Starting from rest at $\mathit{t}\mathbf{=}\mathbf{}\mathbf{0}$, a wheel undergoes a constant angular acceleration. When$\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.0}\mathbf{\text{s}}$, the angular velocity of the wheel is$\mathbf{5}\mathbf{.0}\mathbf{\text{rad/s}}$. The acceleration continues until$\mathit{t}\mathbf{=}\mathbf{20}\mathbf{\text{s}}$, when it abruptly ceases. Through what angle does the wheel rotate in the interval$\mathit{t}\mathbf{=}\mathbf{}\mathbf{0}$to$\mathit{t}\mathbf{=}\mathbf{40}\mathbf{\text{s}}$?

a) The wheel undergoes a constant acceleration state starting from rest at $t=0$.

b) The angular velocity of the wheel at $t=2.0\text{s}$,$\omega =5.0\text{rad/s}$

c) The constant acceleration continues till $t=20\text{s}$before ceasing.

The study of the rotational motion of the body is given as the angular form of kinematics. The angular entities like displacement, velocity, and acceleration are related to the linear kinematics by a radial value. Further, using this radial theorem, we can observe the relatable kinematic equations in angular form.

Formulae:

The final angular velocity of the body in rotational motion,${\omega }_f={\omega }_0+\alpha \text{t}$ (i)

Where,${\omega }_0$is the initial angular velocity of the body,$\alpha$is the angular acceleration of the body,$t$is the time of motion.

The angular displacement of a body in rotational motion analogy to 2^nd law of kinematic equations, $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$ (ii)

Where,data-custom-editor="chemistry" ${\omega }_0$ is the initial angular velocity of the body, $\alpha$is the angular acceleration of the body, $t$is the time of motion.

The angular acceleration of the wheel can be given using the given data in equation (i) as follows:

$\begin{array}{c}\alpha =\frac{({\omega }_f-{\omega }_0)}{t}\\ =\frac{(5\text{rad/s}-0\text{rad/s})}{2.0\text{s}}\\ =2.5\text{\hspace{0.17em}rad}/{\text{s}}^2\end{array}$

So, the initial angular displacement of the body using equation (ii) as follows:

$\begin{array}{c}{\theta }_1=\left(0\text{rad/s}\right)\left(20\text{\hspace{0.17em}s}\right)+\frac{1}{2}\left(2.5{\text{rad/s}}^2\right){\left(20\text{s}\right)}^2\\ =500\text{\hspace{0.17em}rad}\end{array}$

The angular velocity for the time can be given using equation (i) as follows:

$\begin{array}{c}\omega =\alpha t\\ =\left(2.5{\text{rad/s}}^2\right)\left(20\text{\hspace{0.17em}s}\right)\\ =50\text{\hspace{0.17em}rad/s}\end{array}$

The sweep angle nor the angular displacement within the time interval $t=20\text{\hspace{0.17em}s}$to$t=40\text{s}$ which consists of constant angular velocity, i.e., no angular acceleration can be given using equation (ii) as follows:

$\begin{array}{c}{\theta }_2=\omega \text{t}\\ =\left(50\text{rad/s}\right)\left(20\text{\hspace{0.17em}s}\right)\\ =1000\text{\hspace{0.17em}rad}\end{array}$

Thus, the total angular displacement or sweep angle made by the body can be given as:

$\begin{array}{c}\theta ={\theta }_1+{\theta }_2\\ =500\text{\hspace{0.17em}rad}+1000\text{\hspace{0.17em}rad}\\ =1500\text{\hspace{0.17em}rad}\\ =1.5\times {10}^3\text{\hspace{0.17em}rad}\end{array}$

Therefore, the angle through which the wheel rotates in the interval $t=0\text{\hspace{0.17em}s}$to $t=40\text{\hspace{0.17em}s}$is$1.5\times {10}^3\text{\hspace{0.17em}rad}$.