37. II FIGURE EX4.37 shows the angular acceleration graph of a turntable that starts from rest. What is the turntable's angular velocity at (a)$t=1\text{s}$ , (b) $t=2\text{s}$, and (c) $t=3\text{s}$?

The area under the angular acceleration versus time graph gives the change in angular velocity.

$\omega ={\omega }_0+$area under the graph

Here, width="21">ω0is the initial angular velocity.

The angular acceleration versus time graph is shown below.

For : $t=1\text{s}$

The area under the curve is calculated as follows:

Area = Area of rectangle ABCD + Area of triangle CDE

$\begin{array}{l}=\left(\text{AB}\right)\left(\text{CD}\right)+\frac{1}{2}\left(\text{CE}\right)\left(\text{CD}\right)\\ =(1\text{s})\left(2.5\text{rad}/{\text{s}}^2\right)+\left(\frac{1}{2}\right)\left(2.5\text{rad}/{\text{s}}^2\right)(1\text{s})\\ =3.75\text{rad}/\text{s}\end{array}$

The angular velocity

$\omega ={\omega }_0+$area under the graph

$\begin{array}{l}\omega =0\text{rad}/\text{s}+3.75\text{rad}/\text{s}\\ =3.75\text{rad}/\text{s}\end{array}$

Therefore, the angular velocity at time $t=2\text{s}$is$3.75\text{rad}/\text{s}$

For : $t=2\text{s}$

The area under the curve is calculated as follows:

$\begin{array}{l}\text{Area}=\text{AreaoftriangleAEF}\\ =\frac{1}{2}\left(\text{AE}\right)\left(\text{AF}\right)\\ =\left(\frac{1}{2}\right)\left(5\text{rad}/{\text{s}}^2\right)(2\text{s})\\ =5\text{rad}/\text{s}\end{array}$

The angular velocity

$\omega ={\omega }_0+$area under the graph

$\omega =0\text{rad}/\text{s}+5\text{rad}/\text{s}=5\text{rad}/\text{s}$

Therefore, the angular velocity at time $t=2\text{s}$is$5\text{rad}/\text{s}$

The angular acceleration of the turntable after 2s is zero that means after 2s , the turntable moves with constant angular velocity.

Therefore, the angular velocity of the turntable at $t=3\text{s}$is$5\text{rad}/\text{s}$