An electric ceiling fan is rotating about a fixed axis with an initial angular v

Question

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.280 rev/s . The magnitude of the angular acceleration is 0.901 rev/s2 . Both the the angular velocity and angular accleration are directed clockwise. The electric ceiling fan blades form a circle of diameter 0.790 m .

A). Compute the fan's angular velocity magnitude after time 0.209 s has passed. Express your answer numerically in revolutions per second.

_____________ rev/s

B). Through how many revolutions has the blade turned in the time interval 0.209 s from Part A? Express the number of revolutions numerically.

__________________ rev

C). What is the tangential speed vtan(t) of a point on the tip of the blade at time t = 0.209 s ? Express your answer numerically in meters per second.

vtan(t) = ______________ m/s

D). What is the magnitude a of the resultant acceleration of a point on the tip of the blade at time t = 0.209 s ? Express the acceleration numerically in meters per second squared.

a = ___________ m/s2

Explanation / Answer

given data

w1 = 0.28 rev/s = 0.28*2*pi rad/s = 1.76 rad/s

alfa = 0.901 rev/s^2 = 0.901*2*pi rev/s^2 = 5.66 rad/s^2

A) w2 = w1 + alfa*t

= 0.28 + 0.901*0.209

= 0.468 rad/s <<<<<<<<<----------------Answer

B) angular dispalcement, theta = w1*t + 0.5*alfa*t^2

= 0.28*0.209 + 0.5*0.901*0.209^2

= 0.782 rev <<<<<<<<<----------------Answer

C) v_taan = r*w2

= 0.79*0.468*2*pi

= 2.32 m/s <<<<<<<<<----------------Answer

d) a_tan = r*alfa

= 0.79*5.66

= 4.47 m/s^2

a_rad = r*w2^2

= 0.79*(0.468*2*pi)^2

= 6.83 m/s^2

a_total = sqrt(a_tan^2 + a_rad^2)

= sqrt(4.47^2 + 6.83^2)

= 8.16 m/s^2 <<<<<<<<<----------------Answer

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