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

Hint: 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 $.