Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s
The required value of angular velocity of wheel is 75 rad/s.
Given data:
The angular acceleration of the wheel is, [tex]\alpha (t) = 6.0 t^{2} \;\rm rad/s^{4}[/tex].
The turning rate of wheel is, n = 10 rev.
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)
Clearly, the angular velocity is the single integral of angular acceleration. Then,
ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dt dt
Since,
α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴.
Then solve by substituting the values as,
θ(t) = ∫∫α(t) dt dt
θ(t) = ∫∫6t² dt dt
θ(t) =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
Also,
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³ ......................................................(1)
Substitute the value of time in equation (1) as,
ω( t = 3.348) = 2(3.348)³ = 75 rad/s
Thus, we can conclude that the required value of angular velocity of wheel is 75 rad/s.
Learn more about the angular velocity here:
https://brainly.com/question/17592191
