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Question: A wheel is rotating at the rate of 33 revolution / min. if it comes to a stop in 20 seconds. Then th...

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) 11π\pi rad/s2{s^2}
C) π200rad/s2\dfrac{\pi }{{200}}\,\,{\text{rad/}}{{\text{s}}^2}
D) 11π200rad/s2\dfrac{{11\pi }}{{200}}\,\,{\text{rad/}}{{\text{s}}^2}

Explanation

Solution

Hint Since the wheel is going to stop, hence final angular acceleration = 0. Now use the equation ωf=ωi+αt{\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, (ωi{\omega _i} )= 33 rev/min
The Final angular velocity of the wheel, (ωf{\omega _f}) = 0
Time taken by wheel to stop (t) = 20 second = 20s .
We know that,
The initial angular velocity in rad/s (ωi{\omega _i}) is given by =
33×2π60 1.1πrad/s  \Rightarrow 33 \times \dfrac{{2\pi }}{{60}} \\\ \Rightarrow 1.1\pi \,{\text{rad/s}} \\\ (2π60\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. E.g. $v = u + at$ is similar to ${\omega _f} = {\omega _i} + \alpha t$ .