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 0.300 rev/s . The magnitude of the angular acceleration is 0.886 rev/s2 . Both the the angular velocity and angular accleration are directed clockwise. The electric ceiling fan blades form a circle of diameter 0.770 m .

Part A

Compute the fan's angular velocity magnitude after time 0.198 s has passed.

Express your answer numerically in revolutions per second.

2.6

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Part B

Through how many revolutions has the blade turned in the time interval 0.198 s from Part A?

Express the number of revolutions numerically.

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Part C

What is the tangential speed vtan(t) of a point on the tip of the blade at time t = 0.198 s ?

Express your answer numerically in meters per second.

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Part D

What is the magnitude a of the resultant acceleration of a point on the tip of the blade at time t = 0.198 s ?

Express the acceleration numerically in meters per second squared.

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude of the angular acceleration is 0.886 rev/s2 . Both the the angular velocity and angular accleration are directed clockwise. The electric ceiling fan blades form a circle of diameter 0.770 m .

Part A

Compute the fan's angular velocity magnitude after time 0.198 s has passed.

Express your answer numerically in revolutions per second.

2.6

  rev/s  

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Part B

Through how many revolutions has the blade turned in the time interval 0.198 s from Part A?

Express the number of revolutions numerically.

  rev  

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Part C

What is the tangential speed vtan(t) of a point on the tip of the blade at time t = 0.198 s ?

Express your answer numerically in meters per second.

vtan(t) =   m/s  

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Part D

What is the magnitude a of the resultant acceleration of a point on the tip of the blade at time t = 0.198 s ?

Express the acceleration numerically in meters per second squared.

a =   m/s2  

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Explanation / Answer

A) w2 = w1 + alfa*t

= 0.3 + 0.886*0.198

= 0.475 rev/s

B) Use kinematic equation, theta = w1*t + 0.5*alfa*t^2

= 0.3*0.198 + 0.5*0.886*0.198^2

= 0.0768 rev

C)

w2 = 0.475 rev/s

= 0.475*2*pi rad/s

= 2.987 rad/s

tangential speed, v_tan = r*w2

= (0.77/2)*2.987

= 1.15 m/s

d) a_tan = r*alfa

= (0.77/2)*0.886*2*pi

= 2.14 m/s^2

a_rad = v2^2/r

= 1.15^2/(0.77/2)

= 3.44 m/s^2

magnitude a of the resultant acceleration = sqrt(a_tan^2 + a_rad^2)

= sqrt(2.14^2 + 3.44^2)

= 4.05 m/s^2

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