# A disk starts from rest and then rotates so that ? = k t where ? is the angular acceleration,...

## Question:

A disk starts from rest and then rotates so that {eq}\alpha=kt{/eq} where {eq}\alpha{/eq} is the angular acceleration, {eq}t{/eq} is the time, and {eq}k=2{/eq} rad/s{eq}^3{/eq}.

When the disk has completed 2 revolutions, the angular velocity is most nearly

Choose the correct one and show the full work.

## Rotational Kinetics:

Let we have a body which moves in a circle making an angular displacement {eq}\theta (t) {/eq} with time.

Then the angular velocity of the body is obtained using the relation {eq}\omega = \dfrac{d \theta }{dt } {/eq} and the angular acceleration obtained as {eq}\alpha = \dfrac{d \omega }{dt } {/eq}.

The expression of the angular acceleration of the disk is {eq}\alpha=kt{/eq} .

We know the relation of the angular acceleration with angular velocity {eq}\omega {/eq}.

{eq}\begin{align*} \alpha & = \dfrac{d\omega }{dt} \\ kt & = \dfrac{d\omega }{dt} \\ d\omega & = kt \cdot dt \\\\ \text{Integrating both sides we have } \\\\ \int_{0} ^ {\omega} d\omega &=\int_{0} ^ {t} kt \cdot dt \\ \omega & = \frac{kt^2}{2} \Rightarrow Equation(1) \\\\ \omega = \dfrac{d\theta }{dt} & = \frac{kt^2}{2} \\ d\theta & = \frac{kt^2 \cdot dt}{2} \\ \text{Integrating both sides we have } \\\\ \int_{0} ^ {\theta} d\theta &=\int_{0} ^ {t} \frac{kt^2 \cdot dt}{2} \\ \theta & = \frac{kt^3}{6} \\\\ \text{Plugging in the values of k = 2 and } \theta =2 \ rev = 2 \cdot 2 \pi \ rad = 4 \pi \ rad \\\\ 4 \pi &= \frac{2t^3}{6} \\ t &=3.35 \ s\\\\ \text{Now we calculate the angular velocity at 3.35 s} \\ \text{From Equation (1) we have} \\\\ \omega & = \frac{2 \times 3.35^2}{2} \\ & = 11.24 \ rad/s\Rightarrow Option(F) \end{align*} {/eq}