Question

Find the dimension of.
(A) Angular speed $\left( \omega \right)$
(B) Angular acceleration $\left( \alpha \right)$
(C)Torque $\left( \tau \right)$
(D) Moment of Inertia $\left( {{I}} \right)$

Answer 1

Dimensional formula:
The explanation which shows how and which of the base Quantities represents a physical quantity is called the dimensional formula of the given Physical quantity. In the given above Physical Quantity angular speed, angular acceleration and torque are because of the Rotational motion.

Complete Step by step solution:
Here firstly we have to find the dimensional formula of
(A) Angular speed $: \to$ Angular speed of a body moving in a circular path is defined as the rate of change of its angular displacement.
It is represented by $'\omega '$
In mathematical term it is represented as
$\omega = \dfrac{{{{d}}\phi }}{{{{dt}}}}$ ,
where ${{'d}}\phi '$ is the smallest change in angular displacement in a time interval $'{{dt}}'$
The angular speed of a body is measured in radian per second. Where radiation is the unit of angular displacement.
Hence the dimensional formula for $'\omega '$ is.
${{{M}}^0}{{{L}}^0}{{{T}}^{ - 1}}$
(B) Angular acceleration$: \to$Angular acceleration of an object in circular motion is defined as the rate of angular Velocity. It is denoted by $'\alpha '$.
It’s equivalent to the acceleration in linear motion.
In mathematical term it’s denoted as
$\alpha = \dfrac{{{{d}}\omega }}{{{{dt}}}}$
Where ${{d}}\omega$ is the small change in angular Velocity in a time interval $'{{dt}}'$
In SI system it is represented as
$= {{radian se}}{{{c}}^{ - 2}}$
It’s dimensional formula is
${{{M}}^0}{{{L}}^0}{{{T}}^{ - 2}}$
(C) Torque$: -$ It is defined as the product of magnitude of force and perpendicular distance of the line of the action of force from the axis of relation. It is denoted by Tau $'\tau '$.
Torque is that physical Quantity which gives the rotational effect without the actual motion of the body.
$\mathop \tau \limits^ \to = \mathop {{r}}\limits^ \to \times \mathop {{F}}\limits^ \to$
$\mathop {{r}}\limits^ \to$ is the perpendicular distance from the line of Action.
$\mathop {{F}}\limits^ \to$ is the force applied.
In scalar form $'\tau '$ can be written as
$\tau = {{rFsin}}\theta \mathop {{n}}\limits^ \$
${{sin}}\theta {{ }}$ replaces the cross product .
\mathop {{n}}\limits^ \ is a unit vector along $\mathop \tau \limits^ \to$.
The dimensions of Torque is ${{{M}}^1}{{{L}}^2}{{{T}}^{ - 2}}$
(D) Moment of Inertia $: \to$ Moment of inertia of a body about a given axis is the sum of the product of masses of all particles of the body and square of their respective perpendicular distances from the axis of relation.
In mathematical form it is
${{I = M}}{{{R}}^2}$
${{I = }}$ moment of Inertia
${{M = }}$ Mass the body
${{R = }}$ Perpendicular distance from the axis of relation.
The dimensional for the moment of Inertia is
${{{M}}^1}{{{L}}^2}{{{T}}^0}$

Additional Information:
(A) Important points of Inertia:-

• The mass of a body is a measure of inertia of the body in linear motion. Means more will be the mass more be the Inertia.
  *The quantity that measures the inertia of rotational motion of the body is called the moment of Inertia of that body
  (B) Relation between linear and angular velocity:-
  ${{V = r}}\omega$
  Where ${{V}}$ is the velocity of a body in linear motion.
  $'{{r}}'$ is the perpendicular distance from the axis of Rotation.
  $'\omega '$ is the angular Velocity.
  (C) Relation between linear and angular acceleration.
  ${{a = }}\alpha {{r}}$
  ${{a = }}$ linear acceleration.
  ${{r = }}$ perpendicular distance from axis & rotation.
  $\alpha =$ angular acceleration.

Note: To convert any quantity into its dimensional formulae, we should always use the base quantities. For complex or new quantities, it will be easier to find dimensional formulae of the given quantity if we know the source of the given quantity.