Physics > Rotational Dynamics > 1.0 Introduction
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
1.2 Relation between torque and moment of inertia
1.2 Relation between torque and moment of inertia
1.3 Pseudo torque
1.4 Torque equation
1.5 Principal of moments
As we know, $$\begin{equation} \begin{aligned} {\overrightarrow \tau _i} = {\overrightarrow r _i} \times {\overrightarrow F _i} \\ {\overrightarrow \tau _i} = {\overrightarrow r _i} \times {m_i}{\overrightarrow a _i} \\ {\overrightarrow \tau _i} = {m_i}\left[ {{{\overrightarrow r }_i} \times \left( {\overrightarrow \alpha \times {{\overrightarrow r }_i}} \right)} \right] \\\end{aligned} \end{equation} $$
Vector triple product is given by, $\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c $
Therefore, $$\begin{equation} \begin{aligned} \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c \\ {{\vec \tau }_i} = {{\vec r}_i} \times {{\vec F}_i} \\ {{\vec \tau }_i} = {{\vec r}_i} \times {m_i}{{\vec a}_i} \\ {{\vec \tau }_i} = {m_i}\left[ {{{\vec r}_i} \times \left( {\vec \alpha \times {{\vec r}_i}} \right)} \right] \\ {{\vec \tau }_i} = {m_i}\left[ {\left( {\overrightarrow r .\overrightarrow r } \right)\overrightarrow \alpha - \left( {} \right)\overrightarrow r } \right] \\ {{\vec \tau }_i} = {m_i}{r^2}\overrightarrow \alpha \quad \left( {\overrightarrow r .\overrightarrow \alpha = 0,\;As\;\overrightarrow r \bot \overrightarrow \alpha \;} \right) \\\end{aligned} \end{equation} $$ $$\begin{equation} \begin{aligned} {{\vec \tau }_i} = {m_i}\left[ {\left( {{{\overrightarrow r }_i}.{{\overrightarrow r }_i}} \right)\overrightarrow \alpha - \left( {{{\overrightarrow r }_i}.\overrightarrow \alpha } \right){{\overrightarrow r }_i}} \right] \\ {{\vec \tau }_i} = {m_i}r_i^2\overrightarrow \alpha \quad \left( {{{\overrightarrow r }_i}.\overrightarrow \alpha = 0,\;As\;{{\overrightarrow r }_i} \bot \overrightarrow \alpha \;} \right) \\\end{aligned} \end{equation} $$
Summing up for all the particles of the rigid body we get, $$\begin{equation} \begin{aligned} \sum {{{\overrightarrow \tau }_i}} = \sum {{m_i}r_i^2\overrightarrow \alpha } \\ \overrightarrow \tau = I\overrightarrow \alpha \\\end{aligned} \end{equation} $$
where $I$ is the moment of inertia about the axis of rotation
Note:
- Torque $\left( \tau \right)$ is equal to the net torque due to the external forces only because all the internal forces adds to zero.
- Torque produces angular acceleration $\left( {\overrightarrow \alpha } \right)$, so they are in the same direction as shown in the figure
If the axis of rotation pass through $COM$ the above equation becomes, $${\overrightarrow \tau _{COM}} = 
{I_{COM}}\overrightarrow \alpha $$
{I_{COM}}\overrightarrow \alpha $$As shown in the figure, to rotate a body about an axis of rotation with angular acceleration $\left( {{\alpha _1}} \right)$ whose moment of inertia is $I_1$, then torque $\left( {{\tau _1}} \right)$ is need about the axis of rotation which is given by, $${\tau _1} = {I_1}{\alpha _1}$$
Similarly, ${\tau _2} = {I_2}{\alpha _2}$
