Question

Find the dimensions of
(A) Angular speed $ \omega $
(B) Angular acceleration $ \alpha $
(C) Torque $ \tau $
(D) Moment of inertia $ I $

Answer 1

Hint :We have to know about all these four elements. We know that angular speed is characterized as the pace of progress of precise removal and is given by the articulation where the angular displacement is the time is the angular speed. We also know that, At the point when an article moves with a precise speed ω along the circuit of a round way of radius r, at that point we say that it has an angular energy. Angular momentum is what could be compared to straight force and is characterized as the snapshot of Linear momentum.

Complete Step By Step Answer:
Angular speed- $ \dfrac{v}{r} $
Now dimension of angular speed is $ = \dfrac{{\left[ {L{T^{ - 1}}} \right]}}{{\left[ L \right]}} $
Which is equal to $ \left[ {{T^{ - 1}}} \right] $
Angular acceleration- $ \omega /t $
Dimension of angular acceleration is
$ = \dfrac{{\left[ {{T^1}} \right]}}{{\left[ T \right]}} \\
   = \left[ {{T^{ - 2}}} \right] \\ $
So, the dimension of the angular acceleration is $ \left[ {{T^{ - 2}}} \right] $
Torque- $ F.r $
Dimension of torque is – dimension of F. Dimension of r
$ = \left[ {ML{T^{ - 2}}} \right] \times \left[ L \right] \\
   = \left[ {M{L^2}{T^{ - 2}}} \right] \\$
So, the dimension is $ \left[ {M{L^2}{T^{ - 2}}} \right] $
Moment of inertia- $ m.{r^2} $
Dimension of $ I = \left[ M \right] \times \left[ {{L^2}} \right] $ .

Note :
We also have to know about moments of inertia and torque. We know that torque is the proportion of the power that can make an article pivot about a hub. Power is the thing that makes an article speed up in straight kinematics, likewise, force is the thing that causes a precise speed increase. Consequently force can be characterized as what could be compared to direct power. We have to keep this in our mind to solve this question.