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Consider the situation shown in the figure, A cone is set rolling on a cone so there is angular momentum about the vertical and there is an instantaneous angular momentum that is consistently moving in the horizontal plane.

No forces other than gravity, the normal force, and friction are present (but is set rolling without slipping).


Similarly for a cone undergoing rolling on a plane surface about its apex which most probably has the same explanation.

My question is how gravity and the normal force provide the needed torque to consistently change the angular momentum in the horizontal plane.

Which force dominates and provides the torque also why it only dominates when an initial impulse is given and does not spontaneously start rolling due to the imbalanced forces.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Nov 12 '20 at 16:12
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When the motion of a rigid body is known, the equations of dynamics allow the calculation of the (resulting) forces acting on the body

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The contact line of the cones is the axis of instantaneous rotation of the mobile cone and we have the following kinematic relationship:

$$ \overline{\omega} = \overline{\omega}_k + \overline{\Omega} \mspace{8cm} (1)$$

$\overline{\omega} =$ instantaneous angular velocity;

$\overline{\omega}_k=$ spin of the rolling cone;

$\overline{\Omega} =$ precession angular velocity.

On the rolling cone acts the weight applied in its center of gravity G and a system of forces distributed along the contact line. This system of forces is equivalent to a resultant $\overline{T}$ (a priori unknown) applied in a point A (a priori unknown).

The first equation of rigid body dynamics

$$ {d\overline{Q}\over dt} = \overline{R} \qquad\to\qquad -M\Omega^2 \overline{r}_G = M \overline{g} + \overline{T} \mspace{2cm} (2) $$

provides the constraint reaction $\overline{T}$. It is worth to note that $\overline{T}$ lies in the plane of the figure!

The application point A of $\overline{T}$ is given by the second equation (referred to the fixed point O):

$$ {d\overline{L} \over dt} = \overline{M}_o \qquad\to\qquad \overline{L}= I_1 \overline{\Omega} + I_3 \overline{\omega}_k \qquad \overline{M}_o = \overline{r}_G \times M\overline{g} + \overline{r}_A \times \overline{T} \qquad (3)$$

The resulting moment $ \overline{M}_o$ of all forces (with respect to O) is perpendicular to the plane of the figure!

Since in a regular precession $\overline{\Omega}$ is constant, we have

$$ {d\overline{L} \over dt} = {d(I_3 \overline{\omega}_k )\over dt} = I_3 \overline{\Omega} \times \overline{\omega}_k \qquad\to\qquad I_3 \overline{\Omega} \times \overline{\omega}_k = \overline{r}_G \times M\overline{g} + \overline{r}_A \times \overline{T} \qquad (4)$$

It's so possible to calculate the distance $ r_A = \overline{OA}$ (which identifies the point of application A of $\overline{T}$).

From an intuitive insight, it must be understood that the vertical component of $\overline{T} $ balances the weight $M\overline{g}$, while the horizontal component of $\overline{T}$ causes the centripetal acceleration of the center of gravity G of the mobile cone. The resulting moment $ \overline{M}_o$ of all forces (referred to O) is perpendicular to the figure and determines the precession motion of the figure axis of the mobile cone.

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  • $\begingroup$ Question , why is the Instantaneous axis of rotation along the line of contact of the cones, shouldn't it be along the figure axis? $\endgroup$ – JustJohan Nov 12 '20 at 6:41
  • $\begingroup$ "The resulting moment Mo of all forces (with respect to O) is perpendicular to the plane of the figure and determines the precession motion of the figure axis of the mobile cone."Could you elaborate on this part, this is my main query, I am not able to visualize this. $\endgroup$ – JustJohan Nov 12 '20 at 6:53
  • $\begingroup$ @JustJohan The axis of instantaneous rotation of a moving rigid body is the locus of its still points (in the reference system of the fixed cone). In a rolling wheel there is always a fixed point: the one in contact with the ground! $\endgroup$ – Pangloss Nov 12 '20 at 9:17
  • $\begingroup$ @JustJohan The second Eq.(2) shows that the constraint reaction T lies in the plain of the figure. In the third Eq.(3) the cross products prove that $M_o$ is perpendicular to the plain which contains all forces. Therefore the variation of angular momentum $dL = M_o dt$ is perpendicular to the figure too (precession). $\endgroup$ – Pangloss Nov 12 '20 at 9:59
  • $\begingroup$ I think this is a bit mathematical explanation of what R.W Bird said. Could you confirm that? Thanks for your answer. $\endgroup$ – JustJohan Nov 13 '20 at 11:16
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The normal force acting on the rolling cone along the line of contact must must produce a torque about the apex which is slightly greater than that produced by gravity. This excess torque vector is horizontal and perpendicular to, $v_2$, and causes the associated angular momentum to swing in that direction.

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  • $\begingroup$ why this isn't the case when the cone is stationary, shouldn't the excess torque produce rotation from zero angular momentum? $\endgroup$ – JustJohan Nov 7 '20 at 15:37
  • $\begingroup$ If the cone is rolling (in a circle), that requires a centripetal force from friction. But the friction acts below the center of mass. To compensate for this torque, the normal force at the surface must shift outward (increasing its torque about the apex). (Remember that the friction does not produce a torque about the apex.) $\endgroup$ – R.W. Bird Nov 10 '20 at 15:35
  • $\begingroup$ Why would the normal force shift outwards to compensate for friction? Does this mean if friction is neglected this motion is not possible ? Could you elaborate sir,I have many questions. $\endgroup$ – JustJohan Nov 10 '20 at 18:32
  • $\begingroup$ The normal force shifts outward to compensate for the torque which friction produces about the center of mass. Without (static) friction, the cone would not mover in a circle. $\endgroup$ – R.W. Bird Nov 10 '20 at 18:56
  • $\begingroup$ Okay, but for the angular momentum to change in the horizontal plane don't we need a torque perpendicular to the angular momentum vector in the horizontal plane? $\endgroup$ – JustJohan Nov 12 '20 at 6:39
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You can separate the movement into the spinning around the cone's axis of symmetry and orbital movement around a vertical axis going through the top of the cone. The orbital angular momentum is constant (as neither the vertical axis, orbital speed or moment of inertia change), but the angular momentum associated with the spin changes direction, as the cone orbits. By drawing everything out it can be checked that the infinitesimal change of this angular momentum is in the horizontal plane and it has the direction opposite to the orbital velocity. This must be the direction of the moment of the forces that cause this change. That means that the forces that cause this must be vertical.

One can also see the cone as a collection of slices, wheels rolling on the surface. Each slice is tilted towards the vertical axis going through the top of the cone. One can analyze the rolling of such tilted wheel and it can be found that it will also change the direction of its movement just like the whole cone does. That is because its center of gravity is not directly above the point where the wheel touches the surface, which cause an imbalance of the momenta of these forces. The fact that a cone is a solid means just that the movement of all slices must be coordinated.

Therefore it can be concluded that it's the gravity and the normal forces. While the forces negate each other, their momenta don't have to. The situation seems to be somewhat analogous to the gyroscope precession except that for the gyroscope all of the normal force is applied to the top, and for the cone it can be distributed along the line of contact. The condition that all the rolling happens without sliding seem to cause an imbalance between the momenta of the the gravitational force and the normal forces that is necessary to change the axis of the spinning as the cone orbits the vertical axis.

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We can say from the symmetry of the system, that the vertical component of the angular momentum is constant. So $$\tau = \left(\frac{dL_x}{dt}, \frac{dL_y}{dt}, 0 \right)$$

When the cones are at rest, the system is in equilibrium and there is no torque. So the sum of the reaction forces: (friction + normal) must pass through the center of gravity of the movable cone.

When the movable cone is rotating, the friction force that was necessary to avoid slippage must increase. It acts now also as a centripetal force. That $\delta F$ generates now a binary with the weight at CG.

Checking its direction, we can see that the torque is in the plane $xy$ as required by symmetry.

(It is the same answer of R.W.Bird, I haven't note that part of it was in his comments).

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  • $\begingroup$ Why should fricgtion act through com? Shouldn't it act it surface of the which is in contact then roll $\endgroup$ – Buraian Nov 12 '20 at 7:16
  • $\begingroup$ The friction force is parallel to the contact surface, pointing to the cones apex. $\endgroup$ – Claudio Saspinski Nov 12 '20 at 20:43
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enter image description here The only force that produce a torque to rotate the horizontal cone about the y-axis is the constraint force due to the rolling condition, not the normal force.

you can obtain the constraint force $~F_c~$ with these equations:

$$I_1\,\ddot{\varphi}_z=F_c\,r_1+\tau_z$$ $$I_2\,\ddot{\varphi}_y=-F_c\,r_2$$ and the rolling condition

$$r_1\,\varphi_z=r_2\,\varphi_y$$

you have three equations for the three unknowns $\ddot{\varphi}_z~,\ddot{\varphi}_y~,F_c$

$$F_c=-{\frac {I_{{2}}r_{{1}}\tau_{{z}}}{{r_{{1}}}^{2}I_{{2}}+{r_{{2}}}^{2}I _{{1}}}} $$

and the torque about the y-axis if $~F_c\,r_2$

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  • $\begingroup$ Sorry, but what is this constraint force due to rolling condition?I have not advanced that far in physics which is why I wanted a non-equation-ish answer. $\endgroup$ – JustJohan Nov 16 '20 at 15:39
  • $\begingroup$ The rolling condition constraint the rotation between the vertical cone and the horizontal cone. This kinematic equation cause a force between the cones, also the normal force is a „constraint force“. $\endgroup$ – Eli Nov 16 '20 at 15:54
  • $\begingroup$ @JustJohan do you understand the equations that Pangloss wrote ? $\endgroup$ – Eli Nov 16 '20 at 15:59
  • $\begingroup$ So it consists of both friction and normal force? It's not just one force that changes the angular momentum? $\endgroup$ – JustJohan Nov 16 '20 at 16:00
  • $\begingroup$ Not really I'm still in high school I have basic knowledge in rotational dynamics, Is this beyond the scope of my understanding? $\endgroup$ – JustJohan Nov 16 '20 at 16:01
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Intuitively, to me this is a gyroscope that is constrained from falling or slipping. A simple gyroscope that is not spinning rapidly, constrained about only one point, will fall as it tries to precess. Here the initial rotation of the top cone produces an initial angular momentum; the torque from gravity causes a change in angular momentum resulting in precession while the constraint from the lower cone prevents falling downwards or slipping off the lower cone. Not sure about the torque from friction.

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  • $\begingroup$ But the torque from gravity must somehow be greater than the torque from the normal force for it to precess , how does that happen? $\endgroup$ – JustJohan Nov 11 '20 at 1:48
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    $\begingroup$ Not clear to me. Perhaps the effective position of the normal force moves as the object moves. Like when you push on a large solid the effective position of the normal force on the solid moves forward of the center of mass to produce a counter torque to oppose the torque from the force you apply. Look forward to responses from those more knowledgeable. $\endgroup$ – John Darby Nov 11 '20 at 2:08

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