System of two wheels of different sizes with an axle through their centers

[Orodruin]

You do not need to find the angular momentum to conclude that (b) is false. You only need to note that the value given in (b) is the angular momentum it would have if you did not account for the precession. Doing that, the angular momentum will clearly be different.

 

[Orodruin]

t is probably easiest if you resolve the angular velocities into components along the cone axis and normal to it.

I would just compute the moment of inertia tensor for the system and multiply by the angular velocity …

 

[Aurelius120]

Please Check this as well
This is my attempt at Option-A:

Angular Momentum about axle and passing through origin: $L_3=I\omega=\dfrac{17ma^2\omega}{2}$

About an axis perpendicular to axle and through origin:

Velocity of COM of 1st wheel= $\omega_3 ×l=\dfrac{\omega}{5}× l\cos\theta$
Velocity of COM of 2nd wheel=$\omega_4 ×2l=\dfrac{\omega}{5}×2l\cos\theta$
Angular momentum: $L_4=I_3\omega_3+I_4\omega_4$
$$L_3=I\omega=\dfrac{17ma^2\omega}{2}$$ $$L_4=\left(\dfrac{ma^2}{4}+ml^2\right)\omega_3 + \left(\dfrac{4m(2a)^2}{4}+4m(2l)^2\right)\omega_4$$

z-component of angular momentum about origin=
$$L_z=L_4\cos\theta-L_3\sin\theta$$

 

[Aurelius120]

You do not need to find the angular momentum to conclude that (b) is false. You only need to note that the value given in (b) is the angular momentum it would have if you did not account for the precession. Doing that, the angular momentum will clearly be different.

Yes I know that. I was just trying to compute it for the sake of it.
Also I can't use that for Option-A

 

[TSny]

Please Check this as well
This is my attempt at Option-A:

Angular Momentum about axle and passing through origin: $L_3=I\omega=\dfrac{17ma^2\omega}{2}$

You need to use the net angular velocity component along the axle. The precession angular velocity $\omega/5$ has a component along the axle that needs to be included here.

$$L_4=\left(\dfrac{ma^2}{4}+ml^2\right)\omega_3 + \left(\dfrac{4m(2a)^2}{4}+4m(2l)^2\right)\omega_4$$

I think this is correct if you are taking $\omega_3 = \omega_4 = \dfrac{\omega}{5}\cos\theta$.

z-component of angular momentum about origin=
$$L_z=L_4\cos\theta-L_3\sin\theta$$

Looks good.

 

[Aurelius120]

@TSny Can you also check my attempt at Option B in post 30? Thank You

 

[haruspex]

About axis perpendicular to axle:
Velocity of COM of 1st wheel about z-axis=$\dfrac{\omega}{5} l \cos\theta = \omega_1×\dfrac{4l}{5}$
Velocity of COM of 2nd wheel about z-axis= $\omega_2×\dfrac{l}{5}= \dfrac{\omega}{5} × 2l \cos \theta$

I do not understand what you are doing there. For each disc you have a rotation about the z axis, but you need to

• resolve that into a rotation about the axle of the cone and and a perpendicular rotation
• find the moment of inertia of the disc about axes parallel to those through the mass centre of the assembly.

 

[Aurelius120]

I do not understand what you are doing there. For each disc you have a rotation about the z axis, but you need to

• resolve that into a rotation about the axle of the cone and and a perpendicular rotation
• find the moment of inertia of the disc about axes parallel to those through the mass centre of the assembly.

I was going like this
Since linear velocity is same irrespective of axis
I equated the product of angular velocity and perpendicular distance from axis in each case
As in
$$\omega_z × r_z= \omega_{COM}×r_{COM}$$

 

[TSny]

Plz Check this
This is my attempt at Option-B

About axle :
Angular Momentum: $L_1=I\omega= \dfrac{17ma^2\omega}{2}$

You need to include the component of the precessional angular velocity along the axle. That is, the moment of inertia about the axle should be multiplied by the component of the net angular velocity along the axle. See post #26 of @haruspex .

About axis perpendicular to axle:
Velocity of COM of 1st wheel about z-axis=$\dfrac{\omega}{5} l \cos\theta = \omega_1×\dfrac{4l}{5}$
Velocity of COM of 2nd wheel about z-axis= $\omega_2×\dfrac{l}{5}= \dfrac{\omega}{5} × 2l \cos \theta$
Angular Momentum: $L_2=I_1\omega_1 - I_2\omega_2$

So $$L_1=\frac{17ma^2\omega}{2}$$
$$L_2=\left(\dfrac{ma^2}{4}+\dfrac{16ml^2}{25}\right)\omega_1-\left(\dfrac{4m(2a)^2}{4}+\dfrac{4ml^2}{25}\right)\omega_2$$

I get a bit confused with the meaning of your subscripts "1" and "2".

In your expression for $L_2$, I don't understand why you distinguish between $\omega _1$ and $\omega_2$. Also, why the subtraction?

For $L_2$ you need to find the total moment of inertia about the perpendicular axis through the center of mass and multiply by the component of the net angular velocity along this axis.

$$L_{COM}=\sqrt{L_1^2+L_2^2}$$

Ok.

 

[Aurelius120]

I get a bit confused with the meaning of your subscripts "1" and "2".

That represents the angular velocity of wheels 1 and 2 about the axis perpendicular to axle.

In your expression for $L_2$, I don't understand why you distinguish between $\omega _1$ and $\omega_2$. Also, why the subtraction?

The wheels are rotating about z-axis in say ACW sense.
About an axis passing in between the wheels, but perpendicular to axle the rotation of wheels will be in opposite sense to each other, right?

For $L_2$ you need to find the total moment of inertia about the perpendicular axis through the center of mass and multiply by the component of the net angular velocity along this axis.

What is confusing is that angular velocity of both wheels about COM perpendicilar to z-axis can't be same, right?
Because linear velocity is independent of axis and is different for either wheel as is their distance from COM
For the wheels,
$$\omega_{COM}×r_{COM}= \omega_z×r_z$$

 

[TSny]

The wheels are rotating about z-axis in say ACW sense.
About an axis passing in between the wheels, but perpendicular to axle the rotation of wheels will be in opposite sense to each other, right?
What is confusing is that angular velocity of both wheels about COM perpendicilar to z-axis can't be same, right?

Consider a dumbbell rotating about the axis as shown. Both balls have the same angular velocity $\omega$ in the same direction. The linear velocities of the two balls will be in opposite directions with different magnitudes.

 

[Aurelius120]

Consider a dumbbell rotating about the axis as shown. Both balls have the same angular velocity $\omega$ in the same direction. The linear velocities of the two balls will be in opposite directions with different magnitudes.

View attachment 347426

But what if the balls were rotating about an outside axis. Both will have same direction of linear velocity. However when the that system is analysed about an axis passing in-between the balls, the linear velocities being kn the same direction will make angular velocity in opposite directions.

 

[TSny]

But what if the balls were rotating about an outside axis. Both will have same direction of linear velocity. However when the that system is analysed about an axis passing in-between the balls, the linear velocities being kn the same direction will make angular velocity in opposite directions.

Suppose the dumbbell rotates about an "outside" axis. Both balls will have the same angular speed $\omega$. Their linear velocities will be in the same direction (with different magnitudes). Also, their angular velocities will be in the same direction (counter-clockwise as seen from above).

For the inside axis, the balls have the same counter-clockwise angular velocity as seen from above. If one ball makes 5 revolutions per second, so does the other ball. The linear velocities are now opposite in direction and have different magnitudes.

 

[Aurelius120]

View attachment 347427
Suppose the dumbbell rotates about an "outside" axis. Both balls will have the same angular speed $\omega$. Their linear velocities will be in the same direction (with different magnitudes). Also, their angular velocities will be in the same direction (counter-clockwise as seen from above).

For the inside axis, the balls have the same counter-clockwise angular velocity as seen from above. If one ball makes 5 revolutions per second, so does the other ball. The linear velocities are now opposite in direction and have different magnitudes.
View attachment 347429

But doesn't that change the system completely?
Applying that to wheels,
The wheels will stop rotating about the z-axis as cone would have and rather rotate about each other. But that is not the motion described in the question?
Aren't we supposed to analyse the system's motion as it is but about a different axis?

 

[TSny]

Let $\vec {\omega}^{net}$ be the total angular momentum velocity vector of the assembly at some instant of time. This is the vector sum of the "spin" angular velocity $\omega$ along the axle and the "precession" angular velocity $\omega/5$ along the z-axis.

An axis along the axle of the assembly is a "principal" axis of the assembly. Thus, the total angular momentum vector of the system will have a component along the axle given by $L_a = I_a \omega^{net}_a$ where $I_a$ is the moment of inertia of the assembly about the axle and $\omega^{net}_a$ is the component of $\vec {\omega}^{net}$ along the axle. ("$a$" is for "axle".)

Consider another axis that (i) passes through the center of mass of the assembly, (ii) is perpendicular to the axle, and (iii) lies in the plane containing the axle and the z-axis. This axis is also a principal axis. The total angular momentum vector will have a component along this axis given by $L_p = I_p \omega^{net}_p$ where "$p$" indicates components along this perpendicular axis.

The magnitude of the total angular momentum about the center of mass is $L_{cm} = \sqrt{L_a^2 + L_p^2}$.

You already know $I_a$.

You need to calculate $I_p$. This is similar to the dumbbell.

 

[haruspex]

I was going like this
Since linear velocity is same irrespective of axis
I equated the product of angular velocity and perpendicular distance from axis in each case
As in
$$\omega_z × r_z= \omega_{COM}×r_{COM}$$

But the angular momenta of the COMs are inadequate. They are not point masses.
As I wrote, you need to take these steps:

1. Resolve the precession about the z axis into a rotation along the axle of the 'cone' (pair of discs) and a rotation about an axis which is normal to that and lies in the zx plane.
2. For the component along the axle of the cone, you can simply add it to the given $\omega$ (but the sign will be opposite, I believe). From that you can calculate that component of the angular momentum.
3. For the other component of the precession, the axis is parallel to the discs. Find how far each disc is from their common mass centre, compute the two angular momentum components using the parallel axis theorem and add them.
4. Use Pythagoras to combine the two orthogonal angular momentum components.

@TSny has said the same in algebra. Having it expressed in two ways should help.

 

[Aurelius120]

@TSny has said the same in algebra. Having it expressed in two ways should help.

Yes thank you.
What was bugging me was $\vec L=m\vec r\times \vec v$, gave one positive and one negative value for the wheels because the direction of $\vec r$ was opposite.
But

They are not point masses

So this formula failed.

I was considering the system to be still rotating about z-axis while trying to compute it's angular velocity about COM
Unlike this:

You need to calculate Ip. This is similar to the dumbbell.

 

[Aurelius120]

Not to sound disrespectful but
I thought I got it but I didn't.
So forgive my bs here and in 44
@haruspex @TSny
I understand the method you are using and how it is working.

But I don't understand why it is working.
As in why are you considering it to be rotating around the axis through COM and perpendicular to axle? Isn't it actually rotating about z-axis making both their linear velocities in same direction and thus both having different angular velocity about the required COM axis? (which also seems impossible as they are rigidly connected)

 

[haruspex]

why are you considering it to be rotating around the axis through COM and perpendicular to axle?

It is a valid way of decomposing a rotational velocity vector.
I admit it is less obvious than with linear motion. Displacements can be easily shown to add vectorially, and it therefore works for linear velocities and accelerations by differentiation wrt time. But it does not work for rotations through non-infinitesimal angles: the order of two such rotations matters. But again, it does work for infinitesimal angles, so it also works for angular velocities and accelerations.

 

[Aurelius120]

It is a valid way of decomposing a rotational velocity vector.
I admit it is less obvious than with linear motion. Displacements can be easily shown to add vectorially, and it therefore works for linear velocities and accelerations by differentiation wrt time. But it does not work for rotations through non-infinitesimal angles: the order of two such rotations matters. But again, it does work for infinitesimal angles, so it also works for angular velocities and accelerations.

Thanks
Just as a way to visualise:Is the rotation about COM axis what would happen if someone suddenly put a rod at the COM causing it to come to rest immediately while the wheels because of their inertia rotate about that axis?

 

[Orodruin]

The rotation about the CoM is the (instantaneous) motion in the CoM instantaneous rest frame. The separation into CoM angular momentum and angular momentum due to the CoM motion is simply a convenient way of doing the bookkeeping of the angular momentum as the terms separate.

Consider a system of rigidly connected masses $m_i$ with positions $\vec x_i$ and velocities $\vec v_i$ relative to a fixed point of the rotation with angular velocity $\vec \omega$. By assumption of the rigidity:
$$
\vec v_i = \vec\omega \times \vec x_i
$$
and so the total angular momentum relative to the fixed point is
$$
\vec L = \sum_i m_i \vec x_i \times (\vec \omega \times \vec x_i)
= \sum_i m_i [r_i^2 \vec \omega - (\vec\omega\cdot \vec x_i)\vec x_i]
$$
where $r_i$ is the magnitude of $\vec x_i$. This is the usual expression involving what is essentially the definition of the moment of inertia relative to the fixed point.

We can now rewrite this by introducing a reference displacement $\vec x_0$ and the instantaneous rigid velocity $\vec v_0 = \vec\omega \times \vec x_0$ of that point. We introduce $\vec x_i = \vec y_i + \vec x_0$ and $\vec v_i = \vec u_i + \vec v_0$, where $\vec y_i$ and $\vec u_i$ are the position and velocity of mass $i$ relative to the reference displacement and velocity.

The angular momentum about the fixed point now takes the form
$$
\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)
= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]
+ M \vec x_0 \times \vec v_0
+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]
$$
with $R_i$ the magnitude of $\vec y_i$ and $M = \sum_i m_i$ (ie, the total mass). The first term here is now the angular momentum of the system relative to the reference displacement in the instantaneous rest frame of the reference displacement. The second term is the angular momentum of a single particle of mass $M$ at the reference displacement relative to the original fixed point. The last term is quite messy.

However, if you take the CoM as the reference displacement, then $\sum_i m_i \vec y_i = 0$ by definition and the last two terms vanish. Voilà, you have reexpressed the total angular momentum relative to the fixed point as the angular momentum relative to the CoM (in the CoM instantaneous rest frame) plus the angular momentum of a point mass at the CoM moving at the CoM velocity.

The advantage over the first expression is that the moment of inertia relative to the CoM is often much easier to express in terms of standard moments of inertia that have already been computed. Many times simplifying computation.

Iirc, this is also covered in some detail in chapter 10 of my book.

 

[sophiecentaur]

Homework Statement: Two thin circular discs are rigidly fixed by a massless rigid rod passing through centers and laid on on a firm flat surface and set rolling without slipping
Relevant Equations: $$\vec L=\vec r\times m\vec v$$
$$\vec L= I\vec \omega$$

Why can't the wheels be perpendicular to the surface with an axle connecting their centers

They would have equal radii?? It may be a good idea just to look at the 'static' geometry first; you may have given yourself a problem with rotational matters by not looking at the question better (e.g. the figid fixing on the rod dictates what will happen.)

Will a cone rotate on a frictionless surface?

What would make it turn? (Or, rotate about what axis?)
@Orodruin has pointed you in the 'clever clogs' direction (post #32) which would sort out your questions probably in one go, I think.

 

[Aurelius120]

They would have equal radii?? It may be a good idea just to look at the 'static' geometry first; you may have given yourself a problem with rotational matters by not looking at the question better (e.g. the figid fixing on the rod dictates what will happen.)

In the 3rd post I explained the arrangement more.
I meant keep the discs perpendicular to the surface and connect their centers via a rod which is not horizontal.
@Lnewqban described that motion in post 4

What would make it turn? (Or, rotate about what axis?)

Isn't static friction not necessary for rolling? I remember reading (on stackexchange maybe) that when a translating object is placed on a surface kinetic friction causes it to roll after which friction doesn't have a role in the rotation or translation.

@Orodruin has pointed you in the 'clever clogs' direction (post #32) which would sort out your questions probably in one go, I think.

Except unfortunately I haven't been taught what a tensor is. All I know is they are half scalar and half vector or something in between

 

[sophiecentaur]

I meant keep the discs perpendicular to the surface and connect their centers via a rod which is not horizontal.

A sort of ball joint with splines at the centre of each disc? I think that would imply the need for a constant velocity joint, as in car steering arrangements.

when a translating object is placed on a surface kinetic friction

After a transitional period, at startup, there may be lateral slip which will involve energy loss ( I think). That's quite a step-up in difficulty imo. The distinction between static and kinetic friction

 

[Aurelius120]

The rotation about the CoM is the (instantaneous) motion in the CoM instantaneous rest frame. The separation into CoM angular momentum and angular momentum due to the CoM motion is simply a convenient way of doing the bookkeeping of the angular momentum as the terms separate.

Consider a system of rigidly connected masses $m_i$ with positions $\vec x_i$ and velocities $\vec v_i$ relative to a fixed point of the rotation with angular velocity $\vec \omega$. By assumption of the rigidity:
$$
\vec v_i = \vec\omega \times \vec x_i
$$
and so the total angular momentum relative to the fixed point is
$$
\vec L = \sum_i m_i \vec x_i \times (\vec \omega \times \vec x_i)
= \sum_i m_i [r_i^2 \vec \omega - (\vec\omega\cdot \vec x_i)\vec x_i]
$$
where $r_i$ is the magnitude of $\vec x_i$. This is the usual expression involving what is essentially the definition of the moment of inertia relative to the fixed point.

We can now rewrite this by introducing a reference displacement $\vec x_0$ and the instantaneous rigid velocity $\vec v_0 = \vec\omega \times \vec x_0$ of that point. We introduce $\vec x_i = \vec y_i + \vec x_0$ and $\vec v_i = \vec u_i + \vec v_0$, where $\vec y_i$ and $\vec u_i$ are the position and velocity of mass $i$ relative to the reference displacement and velocity.

The angular momentum about the fixed point now takes the form
$$
\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)
= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]
+ M \vec x_0 \times \vec v_0
+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]
$$
with $R_i$ the magnitude of $\vec y_i$ and $M = \sum_i m_i$ (ie, the total mass). The first term here is now the angular momentum of the system relative to the reference displacement in the instantaneous rest frame of the reference displacement. The second term is the angular momentum of a single particle of mass $M$ at the reference displacement relative to the original fixed point. The last term is quite messy.

However, if you take the CoM as the reference displacement, then $\sum_i m_i \vec y_i = 0$ by definition and the last two terms vanish. Voilà, you have reexpressed the total angular momentum relative to the fixed point as the angular momentum relative to the CoM (in the CoM instantaneous rest frame) plus the angular momentum of a point mass at the CoM moving at the CoM

And if COM is the fixed point, the angular momentum is the angular momentum in the COM instantaneous rest frame.
So beautiful.

The advantage over the first expression is that the moment of inertia relative to the CoM is often much easier to express in terms of standard moments of inertia that have already been computed. Many times simplifying computation.

Iirc, this is also covered in some detail in chapter 10 of my book.

Sounds like a great book. Is it for after, before or while engineering?

 

[Aurelius120]

@Orodruin
$$\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)$$ $$= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]+ M \vec x_0 \times \vec v_0+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]$$
In the first term, representing angular momentum about instantaneous COM rest frame,
The first part $\sum_i m_iR_i^2 \vec \omega$ is essentially $I\omega$
What happens to the second part? Does it cancel out ?
As stated by @TSny and @haruspex we only used $I\omega$ about two perpendicular axes to compute angular momentum.

 

[Orodruin]

Is it for after, before or while engineering?

It is a mathematical methods textbook. Chapter 10 discusses some applications of the previous chapters - in this case vector algebra/analysis. Would probably be considered late undergrad level.

 

[Orodruin]

@Orodruin
$$\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)$$ $$= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]+ M \vec x_0 \times \vec v_0+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]$$
In the first term, representing angular momentum about instantaneous COM rest frame,
The first part $\sum_i m_iR_i^2 \vec \omega$ is essentially $I\omega$

It is $I\omega$ (relative to the com) up to the second part.

What happens to the second part? Does it cancel out ?

It is the rest of $I\omega$.

Edit: Note that
$$
I_{ab} = \sum_i m_i (R_i^2 \delta_{ab} - y_{i,a} y_{i,b})
$$ by definition.

 

[Aurelius120]

It is a valid way of decomposing a rotational velocity vector.
I admit it is less obvious than with linear motion. Displacements can be easily shown to add vectorially, and it therefore works for linear velocities and accelerations by differentiation wrt time. But it does not work for rotations through non-infinitesimal angles: the order of two such rotations matters. But again, it does work for infinitesimal angles, so it also works for angular velocities and accelerations.

Order of rotation (as in CW or ACW?) doesn't matter for infinitesimal angles?
Physically it is tough to visualise for sure.

The angular momentum about the fixed point now takes the form
$$\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)$$ $$= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]+ M \vec x_0 \times \vec v_0+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]$$

But mathematically, this formula proves that it is independent of sign of position vector, right?

 

[haruspex]

Order of rotation (as in CW or ACW?) doesn't matter for infinitesimal angles?
Physically it is tough to visualise for sure.

Order, as in which you do first.
Place a die with 1 facing up, 2 towards you, 3 to the right.
If you rotate it 90° around the left-right (3-4) axis, top (1) away from you, then 90° about the vertical axis, right side away from you, the net result is a rotation about a long diagonal. The faces will now be:
1 left
2 top
3 back
If you had done those two rotations in the other order you would have
1 back
2 right
3 bottom

For infinitesimal rotations we need to switch to a sphere. If we draw little arrows on the surface to represent infinitesimal rotations, we can see that the net of two tiny rotations at right angles is almost the same as a rotation along the hypotenuse. So now they add like vectors.