Identical cones rotating while in contact with each other

[Null_Void]

Homework Statement

Two identical uniform cones A and B can rotate freely about their fixed central axis pressing each other uniformly on their entire tapered length as shown in the figure. If the cone A is rotated with a constant angular velocity $ω_{A}$, how much angular velocity will cone B ultimately acquire?

Relevant Equations

Pic Below.

I know that friction will act on both the cones until there is no slipping throughout the length in contact.

Velocity at points M and N should be Zero in the end since
$V = ωR$ and $R = 0$

Total Angular Momentum of the system must be conserved since
$τ_{net} = 0$

But I'm not sure how to find the final angular velocity of cone B from here on.

Any help is appreciated!

 

[haruspex]

friction will act on both the cones until there is no slipping throughout the length in contact.

Are you sure that is possible? Think about tangential velocities.

Total Angular Momentum of the system must be conserved since

Each cone must be supported on a spindle. No matter what axis you choose to take moments about, one or both spindles will be exerting a torque on the pair of cones as a system.

Think about the torques they exert on each other through friction.
(You do have to assume that the normal force per unit length is constant along the line of contact.)

 

[Null_Void]

Are you sure that is possible? Think about tangential velocities

I made a mistake. The velocities at M and N can never be zero since it would imply that
$ω_{A} = ω_{B} = 0$ which is not true. Thus there must be constant slipping.

Each cone must be supported on a spindle. No matter what axis you choose to take moments about, one or both spindles will be exerting a torque on the pair of cones as a system.

Do you mean that $τ_{net} ≠ 0$? implying Angular Momentum is not conserved.
But since there are no external torques shouldn't Angular momentum be conserved? I thought the net torques on both the cones must be equal and opposite which gives $τ = 0$ on the system as a whole.

Think about the torques they exert on each other through friction.

The torques must be equal and opposite and should act on the cones as long as they are rotating.

I think there should be a point on each cone where the friction direction changes, giving rise to friction in two directions over two parts of the cone, whose torques at equilibrium will cancel out giving a constant angular velocity.

 

[jbriggs444]

I made a mistake. The velocities at M and N can never be zero since it would imply that
$ω_{A} = ω_{B} = 0$ which is not true. Thus there must be constant slipping.

There must always be slipping almost everywhere along the line of contact, yes.

Do you mean that $τ_{net} ≠ 0$? implying Angular Momentum is not conserved.

No matter where you choose to place the reference axis, there are multiple sources of external torque.

1. We are told that A rotates with constant angular velocity. That means that it will be driven by some external torque controlled to maintain that constant angular velocity.

2. As @haruspex points out, you have two spindles. At least one must not be on your chosen reference axis. That spindle can be the source of an external torque.

But since there are no external torques shouldn't Angular momentum be conserved? I thought the net torques on both the cones must be equal and opposite which gives $τ = 0$ on the system as a whole.

It might be useful to restrict your attention to spindle B alone and place your reference axis on its axis.

Now you only your only external torque is from friction between the two spindles. At the eventual equilibrium, this torque must be zero.

The torques must be equal and opposite and should act on the cones as long as they are rotating.

Depending on exactly what you mean by this, it could be correct. Or it could be incorrect.

Consider two ordinary (not conical) identical rollers. Or two identical meshing gears. The forces of the gears on each other are equal and opposite. But the torques of each on the other about their respective axes are equal, not opposite (the two moment arms are also equal and opposite). If you give the two rollers or gears different radii then the two torques are not even equal.

However, this point is a distraction from the question at hand. I do not think that considering the angular momentum of the system as a whole is a fruitful approach.

I think there should be a point on each cone where the friction direction changes, giving rise to friction in two directions over two parts of the cone, whose torques at equilibrium will cancel out giving a constant angular velocity.

Yes indeed. This is an insight that you are expected to have.

 

[Lnewqban]

I made a mistake. The velocities at M and N can never be zero since it would imply that
$ω_{A} = ω_{B} = 0$ which is not true. Thus there must be constant slipping.

Again, think about tangential velocities instead of angular ones.
I believe that the angular velocities could be equal at some time, at which the tangential velocities would be of equal value only were radii are equal (mid point).

 

[haruspex]

I believe that the angular velocities could be equal at some time

I assume you mean equal and opposite, but that is not what happens, as it turns out.

 

[Lnewqban]

I assume you mean equal and opposite, but that is not what happens, as it turns out.

Certainly.
Why?

 

[haruspex]

The torques must be equal and opposite

If you mean the net torque each exerts on the other about the other's spindle, that is not a safe assumption.
Think about a small element of the line of contact. What are equal and opposite there?

 

[Null_Void]

There must always be slipping almost everywhere along the line of contact, yes.

I think there must be one point which is at rest where no friction acts. It is at this point that the directions of the forces change.
Using similarity of triangles and equating the velocities at those points, the vertical distance of the point from the tip of Cone-A comes out to be:

$Y = \frac{H}{\frac{ω_{B}}{ω_{A}} + 1}$

Where H is the height of the cone

The torque due to the friction on an element of length dl at a distance r from the axis is given by

$dτ = (μαdl)r$

Where $α$ is the Normal force per unit length which is a constant and $dl$ is an element of the total length

From similarity of triangles:
$dl = dr \frac{l}{R}$
$dτ = krdr$

Therefore $τ = \int kr\,dr$ with limits from $0$ to $x$ and $x$ to $R$

Integrating and Equating both gives the value of $x$ to be $\frac{R}{√2}$ which corresponds to a height of $\frac{H}{√2}$

Equating this height with $(1)$ gives the value $ω_{Β} = {√2 - 1} ω_{Α}$

Which is the right answer.

Just one last question, @haruspex how do I prove that the Normal force per unit length will be constant?

Thanks to everyone!

 

[haruspex]

how do I prove that the Normal force per unit length will be constant?

You can’t, you just have to assume it. In practice, for there to be enough normal force, the contacting surfaces need to be held somewhat pressed against each other. That is likely to cause some deflection of the spindles, resulting in more normal force towards the ends than in the centre. That could be overcome my making the cones a little fatter in the middle, but you'd need a lot of details to figure it all out.

Equating this height with $(1)$ gives the value $ω_{Β} = {√2 - 1} ω_{Α}$

You mean $ω_{Β} = (√2 - 1) ω_{Α}$.
Latex treats curly braces as controls only, unless you escape them.

 

[jbriggs444]

Just one last question, @haruspex how do I prove that the Normal force per unit length will be constant?

The problem statement says "pressing each other uniformly on their entire tapered length"

 

[Lnewqban]

You mean $ω_{Β} = (√2 - 1) ω_{Α}$.

What is the physical reason for the tangential velocities to become equal at 41% of the height of the cone A (measured from the apex) rather than at 50%.
What other factor breaks the symmetry?

 

[haruspex]

What is the physical reason for the tangential velocities to become equal at 41% of the height of the cone A (measured from the apex) rather than at 50%.
What other factor breaks the symmetry?

The symmetry is illusory. In steady state, each cone has zero net torque, but only one cone has a torque applied at its axle.
In general, the torques they exert on each other need not be equal and opposite. If the equal tangential speed point is distance x (along the slope) from one end, out of a total length of L, and the frictional force per unit length is $F$ then the two torques are $F\sin⁡(θ)(\frac 12L^2−x^2)$ and $-F\sin⁡(θ)(\frac 12L^2−2Lx+x^2)$. If the first is zero, as it must be for the cone with no other torque, then the second is $F\sin⁡(θ)L^2(\sqrt 2−1)$.