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
8.1 Geometrical method
1.2 Relation between torque and moment of inertia
1.3 Pseudo torque
1.4 Torque equation
1.5 Principal of moments
Geometrical method is well understood by following three cases,
1. When the velocity $(v)$ of a point and angular velocity $\left( \omega \right)$ of rotation for a rigid body is known
2. When the direction and magnitudes of two non-parallel velocity for any two points on a rigid body is known
3. When the direction and magnitude of two parallel velocity for any two point on a rigid body is known
Case 1: When the velocity $(v)$ of a point and angular velocity $\left( \omega \right)$ of rotation for a rigid body is known
| S. No. | Steps | Diagram |
| a. | Draw a line perpendicular to the velocity $\overrightarrow v $ |  |
| b. | The instantaneous centre can lie anywhere on this perpendicular line passing through point $P$ | |
| c. | Velocity of point $P$ gives the direction of rotation about the axis of rotation |  |
| d. | Calculate the distance $r$ from point $P$ such that $\left( {v = r\omega } \right)$ | |
| e. | Then velocity of any point on the rigid body can be known by applying the relation, $$\left( {v = R\omega } \right)$$ |  |
Case 2: When the direction and magnitudes of two non-parallel velocity for any two points on a rigid body is known
| S. No. | Steps | Diagram |
| a. | Draw perpendicular lines from point $P$ and $Q$ |
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| b. | The point of intersection of these perpendicular lines is known as ICR at that instant | |
| c. | The axis passing through ICR is known as IAR | |
| d. | Find the angular velocity $\left( \omega \right)$ using the relation, $$\begin{equation} \begin{aligned} {\overrightarrow v _1} = \omega {\overrightarrow r _1} \\ {\overrightarrow v _2} = \omega {\overrightarrow r _2} \\\end{aligned} \end{equation} $$ Note: Distance $r_1$ abd $r_2$ can be calculated by simple elementary geometry |
|
| e. | Then velocity of any point on the rigid body can be known by applying the relation, $$\left( {v = R\omega } \right)$$ |
|
Case 3: When the direction and magnitude of two parallel velocity for any two point on a rigid body is known
| S. No. | Steps | Diagram |
| a. | Mark the direction and magnitude of known velocity on the rigid body |
|
| b. | Length of vector should be directly proportional to the magnitude of the body. Here $\left( {{v_1} > {v_2}} \right)$ | |
| c. | Join head to head and tail to tail of the two vectors. Extend these two lines, so that they can intersect |
|
| d. | The point of intersection is known as ICR at that instant and the axis passing through ICR is known as IAR | |
| e. | Find the angular velocity $\left( \omega \right)$ using the relation, $${{\vec v}_1} = \omega {{\vec r}_1}$$$${{\vec v}_2} = \omega {{\vec r}_2}$$ | |
| f. | Then velocity of any point on the rigid body can be known by applying the relation, $$\left( {v = R\omega } \right)$$ |  |
