Question

A grind-stone starts revolving from rest, it its angular acceleration is $4.0\,rad\,{\operatorname{s} ^{ - 2}}$(uniform) then after $4\,\operatorname{s} $. What is its angular displacement & angular velocity respectively-
(A) $32\,rad,\,16\,rad\,{\operatorname{s} ^{ - 1}}$
(B) $16\,rad,\,32\,rad\,{\operatorname{s} ^{ - 1}}$
(C) $64\,rad,\,32\,rad\,{\operatorname{s} ^{ - 1}}$
(D) $32\,rad,\,64\,rad\,{\operatorname{s} ^{ - 1}}$

Answer 1

Hint: To find the angular displacement of a grind-stone which is revolving from rest we are using angular equations of motion which are analogy of linear equations of motion. Here we have to give that the grind-stone starts revolving from rest so the initial angular velocity is zero.

Useful formula
 Here we use the formula for angular displacement $\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}$ where ${\omega _0}$is the initial angular velocity,$t$is time duration,$\alpha $is the angular acceleration.
To find the angular velocity we the formula of final angular velocity $\omega = {\omega _0} + \alpha t$ where ${\omega _0}$is the initial angular velocity, $t$ is time duration,$\alpha $ is the angular acceleration.

Complete step by step solution
Here we have to find the angular acceleration and angular displacement. The angular displacement always happened around the fixed axis and the angular acceleration is the time derivative of angular velocity.
As we know in this problem the initial angular velocity is ${\omega _0} = 0$ and the angular acceleration is $4.0\,rad\,{\operatorname{s} ^{ - 2}}$ and time duration $t = 4\,\operatorname{s} $ So, substitute these value in the formula
$\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}$.
After substitute these values we get,
$
\Rightarrow \theta = 0 \times 4 + \dfrac{1}{2} \times 4 \times {\left( 4 \right)^2}\,rad \\
\Rightarrow \theta = 0 + 2 \times 16\,rad \\
\Rightarrow \theta = 32\,rad \\
$
Thus, the angular displacement is $\theta = 32\,rad$.
Now, to find the angular velocity we use the formula of final angular velocity $\omega = {\omega _0} + \alpha t$.
Substitute the value of initial angular velocity ${\omega _{0\,}} = 0$,angular acceleration $\alpha = 4\,rad\,{\operatorname{s} ^{ - 2}}$and the time duration $t = 4\,s$.
After substitute these values we get,
$
\Rightarrow \omega = 0 + 4 \times 4\,rad\,{s^{ - 1}} \\
\Rightarrow \omega = 16\,rad\,{s^{ - 1}} \\
$
Therefore, the angular acceleration is $\omega = 16\,rad\,{s^{ - 1}}$.

Hence the option (A) $32\,rad,\,16\,rad\,{\operatorname{s} ^{ - 1}}$ is correct.

Note: To solve this problem we have to consider the initial angular velocity is $0$ because this grind-stone is at rest at $t = 0$ and in the angular motion the displacement is also angular and the angular displacement is the $\theta $.