Question

An electric fan has blades of length 60 cm as measured from the axis of rotation. If the fan is rotating at 1200 rpm, the acceleration of a point on the tip of the blade is about:

a.4800 m /s2
b. 9600 m/s2
c. 3600 m/s2
d. 5000 m/s2​

Answer 1

Given,

Length of the blade = 60 cm

Rotation amount = 1200 rpm

To find,

The acceleration of the tip of the fan blade.

Solution,

To solve this mathematical problem, we have to calculate the frequency of the rotation.

Frequency (f) = Total rotation/Total time = 1200/60 = 20hz

(1 minute = 60 seconds)

Now, we have to calculate the centripetal acceleration of the tip of the fan blade by using the following mathematical formula.

Acceleration :

= ω²R

= (2πf)²×R

= (2× 22/7 × 20)² × 0.6 [60cm = 0.6m]

= 9481.80 m/s²

Now, among the given options, the nearest value is 9600 m/s².

Hence, the acceleration will be 9600 m/s².

Answer 2

Given:

An electric fan has blades of length 60 cm as measured from the axis of rotation. The fan is rotating at 1200 rpm.

To find:

Acceleration of tip of blade in fan?

Calculation:

First of all , leta convert the unit of angular velocity to radian/sec.

$\therefore \: 1200 \: rpm = 1200 \times \dfrac{2\pi}{60} = 40\pi \: rad /s$

Now , the centripetal acceleration experienced at tip of fan blade :

$\therefore \: a_{c} = \dfrac{ {v}^{2} }{r}$

$\implies \: a_{c} = {(\dfrac{ v}{r})}^{2} \times r$

$\implies \: a_{c} = { \omega}^{2} \times r$

$\implies \: a_{c} = {(40\pi)}^{2} \times \dfrac{60}{100}$

$\implies \: a_{c} = 1600 \times {\pi}^{2} \times \dfrac{6}{10}$

$\implies \: a_{c} = 960 \times {\pi}^{2}$

Considering π² $\approx$ 10 :

$\implies \: a_{c} = 960 \times 10$

$\implies \: a_{c} = 9600 \: m/ {s}^{2}$

So, acceleration at tip of fan is 9600 m/s².