An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.210 rev/s. The magnitude of the angul

Question

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.210 rev/s. The magnitude of the angular acceleration is 0.900 rev/s^2. Both the angular velocity and angular acceleration are directed counterclockwise. The electric ceiling fan blades form a circle of diameter 0.750 m.
(a) Compute the fan’s angular velocity magnitude after time 0.194s has passed.
(b) Through how many revolutions has the blade turned in the time interval 0.194s from Part (a)?
(c) What is the tangential speed v.tan(t) of a point on the tip of the blade at time t = 0.194s?
(d) What is the magnitude ‘a’ of the resultant acceleration of a point on the tip of the blade at time t = 0.194s?

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Hưng Khoa 5 years 2021-08-16T00:20:45+00:00 2 Answers 308 views 0

Answers ( )

  1. Nick
    0
    2021-08-16T00:22:23+00:00
    Reply

    Answer:

    a) 2.42rad/s

    b) 0.09rev

    c) 0.91m/s

    d) 3.06m/s^2 ;

    Explanation:

    We are given:

    w_0 = 0.210rev/s ;

    (converting to rad we have)

    wo = 0.210rev/s (2πrad/1rev) = 21π/50

    a = 0.900rev/s^2 ==> 9π/5

    d = 0.750m

    a) Let’s use the equation:

    w = w_0 +at

    Substituting figures in the equation:

    w = 21π/50 + (9π/5) (0.194)

    w = 2.42rad/s

    b) we use:

    w_0 t + 1/2 at^2 ;

    = (2.42) (0.194) + 1/2 (9π/5) (0.194)^2

    =0.5752rad

    We need to convert to rev

    = 0.5752rad (1rev/2rad)

    = 0.09rev

    c) t= 0.194s

    V = wr

    But r = d/2

    Therefore,

    V= 2.42 × (0.750/2)

    V= 0.91m/s

    d) t= 0.194s

    a_t_a_n = ar ;

    = (9π/5) (0.750/2)

    =2.121m/s^2 ;

    a_r_a_d = w^2 r ;

    = (2.42)^2 (0.750/2)= 2.2m/s^2 ;

    Therefore

    a= \sqrt*{(a^2_t_a_n) + (a^2_r_a_d)} ;

    =\sqrt*{ (2.121)^2 + (2.2)^2} ;

    = 3.06m/s^2

  2. thucuc
    0
    2021-08-16T00:22:40+00:00
    Reply

    Answer:

    Explanation:

    Given that

    Initial velocity wo=0.210rev/s

    Then, 1rev=2πrad

    wo=0.21×2πrad/s

    wo=0.42π rad/s

    Given angular acceleration of 0.9rev/s²

    α=0.9×2πrad/s²

    α=1.8π rad/s²

    Diameter of blade

    d=0.75m,

    Radius=diameter/2

    r=0.75/2=0.375m

    a. Angular velocity after t=0.194s

    Using equation of angular motion

    wf=wo+αt

    wf=0.42π+ 1.8π×0.194

    wf= 0.42π + 0.3492π

    wf=1.319+1.097

    wf= 2.42rad/s

    If we want the answer in revolution

    1rev=2πrad

    wf= 2.42/2π rev/s

    wf=0.385 rev/s

    b. Revolution traveled in 0.194s

    Using angular motion equation

    θf – θi = wo•t + ½ αt²

    θf – 0= 0.42π•0.194 + ½ × 1.8π•0.194²

    θf = 0.256 + 0.106

    θf = 0.362rad

    Now, to revolution

    1rev=2πrad

    θf=0.362/2π=0.0577rev

    Approximately θf= 0.058rev

    c. Tangential speed? At time 0.194s

    Vt=?

    w=2.42rad/s at t=0.194s

    Using circular motion formulae, relationship between linear velocity and angular velocity

    V=wr

    Vt=wr

    Vt= 2.42×0.375

    Vt=0.9075 m/s

    Vt≈0.91m/s

    d. Magnitude of resultant acceleration

    Tangential Acceleration is given as

    at=αr

    at=1.8π× 0.375

    at=2.12rad/s²

    Now, radial acceleration is given as

    ar=w²r

    ar=2.42²×0.375

    ar=2.196 m/s²

    Then, the magnitude is

    a=√ar²+at²

    a=√2.196²+2.12²

    a=√9.3171

    a=3.052m/s²

    a≈ 3.05m/s²

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