Respuesta :
Answer:
(a) Angular acceleration will be [tex]\alpha =18.66rev/min^2[/tex]
(B) Final angular velocity will be 28 rev/min
Explanation:
We have given time t = 1.5 min
Angular displacement [tex]\Theta =21rev[/tex]
(a) Initial angular speed [tex]\omega _i=0rad/sec[/tex]
From second equation of motion we know that [tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2[/tex]
So [tex]21 =0\times 1.5+\frac{1}{2}\times \alpha\times 1.5^2[/tex]
[tex]42 =\alpha\times 1.5^2[/tex]
[tex]\alpha =18.66rev/min^2[/tex]
(b) Now from first equation of motion
[tex]\omega _f=\omega _i+\alpha t=0+18.66\times 1.5=28rev/min[/tex]
Answer:
(a) 0.0326 rad/s²
(b) 2.93 rad/s
Explanation:
number of revolutions, n = 21
time taken , t = 1.5 minutes = 90 seconds
Angle turned, θ = 2 x π x n = 2 x 3.14 x 21 = 131.88 rad
Let α be the angular acceleration.
initial angular velocity, ωo = 0 rad/s
Use second equation of motion
[tex]\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}[/tex]
131.88 = 0 + 0.5 x α x 90 x 90
α = 0.0326 rad/s²
(b) Let the final angular speed is ω
Use first equation of motion
ω = ωo + αt
ω = 0 + 0.0326 x 90 = 2.93 rad/s
