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The physics concepts here are inertial and non-inertial frames of reference. Also rotational equilibrium.

On the picture, you can see a person performing the stunt in so called "devil's barrel". I want to analyse the scenario as an inertial observer standing near the barrel. The person is speeding up on the bottom of the barrel and then rides on the sides of the barrel. Let's say the person-bike system's center of mass has velocity $V$ (going around the barrel). The person will tilt the bike, as on the second picture. Arrows represent: $W$ - weight, $R$ - reaction force, $F$ - static(?) friction on the biker. The rectangle represent the person-bike system (the distance between contact point and center of mass is $L$).

I have a problem with rotational equilibrum of the described setup.

Firstly, let's choose an axis of rotation around the center of mass (out of the page). The rotational equilibrium gives: $R\cdot L\sin(\alpha)=F\cdot L\cos(\alpha)$. Great, I will get the angle of tilt.

Secondly, let's take an axis of rotation out of the page at the contact point of wheels and walls. There is only one real force acting there (weight) hence it should always rotate and hit the ground/wall. If I choose the non-inertial frame of reference (an ant that sits on the biker's helmet), additional force (centrifugal force) is attached to center of mass and that would allow me to find the angle.

However, I don't understand why the inertial frame of reference is not giving me the same result. What do I omit? I guess there is sth to do with the fact that the CM goes around the circle, so it has some acceleration? Don't know and want some help :)

Free body diagram of the biker

Picture of situation

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    $\begingroup$ Trying to figure out 3D dynamics by looking at single component directions is a recipe for disaster. $\endgroup$
    – JAlex
    Commented Sep 29, 2022 at 14:15

3 Answers 3

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In D'Alembert approach, you forgot the apparent inertia force (the so called centrifugal force), namely a force pointing outward horizontally, with magnitude $F_c = m \dfrac{V^2}{r}$, with the line of application passing through the center of mass, being $r$ the distance of the center of mass from the axis of symmetry of the barrel (that is the axis of rotation).

For the inertial observer standing near the barrel, you need to write equations for translation and rotation, namely

$\dfrac{d \mathbf{Q}}{d t} = \mathbf{R}^e$

$\dfrac{d \mathbf{\Gamma}_H}{dt} = -\mathbf{\dot{x}}_H \times \mathbf{Q} + \mathbf{M}^e_H$

Representing the system with a point mass in the center of mass, and taking the center of mass $G \equiv H$ as the pole of the momentum equation, the dynamic equations become

$m \dfrac{d \mathbf{v}}{dt} = \mathbf{W} + \mathbf{R} + \mathbf{F}$

$\mathbf{0} = \mathbf{0} + \mathbf{\ell} \times \mathbf{R} + \mathbf{\ell} \times \mathbf{F}$,

being $\mathbf{\ell}$ the vector from the center of mass to the point of contact. Now, by means of projection we can get 6 scalar equations from these 2 vector equations in 3D space.

The only information we can get from the last equation corresponds to the component of the angular momentum (or the balance of moments) around an axis perpendicular to the screen, namely

$0 = F \ell \cos \alpha - R \ell \sin \alpha$.

while the projections in orthogonal directions are identically zero.

To determine $F$ and $R$, you can evaluate them from the dynamic equation for translation, projecting the vector equation onto the direction tangent to the trajectory of the bike $\mathbf{\hat{t}}$, the normal direction $\mathbf{\hat{n}}$ pointing towards the axis of the barrel and the binormal direction $\mathbf{\hat{b}}$ pointing in the vertical direction.

$t: m \dfrac{d v}{d t} = 0$
$n: m \dfrac{v^2}{r} = R$$\qquad \rightarrow \qquad$$R = m \dfrac{v^2}{r}$
$b: 0 = F - W$$\qquad \rightarrow \qquad$$F = W = m g$

Thus, $\tan \alpha = \dfrac{F}{R} = \dfrac{g r}{v^2}$.

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  • $\begingroup$ Yes I know that, I suggested that in the approach about the ant sitting at the helmet. My problem is, why the inertial frame of reference doesn't work $\endgroup$ Commented Sep 27, 2022 at 11:03
  • $\begingroup$ What is apparent inertia force in D'Alembert approach is time derivative of momentum (in dynamic equation for translation), and time derivative of angular momentum (in dynamic equation for rotation). I'll edit my answer $\endgroup$
    – basics
    Commented Sep 27, 2022 at 12:50
  • $\begingroup$ random anonymous -1. Old habits never end $\endgroup$
    – basics
    Commented Aug 25, 2023 at 11:21
  • $\begingroup$ Fixed your problem, friend $\endgroup$ Commented Aug 25, 2023 at 16:04
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. . . . let's take an axis of rotation out of the page at the contact point of wheels and walls. There is only one real force acting there (weight) hence it should always rotate and hit the ground/wall.

You have defined a non-inertial frame of reference as that particular axis of rotation is accelerating relative to the ground frame.
To use Newton's second law you must introduce a fictitious force, the centrifugal force acting at the centre of mass.

The reason you were able to use the centre of mass frame, . . . . . let's choose an axis of rotation around the center of mass . . . . . , is that although it is accelerating relative to the ground frame any fictitious force has to act through the centre of mass and hence exerts no torque about the centre of mass.

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  • $\begingroup$ -1 Why? It would be nice not to repeat the mistake that I have made. $\endgroup$
    – Farcher
    Commented Aug 10, 2023 at 16:33
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Let the last figure be projected onto a vertical plane (i.e. the page) so we can define an x - and y - axis laying in this plane with origo at the chosen $fixed$ point, and can give x - and y - coordinates to the moving CM (i.e. of the projection onto the plane of the CMs position vector from the origo). The y - coordinate is constant, say y = h. Let $\alpha (t)$ be the angle defined by the y - and x - coordinate of the CM . Say the rider is at the fixed point at time $t=t_0$ (so $\alpha (t_0)=\alpha$ in the last figure). At a moment after $t_0$ the CM has moved "to the right" so $X_{cm} (t_0 + \Delta t) > X_{cm} (t_0) $, and so $\alpha (t_0 + \Delta t)$ has decreased. Se the illustration below. In fact, by symmetry $\alpha (t_0 - \Delta t)$ is smaller than $\alpha _0$ by the same amount, so the rotation is counter-clockwise from $t_0 - \Delta t$ to $t_0$, and clokwise from $t_0$ to $t_0 + \Delta t$, consistent with an angular acceleration caused by the torque pointing into the page. This movement cannot be attributed to the reaction force alone - if only the reaction force acted, the rider would be flipped up into the wall by intertia without the opposing torque from gravity. (This seems to imply that the CM cant be located at the center of the circle or even further out from the wall to be consistent with the torque. It also rules out $\alpha=0$, but if you ride the devil´s barrel in zero gravity, $\alpha = 0$ is the only posibility). There will also be an increasing moment of inertia from $t_0$ to $t_0 + \Delta t$ (and decreasing from $t_0 - \Delta t$ to $t_0$), so $\omega \frac{dI}{dt}>0$ (actually $\omega = \frac{d\alpha}{dt}= 0$ for $t=t_0$ by continuity). Away from the fixed point there will be a torque from friction opposite that from gravity, in the half - circle as seen from above closest to the point, the torque from gravity is larger because of a longer moment - arm, while in the half - circle furthest from the point, the torque from friction is larger by the same argument.

enter image description here

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