A ceiling fan has diameter (the circle through the outer edge of the blade) of 120cm and 1500rpm at full speed. Consider a particle of mass 1 g stick at the outer edge of the blade. How much force does it experience when the fun runs at full speed?

Hint: We need to understand the relation between the distance of the particle from the center of rotation, the mass of the particle and the rotational speed of the fan to find the force acting on the particle situated on the fan blade while it is rotating.

Complete answer:
We know that any rotating body has a torque acting at each point from the center of rotation, which is dependent on the distance from the center. The force acting on a particle of mass ‘m’ at a distance of ‘r’ from the center of rotation rotating with an angular velocity of \[\omega \] can be given as –
\[F=m{{\omega }^{2}}r\]
We are given a ceiling fan which has a diameter of 120 cm. It is said that it is rotating with a frequency of 1500 rotations per minute at its top speed. We can find the angular velocity of the fan with these data as –
\[\begin{align}
  & f=\dfrac{1500}{60}rps=25rps \\
 & \Rightarrow \omega =2\pi 25rad{{s}^{-1}} \\
 & \therefore \omega =50\pi rad{{s}^{-1}} \\
\end{align}\]
Now, we can find the force acting on the particle due to this rotation of the ceiling fan with the formula given above as –
\[\begin{align}
  & F=m{{\omega }^{2}}r \\
 & \Rightarrow F={{10}^{-3}}kg\times {{(50\pi )}^{2}}\times 0.6 \\
 & \therefore F=14.8N \\
\end{align}\]
The force acting on the particle of 1g placed at the edge of the ceiling fan which rotates at 1500rpm is given as 14.8N.
This is the required solution.

Note:
The force acting on a body due to the rotation is dependent on the distance from the center of rotation. At the center of the ceiling fan, if the same object Is placed, then it experiences no force as the term ‘r’ in the formula reduces the force to zero.