A material point rotates so that its angular displacement…

A material point rotates so that its angular displacement vector changes with time as $\vec\theta(t) = B\,t^{J}\,\vec n_0$, where $B$ is a constant and $\vec n_0$ is the unit vector normal to the plane of the trajectory. Find (a) the angular velocity vector and its magnitude at $t = t_0$; (b) the angular acceleration vector and its magnitude at $t = t_d/6$.

Explanation of Method

Angular velocity and acceleration vectors from θ(t).

Step 1: (a) Angular velocity

$\vec n_0$ is fixed, so only the magnitude changes; differentiate it: $\vec\omega(t) = \frac{d\vec\theta}{dt} = BJ\,t^{J-1}\,\vec n_0$.

\[\vec\omega(t_0) = BJ\,t_0^{J-1}\,\vec n_0, \qquad \omega(t_0) = |BJ|\,t_0^{J-1}\]

Step 2: (b) Angular acceleration

\[ \vec\varepsilon(t) = \frac{d\vec\omega}{dt} = BJ(J-1)\,t^{J-2}\,\vec n_0. \]

\[\vec\varepsilon\!\left(\tfrac{t_d}{6}\right) = BJ(J-1)\left(\tfrac{t_d}{6}\right)^{J-2}\vec n_0, \qquad \varepsilon = |BJ(J-1)|\left(\tfrac{t_d}{6}\right)^{J-2}\]

The likely intended case: $J = 3$

The symbols in the statement seem garbled. If the exponent is 3, i.e. $\vec\theta = Bt^3\,\vec n_0$, everything becomes simple:

$ \vec\omega = 3Bt^2\,\vec n_0 $, so $ \omega(t_0) = 3Bt_0^2 $

$ \vec\varepsilon = 6Bt\,\vec n_0 $, so $ \varepsilon\!\left(\tfrac{t_d}{6}\right) = 6B \cdot \tfrac{t_d}{6} = B\,t_d $

The factor 6 cancels neatly, which suggests $J = 3$ is what was meant.

$\vec n_0$ $\vec\omega$ $\vec\varepsilon$ $\theta$ $\text{point}$ $\text{all along } \vec n_0:$ $\vec\theta \propto t^J$ $\vec\omega \propto t^{J-1}$ $\vec\varepsilon \propto t^{J-2}$

The point moves in a fixed plane, so the angle, the angular velocity and the angular acceleration are all vectors along the same normal $\vec n_0$ (right-hand rule). Only their magnitudes change with time. Drawn for $J > 1$, when $\vec\omega$ and $\vec\varepsilon$ point the same way.

Result

\[\boxed{\,\vec\omega(t_0) = BJ\,t_0^{J-1}\,\vec n_0, \qquad \omega(t_0) = |BJ|\,t_0^{J-1}\,}\]

\[\boxed{\,\vec\varepsilon\!\left(\tfrac{t_d}{6}\right) = BJ(J-1)\left(\tfrac{t_d}{6}\right)^{J-2}\vec n_0, \qquad \varepsilon = |BJ(J-1)|\left(\tfrac{t_d}{6}\right)^{J-2}\,}\]

Remark

Note. Both $\vec\omega$ and $\vec\varepsilon$ point along $\vec n_0$, perpendicular to the plane of motion, as for any rotation about a fixed axis. They point the same way (the rotation speeds up) when $BJ(J-1)$ has the same sign as $BJ$, i.e. for $J \gt 1$.