r \dot{\theta} = \text{constant}$

As you said (paraphrasing\interpretation) "if $v$ itself is a constant of motion needed to maintain the constant height constraint on the cone then the normal (constraint) force cannot depend on $v$, because if it did we wouldn't maintain constant height"

Did I understand you correctly in my overall quote?

 

[kuruman]

Did I understand you correctly in my overall quote?

I never said that $v$ is constant. I said that $v^2/r$ is constant which makes the centripetal force constant and independent of vertical distance of the orbit from the cone's apex.

 

[PhDeezNutz]

I never said that $v$ is constant. I said that $v^2/r$ is constant which makes the centripetal force constant and independent of vertical distance of the orbit from the cone's apex.

Let me take some time to think about this. I don't want to give you a knee jerk response. It's only fair that since you solved in the Newtonian way (and I solved it in the Lagrangian way) that I fully go through your train of thought so that we're on the same page. Hopefully I can reconcile the two.

Regardless

thank you for posting this problem I've enjoyed it very much.

Maybe somewhere down the line I can use this as a chance to practice Hamilton Jacobi Theory to uncover constants of motion if i start from a cartesian stand point without knowing ahead of time to convert to cylindrical.

 

[kuruman]

This is how I would do it with a Lagrangian starting with standard cylindrical coordinates. $$\mathcal L=\frac{1}{2}m\dot r^2+\frac{1}{2}mr^2\dot\theta^2+\frac{1}{2}m\dot z^2-mgz.$$This is also your starting point. I now introduce the constraint that the ball stay on the surface of a cone of half-angle $\alpha$, $~z=r\cot\!\alpha~$ to get (after gathering terms) $$\mathcal L=\frac{1}{2}m(1+\cot^2\!\alpha)\dot r^2+\frac{1}{2}mr^2\dot\theta^2-mg~r\cot\!\alpha.$$From this I get two equations of motion

For $\theta$ the EOM is $$0=\left(\frac{d}{dt}\frac{\partial \mathcal L}{\partial \dot{\theta}}-\frac{\partial \mathcal L}{\partial \theta}\right)=\frac{d}{dt}(mr^2\dot{\theta})-0\implies mr^2\dot{\theta}=\text{const.} $$This says that the angular momentum is conserved (is a constant of the motion) regardless of the trajectory of the ball.

For $r$ the EOM is $$0=\left(\frac{d}{dt}\frac{\partial \mathcal L}{\partial \dot{r}}-\frac{\partial \mathcal L}{\partial r}\right)=(1+\cot^2\!\alpha)m\ddot r-mr\dot{\theta}^2 +mg\cot\!\alpha. $$When the trajectory of the ball is a horizontal circle of constant radius, the equation of motion becomes $$-mr\dot{\theta}^2 +mg\cot\!\alpha=0\implies F_c=mr\dot{\theta}^2=mg\cot\!\alpha=\text{const.}$$ Thus, the centripetal force is constant when the trajectory is a horizontal circle but not in general. You get the same expression from equations (1) and (2) in post #9.

thank you for posting this problem I've enjoyed it very much.

And I thank you for making me realize that I had a blind spot here.

 

[PhDeezNutz]

I really like how you incorporated the constraints directly into the Lagrangian and found how angular momentum is a constant of motion regardless of trajectory within the cone.

With the way I did it (not incorporating the constraints directly but rather through Lagrange multipliers) I would have probably have never come to conclusion that angular momentum is conserved regardless of trajectory.