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A wheel is rotating at the rate of 33 revolution / min. if it comes to a stop in 20 seconds. Then the angular retardation will be:
A) 0
B) 11$\pi $ rad/${s^2}$
C) $\dfrac{\pi }{{200}}\,\,{\text{rad/}}{{\text{s}}^2}$
D) $\dfrac{{11\pi }}{{200}}\,\,{\text{rad/}}{{\text{s}}^2}$

Last updated date: 09th Apr 2024
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MVSAT 2024
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Hint Since the wheel is going to stop, hence final angular acceleration = 0. Now use the equation ${\omega _f} = {\omega _i} + \alpha t$ to get the value of α. Also convert units of ω to rad/sec.

Complete step-by-step answer:
Given that,
The Initial angular velocity of the wheel, (\[{\omega _i}\] )= 33 rev/min
The Final angular velocity of the wheel, (\[{\omega _f}\]) = 0
Time taken by wheel to stop (t) = 20 second = 20s .
We know that,
The initial angular velocity in rad/s (\[{\omega _i}\]) is given by =
   \Rightarrow 33 \times \dfrac{{2\pi }}{{60}} \\
   \Rightarrow 1.1\pi \,{\text{rad/s}} \\
 $ ($\dfrac{{2\pi }}{{60}}$ is the conversion factor from revolution/minute to radian/sec.)
The angular retardation can be calculated by using the formula,
(Since retardation is mentioned, hence we have to take the negative value of alpha and not positive value.) .
  {\omega _f} = {\omega _i} - \alpha t \\
   \Rightarrow 0 = 1.1\pi - \alpha \times 20 \\
   \Rightarrow \alpha = \dfrac{{1.1\pi }}{{20}} = \dfrac{{11\pi }}{{200}} \\
 \] (Putting the values given in the question, we have the following.)

Hence, the angular retardation of the wheel will be \[\dfrac{{11\pi }}{{200}}\] rad/${s^2}$.

Note All the three equations of motion can be used in circular motion just by changing the few notations like “a” (acceleration) with \[\alpha \](angular acceleration), S (displacement) with θ (angular displacement), velocity (v) with omega (ω). But time remains the same.
$v = u + at$ is similar to ${\omega _f} = {\omega _i} + \alpha t$ .