A rotating fan completes 1200 revolutions every minute. Consider a point on the

Question

A rotating fan completes 1200 revolutions every minute. Consider a point on the tip of a blade at a radius of 0.16 m. (A) through what distance does the point move in one revolution in m? (B) what is the speed of the point in m/s? (C) what is te magnitude of its acceleration in m/s squared? (D) what is the period of the motion in s? A rotating fan completes 1200 revolutions every minute. Consider a point on the tip of a blade at a radius of 0.16 m. (A) through what distance does the point move in one revolution in m? (B) what is the speed of the point in m/s? (C) what is te magnitude of its acceleration in m/s squared? (D) what is the period of the motion in s? A rotating fan completes 1200 revolutions every minute. Consider a point on the tip of a blade at a radius of 0.16 m. (A) through what distance does the point move in one revolution in m? (B) what is the speed of the point in m/s? (C) what is te magnitude of its acceleration in m/s squared? (D) what is the period of the motion in s?

Explanation / Answer

1) what distance per revolution

d = 2 x pi x r = 2 x 3.14 x .16 m = 1.0048 m

2) what is the speed

v = 2 x pi x w x r = d x w

1200 rev / min x 1.0048 m / rev = 1205.76 m/min = 20.096 m/s

3) acceleration is due to change in velocity vector. ie tip is not traveling in a straight line so a force must be acting on it. centripetal force in this case. Since F=ma, acceleration is non zero and is found by the following equation...

a = w^2 x r = (v/r)^2 x r = (20.096m/s / .16 m )^2 x (.16 m)
= (125.6 /s)^2 x .14 m = 2524.057 m/s^2

4) period of motion

period = time per revolution = 1 / (1200 rev/min) x (60 sec/min) = .05 sec

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