A fan is rotating at 90 rpm. It is then switched OFF. It stops after 21 rotations. Calculate the time taken by it to stop assuming that the frictional torque is constant.

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> [!TLDR]
> Constant frictional torque → uniform angular deceleration, so the rotational kinematical equations of Table 1.1 give the stopping time from $\omega_0=3\pi$ rad/s and $	heta=42\pi$ rad.

A fan rotating at 90 rpm (i.e. $n_0=1.5$ rps) is switched OFF and stops after completing 21 rotations, under a constant (frictional) retarding torque, so a constant angular deceleration applies -- meaning the rotational kinematics equations of Table 1.1 are valid here. The total angle turned through while stopping is $	heta=2\pi N=2\pi(21)=42\pi$ rad, with final angular velocity $\omega=0$. Using $\omega^2=\omega_0^2+2\alpha	heta$ with $\omega_0=2\pi(1.5)=3\pi$ rad/s first gives the (negative) angular deceleration $\alpha$, and then $\omega=\omega_0+\alpha t$ (with $\omega=0$) gives the stopping time, which works out to 28 s. The example's remark notes that the angle could equally have been measured direct …

Q.A fan is rotating at 90 rpm. It is then switched OFF. It stops after 21 rotations. Calculate the time taken by it to stop assuming that the frictional torque is constant.

Concept understanding — Kinematics of Circular Motion

Q.A fan is rotating at 90 rpm. It is then switched OFF. It stops after 21 rotations. Calculate the time taken by it to stop assuming that the frictional torque is constant.

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