Torque opposite in direction to change in angular momentum?

[Happiness]

Torque $\tau$ should be in the same direction as the change in angular momentum $\Delta L$, but the following example seems to suggest otherwise.

Consider a cone rolling on its side without slipping on a flat surface. Let the apex be the origin and the initial coordinate of the center of mass (CM) of the cone be (-a, 0, c), where a and c are positive constants. (That means the cone, or rather its projection, is in the middle of the 2nd and 3rd quadrant of the x-y plane.) The cone rotates anticlockwise about the z axis. So, when viewed from the apex to the base, the cone rotates anticlockwise around its symmetry axis. That means the angular velocity $\omega$ and hence the angular momentum $L$ points in the +x direction, if we ignore their z components.

At this instant, the CM of the cone will move in the -y direction, making $L$ point into the first quadrant in the next instant. Thus, $\Delta L$ is in the +y direction.

Since the CM moves in a circle, there must be a friction $F$ to provide the centripetal force. Thus, at this instant, the friction acts in the +x direction. The fiction acts along the line of contact, which is vertically below the CM. So the torque $\tau=r\times F$ about the CM is in the -y direction, which is opposite to $\Delta L$.

What's wrong?

 

[Simon Bridge]

So the torque $\tau=r\times F$ about the CM is in the -y direction, which is opposite to $\Delta L$

I think the core misunderstanding is here: you have confuse the overall rate of change with a change in direction.
An object can continue to rotate anticlockwuse while a clockwise torque is applied... it just slows down.

 

[Gmen]

First, this was very interesting and thank you for the researched it triggered to me.

Second, I totally disagree with Simon. OP never implied change in rotational direction.

Now, for my opinion which is certainly incomplete.
We all know a cone will generally do that z-rotation. However, I cannot say for sure what kind of combination of weight, normal force, friction, and internal solid forces do create the centripetal needed (I'd like some enlightenment here). What we do know is that the centripetal, being needed for the solid's motion as-a-whole, is applied to the center of mass.
Now about rotation. The z-rotation is performed around the origin/apex. This is where r is measured from.
And there you have it. The r normal from the apex to the center of mass is +z direction, and the centripetal applied in the center of mass is of course +x. And +z (cross) +x = +y which is the direction of ΔL.

 

[Simon Bridge]

That was my initial reaction... but then I made a more careful reading.
It is tricky to analize because there are two different rotations that are coupled and the angular momentum is not a proper vector. It is total angular momentum that is conserved.

OP points out that there is an Lz and an L ... where the first is the angular momentum due to motion of com of the cone about the z axis (orbital and spin), and the second is the angular momentum of the cone about its own axis. He observes that L changes direction even if Lz does not... ie. at constant angular velocity, for eg, Lz is a constant.

To change the direction of L, you need to do work... like the demo with the bike wheel.
But I could be mistaken: there are quite a lot of issues with the description.

You can look at the interplay of forces with a careful free body diagram.

 

[Happiness]

First, this was very interesting and thank you for the researched it triggered to me.

My pleasure. :)

We all know a cone will generally do that z-rotation. However, I cannot say for sure what kind of combination of weight, normal force, friction, and internal solid forces do create the centripetal needed (I'd like some enlightenment here). What we do know is that the centripetal, being needed for the solid's motion as-a-whole, is applied to the center of mass.

I believe the motion of the cone is unaffected even when the gravitational field strength gets weaker and weaker and becomes zero. So we could let the weight and the normal contact force be zero. When I draw the free body diagram of this cone in zero gravity, there is only one force: the friction along the line of contact. The internal forces of the rigid body will ensure that all the particles of the body experience a net force in the same direction (but different magnitude) as the friction. But we do not need to draw these internal forces in a free body diagram.

Now about rotation. The z-rotation is performed around the origin/apex. This is where r is measured from.
And there you have it. The r normal from the apex to the center of mass is +z direction, and the centripetal applied in the center of mass is of course +x. And +z (cross) +x = +y which is the direction of ΔL.

Friction points along the line of contact. If we take the torque about the apex, it is zero since $r=0$ (because the apex is on the line of contact). Consequently, $\Delta L=0$. But the torque about the CM $\neq 0$, so $\Delta L\neq 0$, a contradiction.

What's wrong?

One way out is to consider that the line of action of the weight and that of the normal contact force do not meet. But I can't think of a reason why they don't.

 

[Simon Bridge]

Your setup is too complicated to see the issue clearly... try simplifying the problem so it has the same features.
ie. Cone rolling over another cone (then the axis is horizontal).
Still, see discussion:
http://physics.stackexchange.com/qu...eous-axis-of-rotation-and-rolling-cone-motion

I thing a rolling disk constrained to a circle by a rod from the center to a pole at the z axis has the features you want. Please clarify: is it the change in direction of the angular momentum of the cone about its axis that you are interested in?

 

[Happiness]

Please clarify: is it the change in direction of the angular momentum of the cone about its axis that you are interested in?

I'm interested to find out if there is a torque acting on the cone as it rolls and also its change in angular momentum.

I found that the angular momentum precesses about the z axis unless $h=\frac{\sqrt{3}}{2}r$, where $h$ and $r$ are the height and radius of the cone respectively, in which case the angular momentum remains constant. But I can't make sense of this result.

 

[hackhard]

What's wrong?

let me clarify some concepts--
concept 1--
torque = dL/dt (where torque and L are about same line)

ΔL\Delta L is in the +y direction

correct You have taken L about x-axis

The fiction acts along the line of contact, which is vertically below the CM.

also correct but torque of this friction force (along +ve x) is 0 about x-axis since both are parallel

torque τ=r×F\tau=r\times F

gives torgue about a point

since you considered L about a line you must calculate torque about a line given by |r x F|cos(m) where r is any point on x-axis and m is angle bet vector r x F and x-axis,. In this case since m =pi/2 so cos(m) =0 so torque about x-axis is 0.
(by the way no friction acts along line of contact since relative velo bet instantaneous point of contact and surface is 0)
(static friction may act( if some other external force acts) but not along line of contact rather normal to it)
then you may ask how L changes with 0 torque
concept 2---
in abscence of ext force a body can rotate only about a line through com(if it is intially rotating)
since cone rotates about z-axis through apex (ie com is accelerated)so there must be an ext force on cone parralel to xy plane acting at normal dist from z-axis greater than that of com and nrmal dist from xy plane greater than that of z coordinate of com

1 more error in post #1 net L of body is not in +ve x dir
since cone rotates about symmetric axis and symmetric axis itself rotates about z-axis , so
net L = L about symmetric axis + L about z axis (+ denotes vector addition operator)

 

[Happiness]

let me clarify some concepts--since you considered L about a line you must calculate torque about a line given by |r x F|cos(m) where r is any point on x-axis and m is angle bet vector r x F and x-axis,. In this case since m =pi/2 so cos(m) =0 so torque about x-axis is 0.

If you add the factor $\cos m$, then what you are basically doing is finding the component of the torque $\tau$ along some direction. It is clear from post #1 that the x components of both $\tau$ and $\Delta L$ are 0. Your argument does not explain why their y components are opposite.

(by the way no friction acts along line of contact since relative velo bet instantaneous point of contact and surface is 0)
(static friction may act( if some other external force acts) but not along line of contact rather normal to it)

Consider an eraser placed on a turntable. Set the turntable in rotation at a low enough $\omega$ such that the eraser doesn't slide. It moves in a circle. The centripetal force for the eraser to move in a circle is provided by static friction, which is always directed radially inwards towards the center of the circle.

Similarly, the friction in the case of a rotating cone is always directed radially inwards towards the center of the circle.

in abscence of ext force a body can rotate only about a line through com(if it is intially rotating)
since cone rotates about z-axis through apex (ie com is accelerated)so there must be an ext force on cone parralel to xy plane acting at normal dist from z-axis greater than that of com and nrmal dist from xy plane greater than that of z coordinate of com

There is no such external force applied. Take a cone and give it an initial push. You'll see that it continues rotating for quite some time even though no force is applied.

1 more error in post #1 net L of body is not in +ve x dir
since cone rotates about symmetric axis and symmetric axis itself rotates about z-axis , so
net L = L about symmetric axis + L about z axis (+ denotes vector addition operator)

That's correct. But I wrote "if we ignore their z components" in post 1.

 

[Happiness]

The mistake is that we cannot take torque about the center of mass but take the angular momentum about the apex.

We have to take both of them about the same point. In this case, it's easier to take them about the apex. We know $\Delta L$ is pointing in the +y direction.¹ So $\tau$ must too. There means the line of action of the normal reaction force and that of the weight of the cone cannot meet. The former must be further away from the apex to create a net torque in the correct direction.

¹Post #1 obtained this result by ignoring the z component of $L$, but it can be shown that the z component is constant.

 

[Happiness]

I found that the angular momentum precesses about the z axis unless $h=\frac{\sqrt{3}}{2}r$, where $h$ and $r$ are the height and radius of the cone respectively, in which case the angular momentum remains constant. But I can't make sense of this result.

I made a mistake¹ in my calculation and hence it produces a nonsensical result. The angular momentum should always precess about the space z axis.

¹I forgot the negative sign in $\dot{\phi}=-(\sin\alpha)\,\dot{\psi}$, where $\alpha$ is the half-angle of the cone.

 

[hackhard]

it can be proved that in this case pure rolling without slipping is impossible
proof by contradiction
ihttps://doc-10-90-docs.googleusercontent.com/docs/securesc/qj2ah0hthn23cer311lsft4dap1meggk/p2cmem31ivpogcub06k30u8sbvs18s0d/1460736000000/17942363115704344375/17942363115704344375/0B5m-h82V0Ej2b1g5YmFhSUk5UEE?e=download&nonce=8amaq9ae1chfs&user=17942363115704344375&hash=f94ubg1s1cttv26lfc50dg85akbjj6ca https://doc-08-90-docs.googleusercontent.com/docs/securesc/qj2ah0hthn23cer311lsft4dap1meggk/io7eeb18uej647t8eq6bjpjc3r1ee575/1460736000000/17942363115704344375/17942363115704344375/0B5m-h82V0Ej2aFRxR2RzRlFmaVk?e=download https://doc-0k-90-docs.googleusercontent.com/docs/securesc/qj2ah0hthn23cer311lsft4dap1meggk/rd4m42j7ia7b294ng8a7fdol5evuvaes/1460736000000/17942363115704344375/17942363115704344375/0B5m-h82V0Ej2aGxLRENyYmdXZVU?e=download

 

[Happiness]

it can be proved that in this case pure rolling without slipping is impossible
proof by contradiction
ihttps://doc-10-90-docs.googleusercontent.com/docs/securesc/qj2ah0hthn23cer311lsft4dap1meggk/p2cmem31ivpogcub06k30u8sbvs18s0d/1460736000000/17942363115704344375/17942363115704344375/0B5m-h82V0Ej2b1g5YmFhSUk5UEE?e=download&nonce=8amaq9ae1chfs&user=17942363115704344375&hash=f94ubg1s1cttv26lfc50dg85akbjj6ca https://doc-08-90-docs.googleusercontent.com/docs/securesc/qj2ah0hthn23cer311lsft4dap1meggk/io7eeb18uej647t8eq6bjpjc3r1ee575/1460736000000/17942363115704344375/17942363115704344375/0B5m-h82V0Ej2aFRxR2RzRlFmaVk?e=download https://doc-0k-90-docs.googleusercontent.com/docs/securesc/qj2ah0hthn23cer311lsft4dap1meggk/rd4m42j7ia7b294ng8a7fdol5evuvaes/1460736000000/17942363115704344375/17942363115704344375/0B5m-h82V0Ej2aGxLRENyYmdXZVU?e=download

Your attachments cannot be seen.

 

[rcgldr]

The cone translates and rotates at constant speed. There's no change in angular momentum ... This would be about an axis vertical to the flat surface, which isn't the question here.

 

[Happiness]

There's no change in angular momentum, so the torque is zero.

The magnitude of the angular momentum is constant but not its direction, so there must be a torque.

 

[Simon Bridge]

ie. For a suspended wheel, there is a torque from it's weight that changes the direction of the spin angular momentum pseudovector.
http://hyperphysics.phy-astr.gsu.edu/hbase/rotv.html

If we rested the wheel on a surface, it's weight is supported, it may still roll in a circle though: how does it do that?

You may like to see:
https://en.m.wikipedia.org/wiki/Angular_momentum#Angular_momentum_.28modern_definition.29
... it may be that you thinking is confused by the vector treatment.

 

[Simon Bridge]

@hackhard And yet we observe cones, in Nature, rolling without slipping... perhaps this rolling is not "pure"?

@Happiness ... the rolling cone is not the samevas a gyroscope as it is not supported on it's point.
Since you won't follow suggestions, I cannot help you.
Good luck.

 

[Orodruin]

Several off-topic posts moved to a new thread. This thread is about helping the OP with his problem.

 

[Orodruin]

I would say the problem is most easily analysed from the apex point of the cone. For the sake of the discussion, let look at the moment the cone is in the x-direction from the apex and that it is rolling into the first quadrant. In this situation, there are two contributions to the angular momentum, one coming from the overall movement of the cone into the first quadrant, which gives an angular momentum in the z-direction, and then the rotation of the cone about its axis, which gives an angular momentum in a direction which is pointing in the same direction as the cone axis, i.e., with negative components in the x- and z-directions. Overall, we have a z-component and a negative x-component. The z-component is obviously conserved during the motion (as long as it is with constant angular velocity), and the other component will precess around the z-axis.

So what are the torques around the apex? The frictional force has zero torque as its line of action is through the apex, the normal force from the table on the other hand gives a torque in the negative y-direction, which is perfectly compatible with the component of the angular momentum perpendicular to the z-axis (being in the negative x-direction at the moment we consider) precessing around the z-axis.

Of course, you could really do everything by actually doing the math, but it would involve the inertia tensor of the cone and I do not think it would add much apart from the math being consistent.

 

[andrewkirk]

I'll throw my analysis in with the others.
As observed, for the axis of rotation (of the cone around its axis of symmetry) to precess anti-clockwise around the origin, we need a torque that is in the positive y direction when the cone is over the negative x axis.
We can imagine the torque being applied at the cone's centre of mass, which is at a point at positive distance above the negative x axis. There are two possible sources for the torque: (1) the normal force of the floor pushing up, and (2) a frictional force pushing in the positive x direction. The cross product of the position vector of the cone's COM with either of these will produce a torque in the necessary direction: the positive y direction. The second one is nonzero because the COM is above the x-y plane.

To work out which of these produces the torque, or whether it's a combination and if so in what proportions, requires more thought than I've put into it. But it's clear that the necessary torque can be generated by available forces.

believe the motion of the cone is unaffected even when the gravitational field strength gets weaker and weaker and becomes zero. So we could let the weight and the normal contact force be zero.

We could not do that. If the gravitational force were zero then the frictional force would be zero (proportionality via coefficient of friction), so the cone would slip-slide in a straight line while maintaining a constant rate of rotation. That leads me to suspect that the frictional force in the positive x direction must form at least part (and maybe all) of the necessary torque for the cone to precess.

 

[Simon Bridge]

More on this sort of topic from same poster:
https://www.physicsforums.com/threa...hether-ellipsoid-is-prolate-or-oblate.866804/

 

[mfb]

I would say the problem is most easily analysed from the apex point of the cone.

This.

I think the confusion is a problem with the coordinate system. If you consider torque around the center of mass, you have to take the non-inertial motion of the center of mass into account - your reference frame is accelerated. That makes the analysis messy, and you probably need fictional torque or something like that.

 

[Happiness]

So what are the torques around the apex? The frictional force has zero torque as its line of action is through the apex, the normal force from the table on the other hand gives a torque in the negative y-direction, which is perfectly compatible with the component of the angular momentum perpendicular to the z-axis (being in the negative x-direction at the moment we consider) precessing around the z-axis.

For some reason, you omitted the torque due to the weight of the cone, about the apex. Since the weight and the normal force are equal in magnitude but opposite in direction, their torque would cancel each other if their lines of action are the same distance away from the apex. Thus, I was led to deduce that the normal force must act at a point horizontally further from the apex than does the weight.

 

[Happiness]

@Happiness ... the rolling cone is not the samevas a gyroscope as it is not supported on it's point.

Yes they are not the same. Perhaps I should have added an explanation when I put in the picture of a gyroscope. The picture is to define the Euler angles $\phi$, $\theta$ and $\psi$, which are used in my calculations. It is not to suggest that a rolling cone is a gyroscope.

 

[mfb]

Thus, I was led to deduce that the normal force must act at a point horizontally further from the apex than does the weight.

It should. The idea is the same as a braking or accelerating bicycle: the circular motion with its corresponding acceleration leads to a stronger normal force on the outside.

 

[Happiness]

It should. The idea is the same as a braking or accelerating bicycle: the circular motion with its corresponding acceleration leads to a stronger normal force on the outside.

Suppose we analyse a braking bicycle such that a sudden brake is applied to its front wheel, making it stop completely. The bicycle will then flip forward. The back wheel will be lifted up and the bicycle will undergo circular motion about the front wheel before it topples. The back wheel is on the outside of this vertical circle (or semi-circle). So do you mean that the normal force on the back wheel, $N_b$ is greater than that on the front wheel, $N_f$?

It seems counterintuitive. I would have thought that initially when the bicycle moves at a constant velocity, $N_b=N_f=c$, where $c$ is a constant. When the brake is applied, $N_b$ decreases from $c$ to 0 until it lifts off the ground, whereas $N_f$ increases beyond $c$ since the bicycle is pushing against the ground through its front wheel as it flips forward. So $N_b<N_f$ instead.

 

[Orodruin]

For some reason, you omitted the torque due to the weight of the cone, about the apex. Since the weight and the normal force are equal in magnitude but opposite in direction, their torque would cancel each other if their lines of action are the same distance away from the apex. Thus, I was led to deduce that the normal force must act at a point horizontally further from the apex than does the weight.

Right, this is a correct deduction. For the same reason you have a larger normal force on the outer wheels when turning your car.

 

[mfb]

Don't lift anything up. Just brake enough to note that the weight shifts towards the front wheel, to cancel the torque coming from the misalignment of friction and the position of the center of mass.To come back to the cone, we have friction force towards the center, so without additional torque the outer parts would go into the surface and the apex would go up. That doesn't happen here (unless we exceed some maximal angular velocity, then the apex will indeed go up), because there is a counter-torque from the normal force (+gravity), which is stronger at the outside compared to a cone that is at rest.

 

[rcgldr]

Don't lift anything up. Just brake enough to note that the weight shifts towards the front wheel, to cancel the torque coming from the misalignment of friction and the position of the center of mass.

To come back to the cone, we have friction force towards the center, so without additional torque the outer parts would go into the surface and the apex would go up. That doesn't happen here (unless we exceed some maximal angular velocity, then the apex will indeed go up), because there is a counter-torque from the normal force (+gravity), which is stronger at the outside compared to a cone that is at rest.

So I'm wondering if the vertical reaction forces of the flat surface to any imbalance in the vertical forces from the cone cancel any net "outward" torque exerted on the cone related to the radial forces, and if so, if it's mathematically possible to show this is the case.

 

[mfb]

I don't understand your post.

 

[andrewkirk]

Suppose we analyse a braking bicycle such that a sudden brake is applied to its front wheel, making it stop completely. The bicycle will then flip forward. The back wheel will be lifted up and the bicycle will undergo circular motion about the front wheel before it topples. The back wheel is on the outside of this vertical circle (or semi-circle). So do you mean that the normal force on the back wheel, $N_b$ is greater than that on the front wheel, $N_f$?

No the normal force on the back wheel doesn't increase. The downwards force exerted by the back wheel decreases to zero as a consequence of the 'upwards' torque exerted on the bicycle frame by the brake pads that are locked on the front wheel.

I'm not sure that that helps with the cone though.

 

[rcgldr]

there is a counter-torque from the normal force (+gravity), which is stronger at the outside compared to a cone that is at rest.

So I'm wondering if the vertical reaction forces of the flat surface to any imbalance in the vertical forces from the cone cancel any net "outward" torque exerted on the cone related to the radial forces, and if so, if it's mathematically possible to show this is the case.

I don't understand your post.

I'm wondering if the normal force related counter-torque exactly cancels the radial (centripetal acceleration) related torque.