Finding Acceleration of a Particle from a Given Velocity Field.

The acceleration of a material particle is the rate of change of its velocity. If the motion of a continuum is given by

$$x_i = x_i (X_1,\ X_2,\ X_3,\ t) \ x_i (X_1,\ X_2,\ X_3,\ 0) = X_i$$

then the velocity $\mathbf{v}$, at time $t$, of a particle $X_j$ is given by

$$v_i = \frac{\partial x_i}{\partial t}\bigg\vert _{X_j-\mbox{fixed}} = \dot x_i,$$

and the acceleration $\mathbf{a}$, at time $t$, of a particle $X_j$ is given by

$$a_i = \frac{\partial v_i}{\partial t}\bigg\vert _{X_j-\mbox{fixed}} = \dot v_i.$$

Thus, if the material description of velocity $\mathbf{v}(X_1,\ X_2,\ X_3,\ t)$ is known, then the acceleration is computed simply by taking the partial derivative with respect to time of the function $\mathbf{v}(X_1,\ X_2,\ X_3,\ t)$. On the other hand, if only the spatial description of velocity (i.e. $v_i = v_i(x_1,\ x_2,\ x_3,\ t)$) is known, then the computation of acceleration involves the use of eqn. (3.5.2). That is,

$$a_i =\ \frac{\partial v_i}{\partial t}\bigg\vert _{X_j-\mbox{fixed}} = \... ...{\partial x_k}\frac{\partial x_k}{\partial t} \bigg\vert _{X_j-\mbox{fixed}}\ ,$$

$$=\ \frac{\partial v_i}{\partial t}\bigg\vert _{x_j-\mbox{fixed}} + \frac{\partial v_i}{\partial x_k} v_k.$$

The part of acceleration given by $v_k\displaystyle\frac{\partial v_i}{\partial x_k}$ is called convective acceleration. When the motion is only along $x_1$-axis, i.e. $v_2 = v_3 = 0$ and $v_1 = v_1(x_1,t)$, then

$$a_1 = \frac{\partial v_1}{\partial t} + v_1\frac{\partial v_1}{\partial x_1}\ .$$

Example: A fluid rotates as a rigid body with a constant angular velocity $\omega\mathbf{e}_3$.

(a)

Write out explicitly the components of the velocity of a material particle in the Lagrangian and Eulerian descriptions of motion.

(b)

Compute the acceleration field in the Eulerian description.

Solution:

(a)

Recall that

$$\mathbf{v} = {\mbox{\boldmath {$\omega$}}}\times \mathbf{r}.$$

Thus

$$v_i =\ \varepsilon_{ijk}\omega_j x_k = \varepsilon_{ijk}\omega\delta_{j3}x_k,$$

$$=\ \omega\varepsilon_{i3k}x_k,$$

is the Eulerian description of velocity.

To convert these expressions into the Lagrangian description, let the fluid particle which presently is at place $P$ be at place $Q$ at $t = 0$. Then, referring to Fig. 3.6.1,

$$\theta - \alpha = \omega t,$$

$$\theta = \alpha + \omega t = \tan^{-1}\left(\frac{X_2}{X_1}\right) + \omega t.$$

Since

$$x_1 =\ OP\cos\theta\ ,$$

$$=\ OQ\cos\theta\ ,$$

$$=\ \sqrt{X^2_1 + X^2_2}\cos\theta\ ,$$

$$v_1 =\ \omega\varepsilon_{132}x_2 = -\omega\sqrt{X^2_1 + X^2_2}\ \sin(\tan^{-1}\left(\frac{X_2}{X_1}\right) + \omega t)\ ,$$

$$v_2 =\ \omega\varepsilon_{231} x_1 = \omega\sqrt{X^2_1 + X^2_2}\ \cos (\tan^{-1}\left(\frac{X_2}{X_1}\right) + \omega t)\ .$$

(b)

$$a_i =\ \frac{\partial v_i}{\partial t} + \frac{\partial v_i}{\partial x_j} v_j\ ,$$

$$=\ 0 + \omega\varepsilon_{i3k}\delta_{kj}v_j\ ,$$

$$=\ \omega\varepsilon_{i3j}(\omega\varepsilon_{j3m}x_m)\ ,$$

$$=\ \omega^2\varepsilon_{i3j}\varepsilon_{3mj}x_m\ ,$$

$$=\ \omega^2(\delta_{i3}\delta_{3m} - \delta_{im}\delta_{33})x_m\ ,$$

$$a_i =\ \omega^2(\delta_{i3}x_3 - x_i)\ .$$

$$\therefore\hspace{.2in} a_1 =\ - \omega^2x_1\ ,$$

$$a_2 =\ -\omega^2x_2\ ,$$

$$a_3 =\ 0\ .$$

Exercise: Given

$$\mathbf{v} = \frac{x_1\mathbf{e}_1 + x_2\mathbf{e}_2}{x^2_1 + x^2_2},\ \theta = 2(x^2_1 + x^2_2)\ ,$$

determine the acceleration and the rate of change of temperature of the material particle which currently is at the place $(1,1)$.

Exercise: For the motion

$$x_i = X_i + \sin (\pi t) \sin (\pi X_1)\delta_{i2}\ ,$$

find the velocity and acceleration in referential and spatial descriptions.