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Question: The angular frequency of a fan increases from \(30rpm\) to \(60rpm\) in \(\pi s.\) A dust particle i...

The angular frequency of a fan increases from 30rpm30rpm to 60rpm60rpm in πs.\pi s. A dust particle is present at a distance of 20cm20cm from the axis of rotation. The tangential acceleration of the particle is?
(a) 0.8 ms20.8\text{ m}{{\text{s}}^{-2}}
(b) 0.34 ms20.34\text{ m}{{\text{s}}^{-2}}
(c) 0.2 ms20.2\text{ m}{{\text{s}}^{-2}}
(d) 1.2 ms21.2\text{ m}{{\text{s}}^{-2}}

Explanation

Solution

Angular frequency ww is a scalar measure of rotation rate. It refers to the angular displacement per unit time or rate of change of the phase of a sinusoidal waveform.
Acceleration is a rate of change of velocity.

Formula used:
Angular speed is given by w=2πb,60w=\dfrac{2\pi b,}{60{}^\circ }
wi=2πt160(rad/sec){{w}_{i}}=\dfrac{2\pi {{t}_{1}}}{60{}^\circ }\text{(rad/sec)}
Angular acceleration:-
α=wfwit\alpha =\dfrac{wf-wi}{t}
Complete solution:
Given:-
N(speed)(30 to 60)rpm.-\left( 30\text{ to 60} \right)rpm.
w=20cm.w=20cm.
\therefore angular speed w=2πb160rad/secw=\dfrac{2\pi {{b}_{1}}}{60}\text{rad/sec}
Angular acceleration
α=wfwitrad/sec2\alpha =\dfrac{{{w}_{f}}-{{w}_{i}}}{t}\text{rad/se}{{\text{c}}^{2}}
wi=2π3060=2πrad/sec{{w}_{i}}=2\pi \dfrac{30}{60}=2\pi \text{rad/sec}
wf=2π6060=2πrad/sec{{w}_{f}}=2\pi \dfrac{60}{60}=2\pi \text{rad/sec}
α=2ππ314\alpha =\dfrac{2\pi -\pi }{314}
α=1rad/sec2\alpha =1\text{rad/se}{{\text{c}}^{2}}

Now the conclusion option (c) is correct.

Additional information:
All the equations Newtonian mechanisms are based on Newton’s Law of Motion. All equations are a derivative and all theories are governed by them.

Note:
The value of tangential acceleration of a particle is given by tangential acceleration formula by calculating angular acceleration.