
The radius of the blade of a fan is \[0.3m\]. It is making \[1200\text{ rev/min}\] . The acceleration of a particle at the tip of the blade is:
(A) \[\text{3733 m/}{{\text{s}}^{\text{2}}}\]
(B) \[\text{2733 m/}{{\text{s}}^{\text{2}}}\]
(C) \[\text{4733 m/}{{\text{s}}^{\text{2}}}\]
(D) \[\text{5733 m/}{{\text{s}}^{\text{2}}}\]
Answer
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Hint: Since the particle at the tip of the blade is exhibiting circular motion, it will experience centripetal acceleration. The angular acceleration of a particle depends on the angular velocity of the body and its radius (which has been specified in the question). We can find the angular velocity of the particle from its frequency of rotation which has been provided to us.
Formula Used:
\[a={{\omega }^{2}}r\]
Complete step by step solution:
We have been giving the frequency of the blade of the fan as \[f=1200\text{ rev/min}\]
But this is in revolutions per minute and our calculations are done in revolutions per seconds
Converting the frequency into revolutions per second, we get
\[\begin{align}
& f=\dfrac{1200}{60}\text{ rev/sec}\left[ \because 1\min =60\sec \right] \\
& \Rightarrow f=20\text{ rev/sec} \\
\end{align}\]
The acceleration of the particle will be centripetal acceleration (as discussed in the hint)
Centripetal acceleration for the particle \[(a)={{\omega }^{2}}r\] where \[\omega \] is the angular velocity of the particle at the tip of the blade and \[r\] is the radius of the blade of the fan
The angular velocity of the particle at the tip can be calculated from the frequency of oscillations as
\[\omega =2\pi f\]
Substituting the value of frequency, we get \[\omega =2\times 3.14\times 20=125.6{{s}^{-2}}\]
Substituting the value of angular velocity in the expression for centripetal acceleration, we get
\[\begin{align}
& a={{(125.6)}^{2}}\times 0.3=4732.6m/{{s}^{2}} \\
& \Rightarrow a=4733m/{{s}^{2}} \\
\end{align}\]
Hence we can say that option (C) is the correct answer.
Note: Students make a common error of considering the frequency of rotations made by the blade as the angular velocity of the blade and this leads them to wrong answers. You have to be clear about the quantities given to you and the quantities needed to reach the answer. The units must be taken care of. We were provided with the frequency expressed in revolutions per minute but we need to express it in revolutions per second. These little things make all the difference in a solution.
Formula Used:
\[a={{\omega }^{2}}r\]
Complete step by step solution:
We have been giving the frequency of the blade of the fan as \[f=1200\text{ rev/min}\]
But this is in revolutions per minute and our calculations are done in revolutions per seconds
Converting the frequency into revolutions per second, we get
\[\begin{align}
& f=\dfrac{1200}{60}\text{ rev/sec}\left[ \because 1\min =60\sec \right] \\
& \Rightarrow f=20\text{ rev/sec} \\
\end{align}\]
The acceleration of the particle will be centripetal acceleration (as discussed in the hint)
Centripetal acceleration for the particle \[(a)={{\omega }^{2}}r\] where \[\omega \] is the angular velocity of the particle at the tip of the blade and \[r\] is the radius of the blade of the fan
The angular velocity of the particle at the tip can be calculated from the frequency of oscillations as
\[\omega =2\pi f\]
Substituting the value of frequency, we get \[\omega =2\times 3.14\times 20=125.6{{s}^{-2}}\]
Substituting the value of angular velocity in the expression for centripetal acceleration, we get
\[\begin{align}
& a={{(125.6)}^{2}}\times 0.3=4732.6m/{{s}^{2}} \\
& \Rightarrow a=4733m/{{s}^{2}} \\
\end{align}\]
Hence we can say that option (C) is the correct answer.
Note: Students make a common error of considering the frequency of rotations made by the blade as the angular velocity of the blade and this leads them to wrong answers. You have to be clear about the quantities given to you and the quantities needed to reach the answer. The units must be taken care of. We were provided with the frequency expressed in revolutions per minute but we need to express it in revolutions per second. These little things make all the difference in a solution.
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