Question

The angular frequency of a fan increases from 30 rpm to 60 rpm in \[\pi \,{\text{s}}\]. A dust particle is present at a distance of 20 cm from the axis of rotation. The tangential acceleration of the fan in \[\pi \,{\text{s}}\] is
A. \[0.8\,{\text{m}}{{\text{s}}^{ - 2}}\]
B. \[0.34\,{\text{m}}{{\text{s}}^{ - 2}}\]
C. \[0.2\,{\text{m}}{{\text{s}}^{ - 2}}\]
D. \[1.2\,{\text{m}}{{\text{s}}^{ - 2}}\]

Answer 1

Hint: Determine the initial and final angular speed of the fan. Then determine the average angular acceleration of the fan. The tangential acceleration is the product of radius of circular motion and angular acceleration.

Formula used:
Acceleration, \[{\alpha _{avg}} = \dfrac{{{\omega _f} - {\omega _i}}}{t}\]
Here, \[{\omega _f}\] is the final angular velocity, \[{\omega _i}\] is the initial angular velocity and t is the time.
Tangential acceleration, \[{a_t} = r\alpha \], where, r is the radius of the circular motion of the particle.

Complete step by step answer:
We know that an average angular acceleration is equal to the rate of change of angular velocity. We have given the angular frequency of the rotation of the fan blade and we have to determine the angular speed of the fan blade. We have the relation,
\[\omega = \dfrac{{2\pi n}}{t}\]
Here, \[\omega \] is the angular velocity and n is the number of rotations per unit time t.

 Let’s calculate the initial angular velocity of the fan as follows,
\[{\omega _i} = 2\pi \dfrac{{{n_i}}}{t}\]
Substituting 30 rpm for \[\dfrac{{{n_i}}}{t}\] in the above equation, we get,
\[{\omega _i} = 2\pi \left( {\dfrac{{30\,{\text{rev}}}}{{60\,{\text{s}}}}} \right)\]
\[ \Rightarrow {\omega _i} = \pi \,{\text{rad/s}}\]

Let’s calculate the final angular velocity of the fan as follows,
\[{\omega _f} = 2\pi \dfrac{{{n_f}}}{t}\]
Substituting 60 rpm for \[\dfrac{{{n_f}}}{t}\] in the above equation, we get,
\[{\omega _i} = 2\pi \left( {\dfrac{{60\,{\text{rev}}}}{{60\,{\text{s}}}}} \right)\]
\[ \Rightarrow {\omega _f} = 2\pi \,{\text{rad/s}}\]

Now, we can calculate the average acceleration of the fan as follows,
\[{\alpha _{avg}} = \dfrac{{{\omega _f} - {\omega _i}}}{t}\]
Substituting \[{\omega _f} = 2\pi \,{\text{rad/s}}\], \[{\omega _i} = \pi \,{\text{rad/s}}\] and \[t = \pi \,{\text{s}}\] in the above equation, we get,
\[{\alpha _{avg}} = \dfrac{{2\pi - \pi }}{\pi }\]
\[ \Rightarrow {\alpha _{avg}} = 1\,{\text{rad/s}}\]

The tangential acceleration of the fan is given as,
\[{a_t} = r{\alpha _{avg}}\]
Substituting \[r = 20\,{\text{cm}} = 0.2\,{\text{m}}\] and \[\omega = 1\,{\text{rad/s}}\] in the above equation, we get,
\[{a_t} = \left( {0.2} \right)\left( 1 \right)\]
\[ \therefore {a_t} = 0.2\,{\text{m}}{{\text{s}}^{ - 2}}\]

So, the correct answer is option C.

Note: Students must note the difference between radial acceleration and tangential acceleration. The radial acceleration is due to centripetal force and it is given as \[a = \dfrac{{{v^2}}}{r}\], where, v is the tangential velocity of the particle. The tangential acceleration is due to angular velocity of the particle.