Question

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.200 rev/s . The magnitude of the angular acceleration is 0.883 rev/s2 . Both the the angular velocity and angular accleration are directed counterclockwise. The electric ceiling fan blades form a circle of diameter 0.760 m .

A. Compute the fan's angular velocity magnitude after time 0.204 ss has passed.

Express your answer numerically in revolutions per second.

B. Through how many revolutions has the blade turned in the time interval 0.204 ss from Part A?

Express the number of revolutions numerically.

C. What is the tangential speed vtvt of a point on the tip of the blade at time ttt = 0.204 ss ?

Express your answer numerically in meters per second.

D. Calculate the magnitude atat of the tangential acceleration of a point on the tip of the blade at time ttt = 0.204 ss .

Express the acceleration numerically in meters per second squared.

Answer 1

Given Data

Initial angular velocity, ω₀ = 0.200 rev/s

Angular acceleration, α = 0.883 rev/s²

Time, t = 0.204 s

Diameter of the circle, d = 0.760 m

Radius of the circle is       r = d/2

= 0.760 m / 2

= 0.38 m

Solution : -

a.    Using the rotational kinematic relation, we have

                    ω = ω₀+ αt

                         = (0.200 rev/s)+(0.883 rev/s² )(0.204 s)

                         = 0.38 rev/s --> Answer

b.     Using the rotational kinematic relation, we have

                           θ = ω₀t+(1/2)αt²

                              = (0.200 rev/s)(0.204 s) + (1/2)(0.883 rev/s² )(0.204 s)²

                              = 0.059 revolutions --> Answer

c.   The final angular velocity is

                 ω = 0.38 rev/s   

                      = (0.38 rev/s)(2π rad/s/1 rev/s)

                      = 2.387 rad/s
      The tangential speed is

                                          vt = rω

                                             = (0.38 m)(2.387 rad/s)

                                             = 0.907 m/s --> Answer

d. The angular acceleration is

                 α  = 0.883 rev/s²   

                      = (0.883 rev/s²)(2π rad/s²/1 rev/s²)

                      = 5.5452 rad/s²       

The tangential acceleration is

                                           a_t= rα

                                               = (0.38 m)(5.5452 rad/s²)

                                              = 2.107 m/s² --> Answer