# Kinetic energy in combined rotational and translational motion

If a body is free to move I have studied that any point can be assumed to be in translational motion with velocity of the centre of mass along with a pure rotational motion with $$\omega$$. For Kinetic Energy we can take the sum(obviously scalar sum) of kinetic energies in both cases, which is equal to $$\frac{1}{2} M (v_{COM})^2 + \frac{1}{2} (I_{COM})\omega ^ 2.$$

Why do we have to take pure rolling about centre of mass & translational motion with velocity of Centre of mass. Can't we take the velocity of any arbitrary point A & pure rolling about that point A too?

And if a rigid body is hinged about, say a massless rod and the rod itself moves with a velocity called $$V_{hinge}$$ , what should be kinetic energy now?

I think it should be $$\frac{1}{2} M (V_{hinge})^2 + \frac{1}{2} I_{hinge} \omega ^ 2.$$

Here also, I have a similar question, why shouldn't pure rolling about any other point along with translational velocity of that point work? Is it so that the hinge force don't produce any torque?

• This post and its answer might be relevant. Commented Feb 22, 2023 at 6:35
• Commented Feb 22, 2023 at 14:31

Angular velocity is a pseudo vector. That has two different interpretations. In discussions of symmetry, it is taken to mean "axial-vector", which is made from the antisymmetric part of a rank-2 tensor:

$$\tilde{\omega}_{ij} = (r_iv_j - r_jv_i)$$

To avoid this complication, we generally use:

$$\omega_i = \frac 1 2\epsilon_{ijk}\tilde{\omega}_{jk}= (\vec r \times \vec v)_i$$

Hence the cross-product. It rotates like a vector, but is even under reflection (parity transformations).

The other meaning of pseudo-vector is that it depends on a choice of origin, while real vectors do not.

Consider a frictionless bead with velocity v confined to a circular hoop with radius r:

$$\vec\omega = \vec r \times \vec v$$

with conserved angular momentum:

$$\vec L = \vec \omega {\bf I} = m(\vec r \vec r) \vec v\propto mvr^2$$

with zero torque:

$$\vec{\tau} = \vec r \times \vec{F}_{\rm radial} = \vec{r} \times -\frac{mv^2}r \hat r = 0$$

Simple enough. We chose the origin at the center of the hoop.

Now put the origin on the hoop. (I'm not going to do the math), but clearly $$L=0$$ when the bead passes through the origin, and is maximized when the bead is furthest from the origin.

Now $$\vec L$$ isn't conserved. How?

Well, the centripetal force is no longer pointed at the origin so the torque is non-zero.

Just by moving the origin we created torques and changing angular momentum (and angular velocity). This is the pseudo-vector.

Such considerations should solve your problem.

When a body is rolling (without slipping) every particle which make up the object have a velocity as shown by the blue arrows in the right-hand diagram, thus, in principle, one could evaluate the kinetic energy of each of the particles and sum them to find the total kinetic energy.

However it is much easier to combine the contributions from rotation about the centre of mass and the translation of the centre of mass.

In terms of your arbitrary point $$A$$ the kinetic energy could be evaluated and in particular note that if $$A$$ is the point of contact between the body and the surface with no slipping the kinetic energy is $$\frac12 I_{\rm A}\omega^2$$ where $$I_{\rm A}$$ is the moment of inertia of the body about $$A$$.
As with you hinge example there is no contribution form $$\frac 12 m v_{\rm A}$$ as $$v_{\rm A}=0$$.
Note that by using the parallel axis theorem $$\frac12 I_{\rm A}\omega^2 = \frac 12 (I_{\rm CoM}+m r^2)\omega^2 = \frac 12 I_{\rm CoM}\omega^2 + \frac 12 m v_{\rm CoM}^2$$ as for no slipping $$v_{\rm CoM}=r\omega$$.

It is likely that when you started the topic of rotational dynamics a lot of these equation were proved from first principles with a view of making it easier in the future to analyse a variety of situations and than these proofs once seen were often moved to the "distant" memory part of the brain.
An example of this sequence is Chapter 16 and later ones of an MIT Mechanics course textbook.

$$\def \b {\mathbf}$$

For arbitrary point A of the rigid body with 6 generalized coordinate, the kinetic energy is:

$$T=\frac 12 m\,\b v_A\cdot \b v_A+\frac 12\b \omega^T\,\b I_A\,\b\omega$$

were $$\b v_A=\b v_{\rm CM}+\b\omega\times\b u\\ \b I_A=\b I_{\rm CM}-m\,\tilde{\b{u}}\,\tilde{\b{u}}\\ \b \omega=[\b J(\b \phi)]_{3\times 3}\,\dot{\b{\phi}}$$

• $$~\b u~$$ vector from the center of mass to point A
• $$~\b\phi=[\phi_1~,\phi_2~,\phi_3]^T~$$ generalized angle coordinate

Rigid body with hinge joint

$$\b v_A=0\quad\Rightarrow\\ T=\frac 12\b \omega^T\,\b I_A\,\b\omega\\ \b\omega=\b J(\phi_1)\dot{\phi_1}$$ $$~\phi_1~$$ is the generalized coordinate

Example 2D Pendulum

\begin{align*} & \b\omega=\begin{bmatrix} 0 \\ 0 \\ \dot{\phi} \\ \end{bmatrix} \quad,\b I_{\rm CM}= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & o & I_z \\ \end{bmatrix}\quad, \b u=\begin{bmatrix} 0 \\ L/2 \\ 0 \\ \end{bmatrix}\quad\Rightarrow\\\\ &\b I_A= \left[ \begin {array}{ccc} 1/4\,m{L}^{2}&0&0\\ 0&0&0 \\ 0&0&1/4\,m{L}^{2}\end {array} \right] \\\\&T=\frac{1}{2}\b\omega^T\,\b I_A\,\b\omega=\frac{1}{2}\dot{\phi}\left(I_z+\frac{1}{4}\,m\,L^2\right) \end{align*}

$$\tilde{\b{u}}=\left[ \begin {array}{ccc} 0&-u_{{z}}&u_{{y}}\\ u_{ {z}}&0&-u_{{x}}\\ -u_{{y}}&u_{{x}}&0\end {array} \right]$$