Question

. A ventilation fan has blades 0.25 m long rotating at 20 rpm (revolutions per minute). What is the centripetal acceleration of a point on the outer tip of a blade

Answer 1

Answer:

The centripetal acceleration of a point on the outer tip of a blade is 1.097 m/s²

Explanation:

Given;

length of the fan blade, r = 0.25 m

angular speed = 20 rpm

The angular speed in rad/s is given as;

$\omega = \frac{20 \ rev}{ \ \ \min} \ \times \ \frac{2\pi}{1 \ rev} \ \times \ \frac{1 \ \min}{60s} = 2.095 \ rad/s$

The centripetal acceleration of a point on the outer tip of a blade is given as;

$\alpha_c = \frac{v^2}{r} = \omega ^2r$

$\alpha _c = (2.095)^2(0.25)\\\\\alpha _c =1.097 \ m/s^2$

Therefore, the centripetal acceleration of a point on the outer tip of a blade is 1.097 m/s²

Answer 2

Answer:

• The centripetal acceleration = $1.1m/s^2$

Explanation:

Given

$rpm = 20$

therefore,

$20rpm = \frac{20*2\pi}{60}\\\\=2.1 rad/s = w$

From,

$V = rw\\\\V = 0.25 * 2.1\\\\V = 0.525m/s$

Centripetal acceleration,

$a_c = \frac{V^2}{r}\\\\a_c = \frac{0.525^2}{0.25}\\\\a_c = 1.1m/s^2$

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