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