Why is angular momentum not conserved in this case?

[leafy]

TL;DR

A mass is traveling at constant speed on the outer ramp. It is then directed to the inner ramp by an intermediate ramp. The mass then circle the inner ramp with the same speed as the outer ramp. Why angular momentum is not conserved?

 

[ergospherical]

During the non-circular "transfer" part of the trajectory, the normal force is not a central force and has a moment about the origin. The normal force is, however, always orthogonal to the velocity which means that the speed is conseved.

 

[jbriggs444]

During the non-circular "transfer" part of the trajectory, the normal force is not a central force and has a moment about the origin. The normal force is, however, always orthogonal to the velocity which means that the speed is conseved.

I would suggest the possibility of experimental error. The junction between the outer ramp and the transfer ramp has a sharp corner which should have the effect of draining kinetic energy. An observation that speed is unchanged is, accordingly, anomalous.

 

[leafy]

We can make the ramp smooth. The question is there isn’t any torque acting on the system, yet angular momentum not conserved.

 

[jbriggs444]

We can make the ramp smooth. The question is there isn’t any torque acting on the system, yet angular momentum not conserved.

Sure. But as pointed out in #2, there is a torque acting on the moving object.

 

[A.T.]

The question is there isn’t any torque acting on the system, yet angular momentum not conserved.

Angular momentum is conserved for isolated systems. What is the isolated system here?

 

[leafy]

Sure. But as pointed out in #2, there is a torque acting on the moving object.

#2 also stated the force is always orthogona. If torque = Frsin(theta) and theta always zero, then torque is always zero?

 

[jbriggs444]

#2 also stated the force is always orthogona. If torque = Frsin(theta) and theta always zero, then torque is always zero?

Orthogonal to the velocity vector, yes. Which means that speed is unchanged. $\vec{F} \cdot \vec{v} = 0$ so no power associated with the force. But that has little to do with torque.

Orthogonal to the radius vector, no. Which means that torque is not maximized. No big deal.

Parallel to the radius vector, not during the transfer. Which means that torque ($\vec{F} \times \vec{r}$) about the reference axis is non-zero. That is a very big deal.

Parallel to the radius vector while the object is on either the outer or the inner circle, yes. Which means that torque ($\vec{F} \times \vec{r}$) is zero during those portions of the trajectory and angular momentum about the reference axis is conserved during those phases.

Parallel to a radius vector toward the center of curvature of the transfer arc during the transfer, yes. Which means that torque about some other reference axis is zero during the transfer and that angular momentum about that other reference axis is temporarily conserved. Not a good argument for conservation of angular momentum about our chosen reference axis.

One can try to cobble up a half-baked argument that angular momentum is conserved about one axis and then another axis and then again about the original axis. But if you keep that argument in the oven long enough to fully bake it, you'll realize that each time you changed reference axis, you needed to add a term to angular momentum to compensate for the cross product of the linear momentum of the system and the displacement of the axis. You did not add that term. So that argument falls flat.

 

[leafy]

Suppose we have the mass and string orbiting like the outer ramp. Then we pull the string to achieve the inner ramp orbit. We can say that the the axis never changed. Can we then make the ramps exactly as the path of the string and consider the axis never changed?

 

[ergospherical]

You can apply a purely radial force to achieve the transfer, but then work will have been done and the speed will not be the same as before.

 

[cjl]

Suppose we have the mass and string orbiting like the outer ramp. Then we pull the string to achieve the inner ramp orbit. We can say that the the axis never changed. Can we then make the ramps exactly as the path of the string and consider the axis never changed?

You can, but then during the spiral inwards portion of the motion, the rope will not be perpendicular to the velocity vector, so you're doing work on the ball and speeding it up, preserving angular momentum in the process.