The angle of a rotating body changes with time as θ(t) = A +…

The angle of a rotating body changes with time as $\theta(t) = A + Bt^2$, with $A = 0.1\ \mathrm{rad}$ and $B = 0.2\ \mathrm{rad/s^2}$. (a) What is its angular velocity $t = 3\ \mathrm{s}$ after the start? (b) What is its angular acceleration?

Explanation of Method

Angular velocity and acceleration from θ = A + Bt².

Step 1: (a) Angular velocity

\[ \omega(t) = \frac{d\theta}{dt} = 2Bt, \]

\[ \omega(3) = 2 \cdot 0.2 \cdot 3. \]

\[\omega(3\ \mathrm{s}) = 1.2\ \mathrm{rad/s}\]

Step 2: (b) Angular acceleration

\[ \varepsilon = \frac{d\omega}{dt} = 2B. \]

\[\varepsilon = 0.4\ \mathrm{rad/s^2} \ \text{(constant)}\]

Check without calculus

Comparing with $\theta = \theta_0 + \omega_0 t + \tfrac12\varepsilon t^2$: $\theta_0 = 0.1$, $\omega_0 = 0$, $\tfrac12\varepsilon = 0.2$, so $\varepsilon = 0.4\ \mathrm{rad/s^2}$ and $\omega = \varepsilon t = 1.2\ \mathrm{rad/s}\ \checkmark$

$\theta\ (\mathrm{rad})$ $t\ (\mathrm{s})$ $0$ $1$ $2$ $3$ $4$ $0.1$ $1$ $1.9$ $\text{slope} = \omega(3) = 1.2\ \mathrm{rad/s}$ $A = 0.1\ \mathrm{rad}\ (t = 0)$ $\theta = 0.1 + 0.2t^2$

The angle grows along a parabola starting at $A = 0.1\ \mathrm{rad}$. The angular velocity at 3 s is the slope of the tangent there; the constant $A$ only shifts the curve up and does not affect $\omega$ or $\varepsilon$.

Result

\[\boxed{\,\omega(3\ \mathrm{s}) = 1.2\ \mathrm{rad/s}\,}\]

\[\boxed{\,\varepsilon = 0.4\ \mathrm{rad/s^2} \ \text{(constant)}\,}\]