Compute the angular velocity after a time of 0.190s .?

Question

A) An electric ceiling fan with blades of diameter 0.800 m is rotating about a fixed axis with an initial angular velocity of 0.300 rev/s. The angular acceleration is 0.914 rev/s^2.Compute the angular velocity after a time of 0.190 .
B) Through how many revolutions has the blade turned in this time interval?
C) What is the tangential speed of a point on the tip of the blade at time T= 0.190 s?
D) What is the magnitude of the resultant acceleration of a point on the tip of the blade at time t = 0.190s ?

Answer 1

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A) Radius of fan blade=r=0.400 m

The initial angular velocity of fan=wi=0.300 rev/s=0.300*2pi radian/s.

The angular acceleration = a! = 0.914 rev/s^2.=0.914*2pi radian/s^2

time 't'=0.190 s

angular velocity after 0.190 s=wf=?

wf=wi+a!t

wf=0. 300*2pi +.0.914*2pi *0.19

wf= 0.47366*2pi rad/s=2.976 rad/s

wf=0.47366 rev./s

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B) number of revolutions=N=angular displacement O/2pi

angular displacement O=[(wf+wi)/2 ]t=[0.3868]*2pi*0.19=0.0735*2pi radian

number of revolutions=N=O/2pi=0.0735

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C) tangential speed v=r wf =0.4*2.976=1.1904 m/s
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D) resultant acceleration is vector sum of centripetal acceleration and tangential acceleration

centripetal acceleration= Ac =r(wf)^2=3.5426 m/s^2

tangential acceleration= At=r*a!=0.4*5.7426=2.297 m/s^2

resultant acceleration = sq rt [Ac^2+At^2]=sq rt 17.8265=4.222 m/s^2

resultant acceleration of a point on the tip at 0.19 s is 4.222 m/s^2
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Answer 2

LOL

a) Angular velocity after time t is .
w= wo + At
w=0.30 + 0.914 x 0.190=
w=0.474 rad/s

b) Angular displacement
Theta= (1/2) At^2
Theta= (1/2) x 0.914 x (0.19)^2
Theta= 0.0165 rad

c) The tangential velocity is
V=w r
V= 0.474 x 0.800= 0.379 m/s

d) a= sqrt([a(tangential)] ^2 + [a(centripetal)]^2)
a(tangential) = A r = 0.914 x 0.8=0.73 m/s^2
a(centripetal)= V^2/r = (0.379)^2/r=0.180 m/s^2

a=sqrt((0.73 )^2 + (0.18)^2)
a=0.752 m/s^2

Have fun!