25. II FIGURE EX4.25 shows the angular-velocity-versus-time graph for a particle moving in a circle, starting from ${\theta }_0=0\text{\hspace{0.05em}\hspace{0.05em}rad}$at $t=0\text{s}$. Draw the angular-position-versus-time graph. Include an appropriate scale on both axes.

In the given time from 0 s to 4s, the angular velocity of the particle is$20\text{rad}/\text{s}$

The angular position of the particle is

$\begin{array}{l}\Delta {\theta }_1={\omega }_1\Delta t_1\\ \Delta {\theta }_1=(20\text{rad}/\text{s})(4\text{s}-0\text{s})\\ \Delta {\theta }_1=80\text{rad}\end{array}$

The angular velocity of the particle is constant but time is changing so the angular position of the particle will increase from$0\text{\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}rad}$ to$80\text{\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}rad}$ linearly.

For 4s to 8s, the angular velocity of the particle is$0\text{\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}rad}/\text{s}$

The angular position of the particle is

$\begin{array}{l}\Delta {\theta }_2={\omega }_2\Delta t_2\\ \Delta {\theta }_2=(0\text{rad}/\text{s})(6\text{s}-4\text{s})=0\text{rad}\end{array}$

The angular velocity of the particle is zero, so the angular position of the particle will remain same at the initial position of $80\text{\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}rad\hspace{0.05em}}$. So the angular position is linear flat.

From 6s to 8s , the angular velocity of the particle is $-10\text{rad}/\text{s}$

$\begin{array}{l}\Delta {\theta }_3={\omega }_3\Delta t_3\\ \Delta {\theta }_3=(-10\text{rad}/\text{s})(8\text{s}-6\text{s})\\ \Delta {\theta }_3=-20\text{rad}\end{array}$

The angular velocity of the particle is constant but time is changing so the angular position of the particle will decrease from 80 rad to 60 rad linearly in time form 8 s to 6 s.

The following graph shows the graph of angular position versus time for a particle moving in circular path.