Question

A rotating fan completes 1160 revolutions every minute. Consider the tip of a blade, at a radius of 19.0 cm. (a) Through what distance does the tip move in one revolution? What are (b) the tip's speed and (c) the magnitude of its acceleration? (d) What is the period of the motion?

Answer 1

For the rotating fan that completes 1160 revolutions every minute, with a tip's radius of 19.0 cm, we have:

a) The tip will move a distance of 1.19 m in one revolution.

b) The tip's speed is 23.08 m/s.

c) The magnitude of the tip's acceleration is 2803.6 m/s².

d) The period of the motion is 0.052 s.  

a) The distance to which the tip move in one revolution is given by the circumference of the circle:

$d = 2 \pi r$

Where:

r: is the radius of the blade = 19.0 cm

$d = 2 \pi r = 2\pi*0.19 m = 1.19 m$                                  

Hence, the tip will move a distance of 1.19 m in one revolution.

b) The speed of the tip can be calculated with the following equation:

$v = \omega r$

Where:

ω: is the angular speed = 1160 rev/min

Then, the tip's speed is:

$v = \omega r = 1160 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.19 m = 23.08 m/s$

c) The acceleration of the tip is given by:

$a = \frac{v^{2}}{r} = \frac{(23.08 m/s)^{2}}{0.19 m} = 2803.6 m/s^{2}$

Therefore, the magnitude of the acceleration is 2803.6 m/s².

d) The period of the motion is:

$T = \frac{1}{f} = \frac{2\pi}{\omega}$

Where:

f: is the frequency

$T = \frac{2\pi}{\omega} = \frac{2\pi}{1160 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}} = 0.052 s$  

Hence, the period of the motion is 0.052 s.

You can find more about the period and angular acceleration here: brainly.com/question/9708010?referrer=searchResults

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