Question

A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1 g sticking at the outer end of a blade.
A. How much force does it experience when the fan runs at full speed?
B. Who exerts this force on a particle?
C. How much force does the particle exert on the blade along its surface?

Answer 1

Hint: The force experienced by the particle would be the centrifugal force. Use Newton’s third law for answering the third part.

Complete answer:
Given-
Diameter of the circle through the outer edges of blades = 120 cm
Rpm = $f = 1500$
Mass of the particle = 1 g

A. The centrifugal force is the outward throwing force which arises as a complement of centripetal force, the force required to keep a body in circular motion.
The formula for centrifugal force = $mr{\omega ^2}$
m = 1 g = 0.001 kgs,
$\omega = 2\pi f = 2 \times 3.14 \times \left( {\dfrac{{1500}}{{60}}rps} \right) = 157$,
r = 120/2 = 60 cm = 0.6 m
$F = 0.001 \times 0.6 \times {\left( {157} \right)^2} = 14.78 = 14.8N$
The force experienced by the particle will be 14.8 Newton.

B. It’s the circular motion of the fan which exerts this outward throwing centrifugal force on the particle.
C. The particle exerts an equal and opposite force of 14.8 Newton the blades of the fan.

Note:
The centrifugal force acts on every particle in circular motion. The inertia of the particle is also responsible for this. It is the fan which exerts this force on the particle.