Question

A flywheel rotating at $600\;{\text{rev/min}}$ is brought uniform deceleration and stopped after $2\;{\text{minutes}}$, then what is angular deceleration in ${\text{rad/se}}{{\text{c}}^{\text{2}}}$?

Answer 1

Hint: To find the angular deceleration of the flywheel first of all convert the angular velocity of the flywheel from ${\text{rev/min}}$ to ${\text{rad/se}}{{\text{c}}^{\text{2}}}$ and then use the formula to calculate angular acceleration.

 Complete step-by-step solution:
The angular acceleration of an object can be determined by taking the ratio of the change in angular velocity of the object and the time taken to complete the rotation.
The angular acceleration of a rotating object can be determined by using the formula as shown below,
$\alpha = \dfrac{{{\omega _f} - {\omega _i}}}{t}$
Here, ${\omega _f}$ is the final angular velocity of the rotating object, ${\omega _i}$ is the initial angular velocity of the rotating object and $t$ time of revolution.
The value of angular acceleration increases when the value of angular velocity of the object increases or value of angular acceleration decreases when the value of angular velocity of the object decreases.
When a flywheel stops after $2\;{\text{minutes}}$, then the final angular velocity of the flywheel becomes 0.
${\omega _f} = 0$
So, the final angular velocity of the rotating object is 0.
Convert the initial angular velocity of the flywheel into ${\text{rad/sec}}$.
$
  {\omega _i} = 600\; \times \dfrac{{2\pi }}{{60}} \\
   = 20\pi \;{\text{rad/sec}} \\
 $
So, the initial angular velocity of the rotating object is $20\pi \;{\text{rad/sec}}$.
The standard unit of the angular acceleration is ${\text{rad/se}}{{\text{c}}^{\text{2}}}$ and it is denoted by the symbol $\alpha $.
The magnitude of angular acceleration depends on the magnitude of change in angular velocity of an object and the time taken to complete the rotation by that object.
Substitute 0 for ${\omega _f}$, $20\pi $ for ${\omega _i}$ and $2\;{\text{minutes}}$ for $t$ in the expression $\alpha = \dfrac{{{\omega _f} - {\omega _i}}}{t}$ to find the angle of deceleration.
$
  \alpha = \dfrac{{{\omega _f} - {\omega _i}}}{t} \\
   = \dfrac{{0 - 20\pi \;{\text{rad/sec}}}}{{2\;\min \times \left( {\dfrac{{60\;\sec }}{{1\;\min }}} \right)}} \\
   = - \dfrac{\pi }{6}{\text{rad/se}}{{\text{c}}^{\text{2}}} \\
   = \dfrac{\pi }{6}{\text{rad/se}}{{\text{c}}^{\text{2}}}\left( {{\text{Deceleration}}} \right) \\
 $
The angle of deceleration of the flywheel is $\dfrac{\pi }{6}{\text{rad/se}}{{\text{c}}^{\text{2}}}$.
So, the option (A) is correct.

Note:- Convert the units carefully. Define properly that there is acceleration or deceleration solved accordingly. Do the calculation carefully and avoid silly mistakes in substituting the values.