Physics > Rotational Dynamics > 8.0 Instantaneous axis of rotation
Rotational Dynamics
1.0 Introduction
1.1 Torque or moment of a force
1.2 Relation between torque and moment of inertia
1.3 Pseudo torque
1.4 Torque equation
1.5 Principal of moments
2.0 Angular momentum or moment of a momentum
3.0 Relation between torque and angular momentum
4.0 Combined translational and rotational motion of a rigid body
5.0 Rotational kinetic energy
6.0 Uniform pure rolling
7.0 Accelerated pure rolling
8.0 Instantaneous axis of rotation
9.0 Toppling
8.2 Relative velocity method
1.2 Relation between torque and moment of inertia
1.3 Pseudo torque
1.4 Torque equation
1.5 Principal of moments
Consider a wheel whose velocities of point $A$ and $B$ are ${\overrightarrow v _A}$ and ${\overrightarrow v _B}$ respectively as shown in the figure.
If ${\overrightarrow v _A} = {\overrightarrow v _B}$, then the wheel is only doing translational motion.
If ${\overrightarrow v _A} \ne {\overrightarrow v _B}$, then the wheel is undergoing plane motion i.e. (rotational + translational) motion.
Let us consider the angular velocity $\left( {\overrightarrow \omega } \right)$ in the clockwise direction about the centre $O$, as shown in the figure.
So,
$\begin{equation} \begin{aligned} \overrightarrow \omega = \omega \left( { - \widehat k} \right) \\ {\overrightarrow v _A} = {v_A}\left( {\widehat i} \right) \\ {\overrightarrow v _B} = {v_B}\left( {\widehat i} \right) \\\end{aligned} \end{equation} $
We can consider either of the point $A$ or $B$ as our instantaneous axis of rotation.
So, as we know that the body undergoes pure rotational motion about ICR, so relative velocity is taken about the IAR.
Let us assume two cases. In case I, assume point $A$ as IAR and in case II, assume point $B$ as IAR.
| Case I | Case II |
| IAR is at point $A$ As we know,\[\left. \begin{gathered} {\overrightarrow v _A} = {v_A}\left( {\widehat i} \right) \hspace{1em} \\ {\overrightarrow v _B} = {v_B}\left( {\widehat i} \right) \hspace{1em} \\ \overrightarrow \omega = \omega \left( { - \widehat k} \right) \hspace{1em} \\ {\overrightarrow r _{AB}} = 2R\left( { - \widehat j} \right) \hspace{1em} \\ \end{gathered} \right\}\quad ...(i)\] $${\overrightarrow v _{BA}} = {\overrightarrow v _B} - {\overrightarrow v _A}\quad ...(ii)$$ Now, point $B$ undergoes pure rotation about point $A$ which is IAR So, $${\overrightarrow v _{BA}} = \overrightarrow \omega \times {\overrightarrow r _{AB}}\quad ...(iii)$$ Fro equation $(ii)$ and $(iii)$ we get, $$\overrightarrow \omega \times {\overrightarrow r _{AB}} = {\overrightarrow v _B} - {\overrightarrow v _A}\quad ...(iv)$$ From equation $(i)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} \left( { - \omega } \right)\widehat k \times \left( { - 2R} \right)\widehat j = \left( {{v_B} - {v_A}} \right)\widehat i \\ - 2\omega R\;\widehat i = \left( {{v_B} - {v_A}} \right)\widehat i \\ \omega = \left( {\frac{{{v_A} - {v_B}}}{{2R}}} \right) \\\end{aligned} \end{equation} $$ or $$\begin{equation} \begin{aligned} \overrightarrow \omega = \omega \left( { - \widehat k} \right) \\ \overrightarrow \omega = \left( {\frac{{{v_A} - {v_B}}}{{2R}}} \right)\left( { - \widehat k} \right) \\ \overrightarrow \omega = \left( {\frac{{{v_B} - {v_A}}}{{2R}}} \right) \\\end{aligned} \end{equation} $$ | IAR is at point $B$ As we know, \[\left. \begin{gathered}{\overrightarrow v _A} = {v_A}\left( {\widehat i} \right) \hspace{1em} \\{\overrightarrow v _B} = {v_B}\left( {\widehat i} \right) \hspace{1em} \\ \overrightarrow \omega = \omega \left( { - \widehat k} \right) \hspace{1em} \\ {\overrightarrow r _{BA}} = 2R\left( {\widehat j} \right) \hspace{1em} \\\end{gathered}\right\}\quad ...(i)\] $${\overrightarrow v _{AB}} = {\overrightarrow v _A} - {\overrightarrow v _B}\quad ...(ii)$$ Now, point $A$ undergoes pure rotation about point $B$ which is IAR So, $${\overrightarrow v _{AB}} = \overrightarrow \omega \times {\overrightarrow r _{BA}}\quad ...(iii)$$ Fro equation $(ii)$ and $(iii)$ we get, $$\overrightarrow \omega \times {\overrightarrow r _{BA}} = {\overrightarrow v _A} - {\overrightarrow v _B}\quad ...(iv)$$ From equation $(i)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} \left( { - \omega } \right)\widehat k \times \left( { 2R} \right)\widehat j = \left( {{v_A} - {v_B}} \right)\widehat i \\ 2\omega R\;\widehat i = \left( {{v_A} - {v_B}} \right)\widehat i \\ \omega = \left( {\frac{{{v_A} - {v_B}}}{{2R}}} \right) \\\end{aligned} \end{equation} $$ or $$\begin{equation} \begin{aligned} \overrightarrow \omega = \omega \left( { - \widehat k} \right) \\ \overrightarrow \omega = \left( {\frac{{{v_A} - {v_B}}}{{2R}}} \right)\left( { - \widehat k} \right) \\ \overrightarrow \omega = \left( {\frac{{{v_B} - {v_A}}}{{2R}}} \right) \\\end{aligned} \end{equation} $$ |
