A wheel starts from rest and has an angular acceleration that is given by α (t) = (6.0 rad/s4)t2. After it has turned through 10 rev its angular velocity is:

The required value of angular velocity of wheel is 75 rad/s.

Given data:

The angular acceleration of the wheel is, .

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.

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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

d eras and periods