Question

A wheel is rotating at 900 rpm about its axis. When the power is cut off, it comes to rest in 1 min. The angular retardation (in $rad{\text{ }}{s^{ - 1}}$) is
A. $\dfrac{\pi }{2}$
B. $\dfrac{\pi }{4}$
C. $\dfrac{\pi }{6}$
D. $\dfrac{\pi }{8}$

Answer 1

If a body moves from one position to other position then subtracting the initial position vector from the final position vector gives us displacement. Rate of change of displacement with respect to time gives us velocity. The rate of change of velocity gives us acceleration. We will define these in rotational terms to solve this question.

Formula used:
$\alpha = \dfrac{{d\omega }}{{dt}}$

Complete step by step answer:
Just like the linear displacement in case of translational motion there will be angular displacement in case of rotational motion. In translational motion there will be linear displacement which is measured in meters and in case of rotational motion there will be angular displacement which will be measured as angle rotated in radians. Rate of change of linear displacement is called linear velocity while the rate of change of angular displacement is called angular velocity. Linear acceleration is considered as the rate of change of linear velocity while angular acceleration is considered as rate of change of angular velocity.
Angular velocity is generally denoted as $\omega$ while angular acceleration is denoted as $\alpha$ and the relation between them is given as
$\alpha = \dfrac{{d\omega }}{{dt}}$
$\Rightarrow \alpha = \dfrac{{{\omega _f} - {\omega _i}}}{t}$
Here ${\omega _i} = 900rpm = \dfrac{{900 \times 2\pi }}{{60}}rad{s^{ - 1}}$ and ${\omega _f} = 0$ and $t = 1\min = 60\sec$
So by substituting all the above values in angular acceleration equation we will get
\eqalign{ & \alpha = \dfrac{{0 - \left( {\dfrac{{900 \times 2\pi }}{{60}}} \right)}}{{60}} \cr & \Rightarrow \alpha = \dfrac{{ - 900 \times 2\pi }}{{3600}} \cr & \therefore \alpha = \dfrac{{ - \pi }}{2}rad{s^{ - 1}} \cr}
Hence the magnitude of angular retardation will be $\dfrac{\pi }{2}rad{s^{ - 1}}$

So option A will be the answer.

Note:
If the acceleration doesn’t vary with time then that is called uniform acceleration. The negative sign which we had got after solving for the angular acceleration denotes that it is under angular retardation i.e its angular velocity is decreasing with the time and we should convert the rotations per minute to radians per sec to continue the calculations.