Angular displacement (θ) of a flywheel varies with time as θ = at + bt² + ct³ then angular acceleration is given by

To find the angular acceleration of the flywheel, we need to understand the relationship between angular displacement ($\theta$), angular velocity ($\omega$), and angular acceleration ($\alpha$).

1. **Angular Displacement ($\theta$)**:
The angular displacement as a function of time $t$ is given by the equation:
$\theta =at+b{t}^2+c{t}^3$

2. **Angular Velocity ($\omega$)**:
Angular velocity is the rate of change of angular displacement with respect to time. Mathematically, it is the first derivative of $\theta$ with respect to $t$:
$\omega =\frac{d\theta }{dt}$
Differentiating each term of the displacement equation with respect to $t$:
$\frac{d}{dt}\left(at\right)=a$
$\frac{d}{dt}\left(b{t}^2\right)=2bt$
$\frac{d}{dt}\left(c{t}^3\right)=3c{t}^2$
Therefore, the angular velocity is:
$\omega =a+2bt+3c{t}^2$

3. **Angular Acceleration ($\alpha$)**:
Angular acceleration is the rate of change of angular velocity with respect to time. Mathematically, it is the derivative of $\omega$ with respect to $t$ (or the second derivative of $\theta$ with respect to $t$):
$\alpha =\frac{d\omega }{dt}$
Differentiating the angular velocity equation with respect to $t$:
$\frac{d}{dt}\left(a\right)=0$ (since $a$ is a constant)
$\frac{d}{dt}\left(2bt\right)=2b$
$\frac{d}{dt}\left(3c{t}^2\right)=2\times 3ct=6ct$
Combining these terms, we get:
$\alpha =2b+6ct$

Thus, the angular acceleration is given by the expression 2b + 6ct.