Question 539: Centre of Mass and Rotational Motion

Question

In case of rotational dynamics, which one of the following statements is correct?
[$\vec{\omega}=$ angular velocity, $\overrightarrow{\mathrm{v}}=$ linear velocity
$\overrightarrow{\mathbf{r}}=$ radius vector, $\vec{\alpha}=$ angular acceleration
$\overrightarrow{\mathrm{a}}=$ linear acceleration, $\overrightarrow{\mathrm{L}}=$ angular momentum
$\overrightarrow{\mathrm{p}}=$ linear momentum, $\bar{	au}=$ torque,
$\overrightarrow{\mathrm{f}}=$ centripetal force]

Accepted Answer

The correct statement in rotational dynamics is:
Option C:
$$    \overrightarrow{\mathrm{v}} = \vec{\omega} 	imes \overrightarrow{\mathrm{r}}, \quad    \vec{\alpha} = \overrightarrow{\mathrm{a}} 	imes \overrightarrow{\mathrm{r}}, \quad    \overrightarrow{\mathrm{L}} = \overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{p}}, \quad    \vec{	au} = \overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{f}}   $$
Explanation:
Linear velocity ($\overrightarrow{\mathrm{v}}$)
:
The linear velocity of a point in rotational motion is given by the cross product of the angular velocity vector ($\vec{\omega}$) and the radius vector ($\overrightarrow{\mathrm{r}}$):
$$   \overrightarrow{\mathrm{v}} = \vec{\omega} 	imes \overrightarrow{\mathrm{r}}   $$
This shows the perpendicular relationship between linear velocity and both the radius vector and the angular velocity.
Angular acceleration ($\vec{\alpha}$)
:
The angular acceleration and its relation to linear acceleration can often be represented with:
$$   \vec{\alpha} = \overrightarrow{\mathrm{a}} 	imes \overrightarrow{\mathrm{r}}   $$
Though, if specified to mean tangential component in a circular path, it directly relates to the tangential acceleration.
Angular momentum ($\overrightarrow{\mathrm{L}}$)
:
The angular momentum of a particle with respect to a point is defined as:
$$   \overrightarrow{\mathrm{L}} = \overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{p}}   $$
where $\overrightarrow{\mathrm{p}}$ is the linear momentum ($mv$) of the particle.
Torque ($\vec{	au}$)
:
Torque is defined as the cross product of the radius vector ($\overrightarrow{\mathrm{r}}$) and the force vector ($\overrightarrow{\mathrm{f}}$):
$$   \vec{	au} = \overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{f}}   $$
This defines the rotational effect of a force applied at a distance from a pivot.
Each component is consistent with the right-hand rule and the classical definitions in rotational dynamics.

Answer

$\overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{r}} 	imes \vec{\omega}, \overrightarrow{\boldsymbol{\alpha}}=\overrightarrow{\mathbf{r}} 	imes \vec{a}, \vec{L}=\overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{p}}, \vec{	au}=\overrightarrow{\mathrm{f}} 	imes \overrightarrow{\mathrm{r}}$

Answer

$\overrightarrow{\mathrm{v}}=\vec{\omega} 	imes \overrightarrow{\mathrm{r}}, \vec{\alpha}=\overrightarrow{\mathrm{a}} 	imes \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{p}} 	imes \overrightarrow{\mathrm{r}}, \vec{	au}=\overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{f}}$

Answer

$\overrightarrow{\mathrm{v}}=\vec{\omega} 	imes \overrightarrow{\mathrm{r}}, \vec{\alpha}=\overrightarrow{\mathrm{a}} 	imes \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{p}}, \vec{	au}=\overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{f}}$

Answer

$\overrightarrow{\mathrm{v}}=\vec{\omega} 	imes \overrightarrow{\mathrm{r}}, \vec{\alpha}=\overrightarrow{\mathrm{a}} 	imes \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}, \vec{	au}=\overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{f}}$

Centre of Mass and Rotational Motion

Choose the correct answer: