Rotational kinetic energy of rigid bar

Question

Consider a rigid bar (infinitely thin and with uniform mass density) of length $L$ with $x_1(t), x_2(t) \in \mathbb{R}^3$ each describing the positions of an endpoint of the bar in some fixed inertial frame at time $t$.

The vector $q(t) := \frac{x_2(t) - x_1(t)}{L}$ is a curve on the unit sphere $S^2$ describing the orientation of the bar. The vectors $\dot{q}(t)$ and $\omega(t) := q(t) \times \dot{q}(t)$ describe the velocity and angular velocity of the bar with respect to the inertial frame.

My question is, what is the rotational kinetic energy of the bar in terms of these variables? From the formulas I know, I would like to say it's just:

$$KE_{rot} = \frac13 mL^2 ||\omega||^2$$

But I haven't worked with this material in a number of years, and am quite rusty.

Answer 1

Let $K$ denote the total rotational kinetic energy of the rotating rod (with mass density $\lambda$). Take $x_1(t)$ to be the (fixed) axis of rotation of the rod. For simplicity, we assume that the magnitude of the angular velocity of the rod, $|\omega(t)|$, is a constant.

First of all, note that since $q(t)$ is a unit vector, we have that

\begin{equation} \frac{d |q(t)|^2}{dt} = \frac{d (q(t) \cdot q(t))}{dt} = \frac{d(q(t))}{dt} \cdot q(t) + \frac{d(q(t))}{dt} \cdot q(t) = 2 \frac{d(q(t))}{dt} \cdot q(t) = 0 \;\; \; (\textrm{i}) \end{equation}

That is, $ \frac{d(q(t))}{dt} \cdot q(t) = 0$, so $\frac{d(q(t))}{dt}$ and $q(t)$ are perpendicular.

Since $\omega(t) = q(t) \times \frac{d(q(t))}{dt}$, we have that

\begin{equation} |\omega(t)| = |q(t)| * |\frac{d(q(t))}{dt}| * \sin(\frac{\pi}{2})= |\frac{d(q(t))}{dt}| \;\; \; (\textrm{ii}) \end{equation}

To calculate $K$, we imagine dividing the rod up into infinitesimal chunks of mass $dm$. Assume that this mass element has a displacement vector $x(t)$ from the axis of rotation. Now, we can calculate $x(t)$ using simple vector geometry:

\begin{equation} x(t) = x_1(t) + |x(t)-x_1(t)| q(t) \;\; \; (\textrm{iii}) \end{equation}

Using equation (iii), we have that the velocity $v(t)$ of this chunk is given by

\begin{equation} v(t) = \frac{dx(t)}{dt} = \frac {d(x_1(t))}{dt} + \frac{d(|x(t)-x_1(t)|)}{dt} q(t) + |x(t)-x_1(t)|\frac{d(q(t))}{dt} \;\; \; (\textrm{iv}) \end{equation}

But $\frac{d(|x(t)-x_1(t)|)}{dt} = 0$ for our mass element (why). Furthermore, $\frac{d(x_1(t))}{dt} = 0$, for the axis of rotation is fixed. Hence, equation (iv) becomes

\begin{equation} v(t) = |x(t)-x_1(t)|\frac{d(q(t))}{dt} \; \; \; (\textrm{iv}) \end{equation}

We can easily calculate the magnitude of $v(t)$:

\begin{equation} |v(t)| = |x(t)-x_1(t)| * |\frac{d(q(t))}{dt}| = |x(t)-x_1(t)| * |\omega(t)| \; \; \; (\textrm{v}) \end{equation}

The kinetic energy $dK$ of the mass element is then given by \begin{equation} dK = \frac{1}{2} dm |v(t)|^2 = \frac{1}{2} dm |x(t)-x_1(t)|^2 * |\omega(t)|^2 \; \; \; (\textrm{vi}) \end{equation}

We now impose a new coordinate system on the rod. Imagine we place the real line $r$ along the rod, with $r = 0$ being the axis of rotation. It then makes sense to set $|x(t) - x_1(t)| = r$ (for the mass element is an arbitrary distance $r$ from the axis of rotation). With these substitutions, equation (vi) reads \begin{equation} dK = \frac{1}{2} dm r^2 \omega^2 \; \; \; (\textrm{vii}) \end{equation}

(Here, $\omega = |\omega(t)|$).

Now, we can finally calculate $K$ by integrating the kinetic energies of each mass chunk along the rod.

\begin{equation} K = \int dK = \int_{0}^{L} \frac{1}{2} dm r^2 \omega^2 = \int_{0}^{L} \frac{1}{2} (\lambda dr) r^2 \omega^2 = \frac{1}{2} \lambda \omega ^2 \int_{0}^{L} r^2 dr =\frac{1}{6} \lambda \omega^2 L^3 \; \; \; (\textrm{viii}) \end{equation}

Since $\lambda L = m$, we have that

\begin{equation} K = \frac{1}{6} m \omega^2 L^2 \end{equation}

Answer 2

So you have the two endpoint vectors and you need to describe the motion in terms of the average and difference of their positions

$$\begin{aligned}\boldsymbol{c} & =\frac{\boldsymbol{x}_{2}+\boldsymbol{x}_{1}}{2}\\ \boldsymbol{q} & =\frac{\boldsymbol{x}_{2}-\boldsymbol{x}_{1}}{\ell} \end{aligned}$$

Now you can describe the translational and rotational velocities of the center of mass as

$$\begin{aligned}\boldsymbol{v} & =\dot{\boldsymbol{c}}\\ \boldsymbol{\omega} & =\boldsymbol{q}\times\dot{\boldsymbol{q}} \end{aligned}$$

and translational and rotational momentum as

$$\begin{aligned}\boldsymbol{p} & =m\,\boldsymbol{v}=m\,\dot{\boldsymbol{c}}\\ \boldsymbol{L} & ={\rm I}\,\boldsymbol{\omega}={\rm I}\,\left(\boldsymbol{q}\times\dot{\boldsymbol{q}}\right) \end{aligned}$$

where ${\rm I}$ is the mass moment of inertia tensor. For the case of a thin slender rod of uniform mass $m$ and length $\ell$, you have ${\rm I} = \tfrac{m}{12} \ell^2$ and it is a scalar value.

Kinetic energy in vector form is

$$\begin{aligned}K & =\tfrac{1}{2}\left(\boldsymbol{v}\cdot\boldsymbol{p}\right)+\tfrac{1}{2}\left(\boldsymbol{\omega}\cdot\boldsymbol{L}\right)\\ & =\tfrac{1}{2}m\left(\dot{\boldsymbol{c}}\cdot\dot{\boldsymbol{c}}\right)+\tfrac{1}{2}\left(\boldsymbol{q}\times\dot{\boldsymbol{q}}\right)\cdot{\rm I}\left(\boldsymbol{q}\times\dot{\boldsymbol{q}}\right) \end{aligned}$$

which is described in terms of both $\boldsymbol{v}$ and $\boldsymbol{\omega}$ in the fist line, and in terms of $\boldsymbol{c}$ and $\boldsymbol{q}$ in the second line.

Answer 3

The vector to the center of mass is

$$\mathbf R=\frac 12 L\,\mathbf q+\mathbf x_1=\frac 12 (\mathbf x_1+\mathbf x_2)$$

form here the kinetic energy

$$T=\frac 12 m\,\mathbf v\cdot \mathbf v+ \frac 12 \mathbf \omega ^T\,\mathbf I_{\rm CM}\,\mathbf \omega $$

where $$\mathbf v=\frac{d}{dt}\mathbf R\quad, \mathbf\omega=\mathbf q\times\dot{\mathbf{q}}\\ \mathbf I_{\rm CM}= \begin{bmatrix} I_x & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z \\ \end{bmatrix}\quad,I_x=I_z=\frac{1}{12}\,m \left( 4\,{L}^{2}+3\,{r}^{2} \right)\quad,I_y=\frac 12 m\,r^2 $$

• $~r~$ bar radius
• $~L~$ bar length
• $~m~$ bar mass

Notice

the angular velocity $~(\mathbf\omega=\mathbf\omega(\mathbf x_i))~$ is given in the bar fixed coordinate system , thus the vector components $~\mathbf x_i~$ must be also given in the bar system.

to obtain the equations of motion , you need additional the pseudo forces

Answer 4

The rotational energy $E_{rot}$ is defined as:

$$ E_{\text{rot}} =\frac{1}{2} \int_C (v_{\text{particle}}-v_{CM})^2 dm $$

We know that the position of a particle in a rigid body can be expressed as a rotation matrix $R(t)$ multiplied by the initial position of the particle plus a time-dependent vector:

$$ r(t) = R(t)r_0 + v(t), $$

The velocity of the particle is the time derivative of this expression.

Parametrize the bar as:

$$ r(h) = A + h(B - A), $$

where $h \in [0, 1]$.

The mass of each particle is $dm = M dh $.

So:

$$ E_{\text{rot}} = \frac{1}{2} \int_0^1 \left| \dot{R}(t) \left( A + h(B - A) - \frac{A + B}{2} \right) \right|^2 M dh=\\ \frac{M}{2} \int_0^1 \left| \dot{R}(t)(B-A) \left( h - \frac{1}{2} \right) \right|^2 \, dh=\\ \frac{M}{2} \int_0^1 \left|L\dot{q}(t) \left( h - \frac{1}{2} \right) \right|^2 \, dh=\frac{ML^2}{24}|\dot{q}(t)|^2 $$