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 . The magnitude of the angular acceleration is 0.912 . Both the the angular velocity and angular accleration vectors are in the same direction. The electric ceiling fan blades form a circle of diameter 0.710 .
Compute the fan's angular velocity magnitude after time 0.198 has passed.
Express your answer numerically in revolutions per second.

Compute the fan's angular velocity magnitude after time 0.198 has passed.
Express your answer numerically in revolutions per second.

Through how many revolutions has the blade turned in the time interval 0.198?
Express the number of revolutions numerically.

What is the tangential speed of a point on the tip of the blade at time = 0.198 ?
Express your answer numerically in meters per second.

What is the magnitude of the resultant acceleration of a point on the tip of the blade at time = 0.198 ?
Express the acceleration numerically in meters per second squared.

Explanation / Answer

           The initial angular velocity, 0 = 0.280 rev/s       The angular acceleration, = 0.912 rev/s2       The time, t = 0.198 s       The diameter of the circle, d = 0.710 m       The radius of the circle is                                               r = d/2                                                  = 0.710 m/2                                                 = 0.355 m __________________________________________________________       The time, t = 0.198 s       The diameter of the circle, d = 0.710 m       The radius of the circle is                                               r = d/2                                                  = 0.710 m/2                                                 = 0.355 m __________________________________________________________ a.  
      Using the rotational kinematic relation, we have                      = 0+ t                          = (0.280 rev/s)+(0.912 rev/s2 )(0.198 s)                          = 0.460 rev/s __________________________________________________________ __________________________________________________________ b.           Using the rotational kinematic relation, we have                             = 0t+(1/2)t2                               = (0.280 rev/s)(0.198 s)+(1/2)(0.912 rev/s2 )(0.198 s)2                               = 0.0733 revolutions _____________________________________________________________ _____________________________________________________________                               = 0.0733 revolutions _____________________________________________________________ _____________________________________________________________ c.          The final angular velocity is                   = 0.460 rev/s                          = (0.460 rev/s)(2 rad/s/1 rev/s)                       = 2.8888 rad/s
      The tangential speed is                                           v = r                                              = (0.355 m)(2.8888 rad/s)                                              = 1.025 m/s ____________________________________________________________ ____________________________________________________________                                              = (0.355 m)(2.8888 rad/s)                                              = 1.025 m/s ____________________________________________________________ ____________________________________________________________ d.       The angular acceleration is                    = 0.912 rev/s2                          = (0.912 rev/s2)(2 rad/s2/1 rev/s2)                       = 5.72736 rad/s2       The tangential acceleration is                                            at = r                                                = (0.355 m)(5.72736 rad/s2)                                               = 2.033 m/s2 Note:        The radial acceleration is Note:        The radial acceleration is                                        ac= r2                                             = (0.355 m)(2.8888 rad/s)2                                             = 2.96 m/s2
       The resultant acceleration is                                          a = (at2+ac2)                                             = (2.0332 m/s2+2.962 m/s2)                                             = 3.59 m/s2                                            = 3.6 m/s2                                             = (0.355 m)(2.8888 rad/s)2                                             = 2.96 m/s2
       The resultant acceleration is                                          a = (at2+ac2)                                             = (2.0332 m/s2+2.962 m/s2)                                             = 3.59 m/s2                                            = 3.6 m/s2                                            = 3.6 m/s2

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