General Mechanics/The Uneven Dumbbell

The kinetic energy and angular momentum of the dumbbell may be split into two parts, one having to do with the motion of the center of mass of the dumbbell, the other having to do with the motion of the dumbbell relative to its center of mass.

To do this we first split the position vectors into two parts. The centre of mass is at.

$\mathbf {R} ={\frac {M_{1}\mathbf {r} _{1}+M_{2}\mathbf {r} _{2}}{M_{1}+M_{2}}}$

so we can define new position vectors, giving the position of the masses relative to the centre of mass, as shown.

$\mathbf {r} _{1}^{*}=\mathbf {r} _{1}-\mathbf {R} \quad \mathbf {r} _{2}^{*}=\mathbf {r} _{2}-\mathbf {R}$

The total kinetic energy is

${\begin{matrix}K&=&{\frac {1}{2}}M_{1}V_{1}^{2}&+&{\frac {1}{2}}M_{2}V_{2}^{2}\\&=&{\frac {1}{2}}M_{1}|\mathbf {V} -\mathbf {v} _{1}^{*}|^{2}&+&{\frac {1}{2}}M_{2}|\mathbf {V} -\mathbf {v} _{2}^{*}|^{2}\\&=&{\frac {1}{2}}(M_{1}+M_{2})V^{2}&+&{\frac {1}{2}}M_{1}{v_{1}^{*}}^{2}+{\frac {1}{2}}M_{2}{v_{2}^{*}}^{2}\\&=&K_{ext}&+&K_{int}\\\end{matrix}}$

which is the sum of the kinetic energy the dumbbell would have if both masses were concentrated at the center of mass, the translational kinetic energy and the kinetic energy it would have if it were observed from a reference frame in which the center of mass is stationary, the rotational kinetic energy.

The total angular momentum can be similarly split up

${\begin{matrix}\mathbf {L} &=&\mathbf {L} _{orb}&+&\mathbf {L} _{spin}\\&=&(M_{1}+M_{2})\mathbf {R} \times \mathbf {V} &+&(M_{1}\mathbf {r} _{1}^{*}\times \mathbf {v} _{1}^{*}+M_{2}\mathbf {r} _{2}^{*}\times \mathbf {v} _{2}^{*})\end{matrix}}$

into the sum of the angular momentum the system would have if all the mass were concentrated at the center of mass, the orbital angular momentum, and the angular momentum of motion about the center of mass, the spin angular momentum.

We can therefore assume the centre of mass to be fixed.

Since ω is enough to describe the dumbbells motion, it should be enough to determine the angular momentum and internal kinetic energy. We will try writing both of these in terms of ω

First we use two results from earlier

$\mathbf {L} =\mathbf {r} \times \mathbf {v} \quad {\mbox{and}}\quad \mathbf {v} ={\boldsymbol {\omega }}\times \mathbf {r}$

to write the angular momentum in terms of the angular velocity

${\begin{matrix}\mathbf {L} &=&M_{1}\mathbf {r} _{1}^{*}\times \mathbf {v} _{1}^{*}&+&M_{2}\mathbf {r} _{2}^{*}\times \mathbf {v} _{2}^{*}\\&=&M_{1}\mathbf {r} _{1}^{*}\times ({\boldsymbol {\omega }}\times \mathbf {r} _{1}^{*})&+&M_{2}\mathbf {r} _{2}^{*}\times ({\boldsymbol {\omega }}\times \mathbf {r} _{2}^{*})\\&=&M_{1}\left({\boldsymbol {\omega }}\left(\mathbf {r} _{1}^{*}\cdot \mathbf {r} _{1}^{*}\right)-\mathbf {r} _{1}^{*}\left(\mathbf {r} _{1}^{*}\cdot {\boldsymbol {\omega }}\right)\right)&+&M_{2}\left({\boldsymbol {\omega }}\left(\mathbf {r} _{2}^{*}\cdot \mathbf {r} _{2}^{*}\right)-\mathbf {r} _{2}^{*}\left(\mathbf {r} _{2}^{*}\cdot {\boldsymbol {\omega }}\right)\right)\\&=&M_{1}\left({\boldsymbol {\omega }}d_{1}^{2}-\mathbf {r} _{1}^{*}\left(\mathbf {r} _{1}^{*}\cdot {\boldsymbol {\omega }}\right)\right)&+&M_{2}\left({\boldsymbol {\omega }}d_{2}^{2}-\mathbf {r} _{2}^{*}\left(\mathbf {r} _{2}^{*}\cdot {\boldsymbol {\omega }}\right)\right)\\&=&(M_{1}d_{1}^{2}+M_{2}d_{2}^{2}){\boldsymbol {\omega }}&-&\left(M_{1}\mathbf {r} _{1}^{*}\left(\mathbf {r} _{1}^{*}\cdot {\boldsymbol {\omega }}\right)+M_{2}\mathbf {r} _{2}^{*}\left(\mathbf {r} _{2}^{*}\cdot {\boldsymbol {\omega }}\right)\right)\end{matrix}}$

The first term in the angular momentum is proportional to the angular velocity, as might be expected, but the second term is not.

What this means becomes clearer if we look at the components of L For notational convenience we'll write

$\mathbf {r} _{1}^{*}=(x_{1},y_{1},z_{1})\quad \mathbf {r} _{2}^{*}=(x_{2},y_{2},z_{2})$

These six numbers are constants, reflecting the geometry of the dumbbell.

${\begin{matrix}L_{x}&=&(M_{1}d_{1}^{2}+M_{2}d_{2}^{2})\omega _{x}-(M_{1}x_{1}^{2}+M_{2}x_{2}^{2})\omega _{x}\\&&-(M_{1}x_{1}y_{1}+M_{2}x_{2}y_{2})\omega _{y}-(M_{1}x_{1}z_{1}+M_{2}x_{2}z_{2})\omega _{z}\\L_{y}&=&(M_{1}d_{1}^{2}+M_{2}d_{2}^{2})\omega _{y}-(M_{1}x_{1}^{2}+M_{2}x_{2}^{2})\omega _{y}\\&&-(M_{1}x_{1}y_{1}+M_{2}x_{2}y_{2})\omega _{x}-(M_{1}y_{1}z_{1}+M_{2}y_{2}z_{2})\omega _{z}\\L_{z}&=&(M_{1}d_{1}^{2}+M_{2}d_{2}^{2})\omega _{z}-(M_{1}x_{1}^{2}+M_{2}x_{2}^{2})\omega _{z}\\&&-(M_{1}x_{1}z_{1}+M_{2}x_{2}z_{2})\omega _{x}-(M_{1}y_{1}z_{1}+M_{2}y_{2}z_{2})\omega _{y}\\\end{matrix}}$

This, we recognise as being a matrix multiplication.

${\begin{pmatrix}L_{x}\\L_{y}\\L_{z}\end{pmatrix}}={\begin{pmatrix}I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\end{pmatrix}}{\begin{pmatrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{pmatrix}}$

where

${\begin{pmatrix}I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\end{pmatrix}}={\begin{pmatrix}M_{1}(d_{1}^{2}-x_{1}^{2})+M_{2}(d_{2}^{2}-x_{2}^{2})&-(M_{1}x_{1}y_{1}+M_{2}x_{2}y_{2})&-(M_{1}x_{1}z_{1}+M_{2}x_{2}z_{2})\\-(M_{1}x_{1}y_{1}+M_{2}x_{2}y_{2})&M_{1}(d_{1}^{2}-y_{1}^{2})+M_{2}(d_{2}^{2}-y_{2}^{2})&-(M_{1}y_{1}z_{1}+M_{2}y_{2}z_{2})\\-(M_{1}x_{1}z_{1}+M_{2}x_{2}z_{2})&-(M_{1}y_{1}z_{1}+M_{2}y_{2}z_{2})&M_{1}(d_{1}^{2}-z_{1}^{2})+M_{2}(d_{2}^{2}-z_{2}^{2})\end{pmatrix}}$

The nine coefficients of the matrix I are called moments of inertia.

By choosing our axis carefully we can make this matrix diagonal. E.g. if

$\mathbf {r} _{1}^{*}=(d_{1},0,0)\quad \mathbf {r} _{2}^{*}=(-d_{2},0,0)$

then

${\begin{pmatrix}I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\end{pmatrix}}={\begin{pmatrix}0&0&0\\0&M_{1}d_{1}^{2}+M_{2}d_{2}^{2}&0\\0&0&M_{1}d_{1}^{2}+M_{2}d_{2}^{2}\end{pmatrix}}$

Because the dumbbell is aligned along the x-axis, rotating it around that axis has no effect.

The relationship between the kinetic energy, T, and ω quickly follows.

${\begin{matrix}2T&=&M_{1}{v_{1}^{*}}^{2}&+&M_{2}{v_{2}^{*}}^{2}\\&=&M_{1}\mathbf {v} _{1}^{*}\cdot ({\boldsymbol {\omega }}\times \mathbf {r} _{1}^{*})&+&M_{1}\mathbf {v} _{2}^{*}\cdot ({\boldsymbol {\omega }}\times \mathbf {r} _{2}^{*})\\&=&M_{1}{\boldsymbol {\omega }}\cdot (\mathbf {r} _{1}^{*}\times \mathbf {v} _{1}^{*})&+&M_{2}{\boldsymbol {\omega }}\cdot (\mathbf {r} _{2}^{*}\times \mathbf {v} _{2}^{*})\end{matrix}}$

On the right hand side we immediately recognise the definition of angular momentum.

$T={\frac {1}{2}}{\boldsymbol {\omega }}\cdot \mathbf {L}$

Substituting for L gives

$T={\frac {1}{2}}{\boldsymbol {\omega }}\mathbf {I} {\boldsymbol {\omega }}$

Using the definition

${\boldsymbol {\omega }}=\omega \mathbf {n} \quad \mathbf {n} \cdot \mathbf {n} =1$

this reduces to

$T={\frac {1}{2}}I\omega ^{2}$

where the moment of inertia around the axis n is

$I=\mathbf {n} \mathbf {I} \mathbf {n}$

a constant.

If the dumbbell is aligned along the x-axis as before we get

$T={\frac {1}{2}}(M_{1}d_{1}^{2}+M_{2}d_{2}^{2})(\omega _{y}^{2}+\omega _{z}^{2})$

These equations of rotational dynamics are similar to those for linear dynamics, except that I is a matrix rather than a scalar.

Further reading

• Modern Physics/The Uneven Dumbbell