Question about the Product of Inertia for a Rolling Disk

[Anubhav singh]

Homework Statement

If an uniformly mass distributed perfectly symmetric circular disk perform pure rolling motion or trans rotational motion on stationary plane and alpha angular acceleration about centre of mass in this case product of inertia is always zero is it true or not

Relevant Equations

I=Mr^2/2

I think product of inertia always zero because rolling motion of disk is fully balanced

 

[kuruman]

When you say "product of inertia" what exactly do you mean? If you mean moment of inertia, then the moment of inertia of the disk is what you have as your relevant equation. If you don't mean moment of inertia, then please explain what you mean.

 

[Anubhav singh]

Product of inertia means off diagonal elements of inertia tensor

 

[Anubhav singh]

When you say "product of inertia" what exactly do you mean? If you mean moment of inertia, then the moment of inertia of the disk is what you have as your relevant equation. If you don't mean moment of inertia, then please explain what you mean.

Product of inertia means off diagonal elements of inertia tensor

 

[Anubhav singh]

When you say "product of inertia" what exactly do you mean? If you mean moment of inertia, then the moment of inertia of the disk is what you have as your relevant equation. If you don't mean moment of inertia, then please explain what you mea

 

[Anubhav singh]

Sir i am talking about inertia tensor which is 3*3 matrix

 

[Anubhav singh]

Sir i am talking about inertia tensor which is 3*3 matrix

 

[haruspex]

Product of inertia means off diagonal elements of inertia tensor

Doesn't it depend on the choice of coordinates? And what has the state of motion got to do with it?

 

[kuruman]

I think product of inertia always zero because rolling motion of disk is fully balanced

The off-diagonal elements of the inertia tensor are zero in the principal frame where the inertia tensor is diagonal. Is this what you are trying to say? If so, it is true by definition of the principal frame.

 

[Anubhav singh]

The off-diagonal elements of the inertia tensor are zero in the principal frame where the inertia tensor is diagonal. Is this what you are trying to say? If so, it is true by definition of the principal frame.

I think product of inertia measures imbalance and if symmetric uniform mass distributed circular disk rolls on surface then no imbalance because rotation of disk fully balanced

 

[kuruman]

You need to define imbalance better. As @haruspex remarked, the moment of inertia tensor depends on your choice of coordinate axes. Suppose you have a tensor with off-diagonal elements in some coordinate system. According to you ii is imbalanced. Now you know that you can always diagonalize the 3×3 moment of inertia matrix which is another way of saying that you can find a new coordinate system in which the matrix is diagonal. Do you really believe that merely by choosing a different coordinate system the rigid body stops being imbalanced and becomes balanced? That is why I am saying that you need a better definition for "imbalance."

 

[haruspex]

I think product of inertia measures imbalance and if symmetric uniform mass distributed circular disk rolls on surface then no imbalance because rotation of disk fully balanced

As I wrote, the inertia tensor is a function of the object's mass distribution and the coordinate system used to represent it. It does not depend on the motion.
Are you assuming the coordinate system is aligned with the motion?

 

[Anubhav singh]

I think product of inertia measures imbalance and if symmetric uniform mass distributed circular disk rolls on surface then no imbalance because rotation of disk fully balanced

 

[Anubhav singh]

Sir i

You need to define imbalance better. As @haruspex remarked, the moment of inertia tensor depends on your choice of coordinate axes. Suppose you have a tensor with off-diagonal elements in some coordinate system. According to you ii is imbalanced. Now you know that you can always diagonalize the 3×3 moment of inertia matrix which is another way of saying that you can find a new coordinate system in which the matrix is diagonal. Do you really believe that merely by choosing a different coordinate system the rigid body stops being imbalanced and becomes balanced? That is why I am saying that you need a better definition for "imbalance."

Sir i think in rolling motion if disk is symmetric and unique and if disk perform trans rotational motion then product of inertia is zero because fully symmetric body and also in real life application if a circular wheel perform trans rotational motion and mass distributed uniformly then no imbalance and if no imbalance means the term of product of inertia also zero and in this case we only consider moment of inertia about centre of mass

 

[Anubhav singh]

Sir i think in rolling motion if disk is symmetric and unique and if disk perform trans rotational motion then product of inertia is zero because fully symmetric body and also in real life application if a circular wheel perform trans rotational motion and mass distributed uniformly then no imbalance and if no imbalance means the term of product of inertia also zero and in this case we only consider moment of inertia about centre of mass no need to consider product of inertia

As I wrote, the inertia tensor is a function of the object's mass distribution and the coordinate system used to represent it. It does not depend on the motion.
Are you assuming the coordinate system is aligned with the motion?

 

[Anubhav singh]

Sir i think in rolling motion if disk is symmetric and unique and if disk perform trans rotational motion then product of inertia is zero because fully symmetric body and also in real life application if a circular wheel perform trans rotational motion and mass distributed uniformly then no imbalance and if no imbalance means the term of product of inertia also zero and in this case we only consider moment of inertia about centre of mass no need to consider product of inert

As I wrote, the inertia tensor is a function of the object's mass distribution and the coordinate system used to represent it. It does not depend on the motion.
Are you assuming the coordinate system is aligned with the motion?

You need to define imbalance better. As @haruspex remarked, the moment of inertia tensor depends on your choice of coordinate axes. Suppose you have a tensor with off-diagonal elements in some coordinate system. According to you ii is imbalanced. Now you know that you can always diagonalize the 3×3 moment of inertia matrix which is another way of saying that you can find a new coordinate system in which the matrix is diagonal. Do you really believe that merely by choosing a different coordinate system the rigid body stops being imbalanced and becomes balanced? That is why I am saying that you need a better definition for "imbalance."

Imbalance means no wobble during rolling or trans rotational motion

 

[haruspex]

Sir i think in rolling motion if disk is symmetric and unique and if disk perform trans rotational motion then product of inertia is zero because fully symmetric body and also in real life application if a circular wheel perform trans rotational motion and mass distributed uniformly then no imbalance and if no imbalance means the term of product of inertia also zero and in this case we only consider moment of inertia about centre of mass no need to consider product of inert
Imbalance means no wobble during rolling or trans rotational motion

Please make some attempt to understand what @kuruman and I are telling you.

• The products of inertia depend on the mass distribution of the body and on the coordinate system chosen.
• They do not depend on the state of motion.
• With certain choices of coordinates ('principal axes') they are all zero.
• Stability depends on the relationship between the principal axes and the state of motion.

See https://en.wikipedia.org/wiki/Moment_of_inertia#Principal_axes

 

[Anubhav singh]

Please make some attempt to understand what @kuruman and I are telling you.

• The products of inertia depend on the mass distribution of the body and on the coordinate system chosen.
• They do not depend on the state of motion.
• With certain choices of coordinates ('principal axes') they are all zero.
• Stability depends on the relationship between the principal axes and the state of motion.

See https://en.wikipedia.org/wiki/Moment_of_inertia#Principal_axes

Sir i am taking rolling motion of disk about centre of mass and principal axis always passed through centre of mass then product of inertia always zero in this case

 

[haruspex]

i am taking rolling motion of disk about centre of mass

See my second bullet. The products of inertia do not depend on the motion.

principal axis always passed through centre of mass then product of inertia always zero in this case

Merely using an axis passing through the mass centre is not enough to make the products zero. Consider an axis at 45° to the plane of the disc.

 

[Anubhav singh]

See my second bullet. The products of inertia do not depend on the motion.

Merely using an axis passing through the mass centre is not enough to make the products zero. Consider an axis at 45° to the plane of the disc.

Sir i am taking an axis perpendicular to plane of disk passed through centre of mass and rolling motion about this axis in this case definitely product of inertia always zero sir please reply fast

 

[haruspex]

i am taking an axis perpendicular to plane of disk passed through centre of mass

That is one principal axis. Assuming you are taking three perpendicular axes, all three will be principal axes because of the symmetry. Consequently the products of inertia will be zero.

rolling motion about this axis

Irrelevant in determining the products of inertia, but relevant to stability.

 

[Anubhav singh]

That is one principal axis. Assuming you are taking three perpendicular axes, all three will be principal axes because of the symmetry. Consequently the products of inertia will be zero.

Irrelevant in determining the products of inertia, but relevant to stability.

Sir at the end if a symmetric uniform mass distributed circular disk rolls about centre of mass and axis is passed perpendicular to the plane of disk and also this axis is principal axis on plane inclined stationary surface without slipping and pure rolling motion or trans rotational motion is Fully stable and balance and no wobble it means product of inertia is always zero
Sir please give final answer in short time

 

[Anubhav singh]

Sir at the end if a symmetric uniform mass distributed circular disk rolls about centre of mass and axis is passed perpendicular to the plane of disk and also this axis is principal axis on plane inclined stationary surface without slipping and pure rolling motion or trans rotational motion is Fully stable and balance and no wobble it means product of inertia is always zero
Sir please give final answer in short time

 

[haruspex]

Sir at the end if a symmetric uniform mass distributed circular disk rolls about centre of mass and axis is passed perpendicular to the plane of disk and also this axis is principal axis on plane inclined stationary surface without slipping and pure rolling motion or trans rotational motion is Fully stable and balance and no wobble it means product of inertia is always zero
Sir please give final answer in short time

It doesn’t matter how often you repeat yourself, you are wrong.
If you express the tensor in terms of the principal axes then the products of inertia are zero. That's it. Nothing to add about the state of motion or whether it is stable.

 

[kuruman]

I will summarize what we are trying to say with an example. Start with a disk of mass ##M## and radius ##R##. Consider Cartesian coordinate axes with the origin at the center of the disk. Let the ##z## axis be perpendicular to the plane of the disk. The ##x## and ##y## axes are, respectively, along "East" and "North." We know that there are no off-diagonal elements and that ##~I_{zz}=\frac{1}{2}MR^2~## and by the perpendicular axes theorem, ##I_{xx}=I_{yy}=\frac{1}{2}I_{zz}.## Thus, the inertia tensor is $$I_{\text{disk}}=
\frac{1}{2}R^2\begin{pmatrix}
\frac{M}{2} & 0 & 0 \\
0 &\frac{M}{2} & 0 \\
0 & 0 & M \end{pmatrix}.$$ We all agree that this is a "balanced" disk that does not "wobble" when allowed to roll. You say that it is balanced because it has no off-diagonal elements. OK, let's introduce off-diagonal elements by adding mass ##m## on the rim of the disk at 45° "North" of "East", i.e. at point ##\frac{R}{\sqrt{2}}\{1,1,0\}.## The inertia tensor of this mass is
$$I_{\text{mass}}=
\frac{1}{2}R^2\begin{pmatrix}
m & m & 0 \\
m &m & 0 \\
0 & 0 & 2m \end{pmatrix}.$$ The inertia tensor of the composite object is the sum of the two, $$
I_{\text{comp}}=I_{\text{disk}}+I_{\text{mass}}=
\frac{1}{2}R^2\begin{pmatrix}
\frac{M}{2}+m & m & 0 \\
m &\frac{M}{2}+m & 0 \\
0 & 0 & M +2m\end{pmatrix}.$$We all agree that the composite object is unbalanced and will wobble when allowed to roll because of the added mass ##m## on the rim. You say that the off-diagonal elements are an indication that it is so. We say not necessarily and here is why.

Suppose we rotate the coordinate axes about the ##z##-axis by 45° so that the added mass is on the new ##x##-axis. This is equivalent to diagonalizing the matrix. In the new frame the inertia tensor of the added mass is
$$I'_{\text{mass}}=
\frac{1}{2}R^2\begin{pmatrix}
0 & 0 & 0 \\
0 &2m & 0 \\
0 & 0 & 2m \end{pmatrix}$$ whilst the inertia tensor of the disk is the same. Thus, in the new frame the inertia tensor of the composite is
$$
I'_{\text{comp}}=I_{\text{disk}}+I'_{\text{mass}}=
\frac{1}{2}R^2\begin{pmatrix}
\frac{M}{2} & 0 & 0 \\
0 &\frac{M}{2}+2m & 0 \\
0 & 0 & M +2m\end{pmatrix}.$$Here we have an inertia tensor that has zero off-diagonal elements (zero product of inertia) for a disk that we have all agreed is unbalanced because of the added mass on the rim. Clearly, your assertion that "unbalance" is indicated by non-zero product of inertia has no merit.

What you can say merely by looking at the matrix for ##I'_{\text{comp}}## is that the new ##y##-axis is unstable, meaning that if you start this object spinning in free space with initial angular velocity ##\vec{\omega}=\omega~\mathbf{\hat y}##, the direction of the angular velocity will change with respect to time. That's because ##I_{yy}## has an intermediate value ##I_{xx}<I_{yy}<I_{zz}.## If you've studied Euler's equations for rigid body dynamics, you might have encountered this.