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?

Answer 1

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

Answer 2

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

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

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

$= 3.06m/s^2$