The angular position of a flywheel is given by theta=(2+5t+2t^2)rad . find the angular speed and acceleration at 3^rd second after the motion had been started.

To find the angular speed and acceleration at the 3rd second, we need to differentiate the given equation for angular position (θ) with respect to time (t).

Given: θ = 2 + 5t + 2t^2

1. Differentiate θ with respect to t to find the angular speed (ω):
ω = dθ/dt

Differentiating θ = 2 + 5t + 2t^2:
ω = d(2 + 5t + 2t^2)/dt

ω = 5 + 4t

The angular speed (ω) is given by ω = 5 + 4t rad/s.

2. To find the angular acceleration (α), differentiate ω with respect to t:
α = dω/dt

Differentiating ω = 5 + 4t:
α = d(5 + 4t)/dt

α = 4

The angular acceleration (α) is constant and given by α = 4 rad/s^2.

Therefore, at the 3rd second after the motion started:
Angular speed (ω) = 5 + 4t = 5 + 4(3) = 17 rad/s
Angular acceleration (α) = 4 rad/s^2

To find the angular speed and acceleration at the 3rd-second mark, we need to differentiate the given angular position equation with respect to time.

Given: θ(t) = 2 + 5t + 2t^2 (where θ represents the angular position and t represents time)

To find the angular speed, we need to differentiate θ(t) with respect to time (t).

Differentiating θ(t) will give us the angular velocity (ω):

ω(t) = dθ(t)/dt

Applying the power rule of differentiation, we can differentiate each term of θ(t) one by one:

d(2)/dt = 0 (since 2 is a constant, its derivative is zero)
d(5t)/dt = 5
d(2t^2)/dt = 4t (using the power rule: nx^(n-1))

Therefore, the derivative of θ(t) with respect to t is:

ω(t) = 5 + 4t

To find the angular acceleration, we now need to differentiate the angular velocity (ω(t)) equation with respect to time (t):

a(t) = dω(t)/dt

Differentiating ω(t) will give us the angular acceleration (α):

a(t) = d(5 + 4t)/dt

Applying the power rule of differentiation:

d(5)/dt = 0 (since 5 is a constant, its derivative is zero)
d(4t)/dt = 4

Therefore, the derivative of ω(t) with respect to t is:

a(t) = 4

To find the angular speed at the 3rd-second mark, substitute t = 3 into the ω(t) equation:

ω(3) = 5 + 4(3) = 5 + 12 = 17 rad/s

To find the angular acceleration at the 3rd-second mark, substitute t = 3 into the a(t) equation:

a(3) = 4

Thus, the angular speed at the 3rd-second mark is 17 rad/s, and the angular acceleration is 4 rad/s^2.