Question

If $\alpha$ of a body is given by $\alpha \left( \theta \right) = 3\theta - 2\left( {\dfrac{{rad}}{{{s^2}}}} \right)$ . Find $\omega$ of a body after it covered 3 rad, if it was at rest at the start of the circular motion.
(A) $\sqrt {11} {\text{ rad/s}}$
(B) $\sqrt {12} {\text{ rad/s}}$
(C) $\sqrt {15} {\text{ rad/s}}$
(D) $\sqrt {14} {\text{ rad/s}}$

Answer 1

The angular acceleration can be defined as the rate of change of the angular velocity $\omega$ with respect to time. The statement “it was at rest at the start of circular motion” signifies that at the angular displacement $\theta$ or time $t$ equal to zero angular velocity is equal to zero too.

Formula used: In this solution we will be using the following formulae;
$d\omega = \alpha dt$ where $\omega$ is the instantaneous angular velocity of a body, $\alpha$ is the instantaneous angular acceleration, and $t$ is time.
$\omega = \dfrac{{d\theta }}{{dt}}$ where $\theta$ is angular displacement.

Complete Step-by-Step solution:
To solve the above question, we note that we are given the equation of the angular acceleration of a body with respect to its angular displacement.
The equation is
$\alpha \left( \theta \right) = 3\theta - 2\left( {\dfrac{{rad}}{{{s^2}}}} \right)$
We also note that we can write that
$d\omega = \alpha dt$ where $\omega$ is the instantaneous angular velocity of a body, $\alpha$ is the instantaneous angular acceleration, and $t$ is time.
From mathematical principles, this can be written as
$d\omega \dfrac{{d\theta }}{{dt}} = \alpha d\theta$
But we know that
$\omega = \dfrac{{d\theta }}{{dt}}$ where $\theta$ is angular displacement.
Hence, we have
$\omega d\omega = \alpha d\theta$
Integrating both sides, we have that
$\dfrac{{{\omega ^2}}}{2} = \int {\alpha d\theta }$
Hence, inserting the given equation, we get
$\dfrac{{{\omega ^2}}}{2} = \int {\left( {3\theta - 2} \right)d\theta }$
$\Rightarrow \dfrac{{{\omega ^2}}}{2} = \dfrac{3}{2}{\theta ^2} - 2\theta + C$
Now, we were told that at the beginning of the motion, hence,
We have
${\omega ^2} = 2\left( {\dfrac{3}{2}{\theta ^2} - 2\theta } \right) = 3{\theta ^2} - 4\theta$
Now, at angular displacement of 3 rad, we have
${\omega ^2} = 3{\left( 3 \right)^2} - 4\left( 3 \right) = 15$
$\Rightarrow \omega = \sqrt {15} rad/s$
Hence, the correct option is C.

Note:
For clarity, we have simply prove the validity of the equation $d\omega \dfrac{{d\theta }}{{dt}} = \alpha d\theta$ by cancelling out common terms and multiplying both sides by $dt$ in which we get $d\omega = \alpha dt$ .
Also, the constant $C$ is equal to zero in the equation $\dfrac{{{\omega ^2}}}{2} = \dfrac{3}{2}{\theta ^2} - 2\theta + C$ because, if we write make $\theta$ and $\omega$ zero, we have
$\dfrac{{\left( 0 \right)}}{2} = \dfrac{3}{2}{\left( 0 \right)^2} - 2\left( 0 \right) + C$
$\Rightarrow C = 0$.