Question

The angular displacement of a particle is given by $\theta = {t^3} + {t^2} + t + 1$ then, its angular velocity at t = 2sec is ………………. $rad{\text{ }}{{\text{s}}^{ - 1}}$
$
  A. 27 \\
  B. 17 \\
  C. 15 \\
  D. 16 \\
 $

Answer 1

- Hint- Here we will proceed by using the formula of angular velocity i.e. $\dfrac{{\vartriangle D}}{{\vartriangle T}}or\dfrac{{d\theta }}{{dt}}$ where D is change in displacement calculated by differentiating given displacement and T is time taken to cover the distance to get the required result.

Complete step-by-step solution -
As we are given that-
Angular displacement of a particle is $\theta = {t^3} + {t^2} + t + 1$
And t = 2sec
Also we know that formula of angular velocity is$\dfrac{{\vartriangle D}}{{\vartriangle T}}or\dfrac{{d\theta }}{{dt}}$
Where displacement$\left( \theta \right)$ is final position – initial position or change in position
And time is time taken to cover the distance.
Now we will substitute the values of displacement and time in the formula of angular velocity i.e. $\omega = \dfrac{{\vartriangle D}}{{\vartriangle T}}or\dfrac{{d\theta }}{{dt}}$
Differentiating displacement $\theta = {t^3} + {t^2} + t + 1$,
We get- $\dfrac{{d\theta }}{{dt}} = 3{t^2} + 2t + 1$
Now angular velocity at given time t = 2 in Displacement $3{t^2} + 2t + 1$,
We get- $3{\left( 2 \right)^2} + 2\left( 2 \right) + 1$
$\Rightarrow$ 3(4) + 4 + 1
$\Rightarrow$ 12 + 5
$\Rightarrow 17rad{\text{ }}{s^{ - 1}}$
Therefore, the angular velocity at t = 2 is $17rad{\text{ }}{{\text{s}}^{ - 1}}$.
Hence, option B is correct.

Note- While solving this type of question, we should not forget to put the SI unit with the answer as here we kept $rad{\text{ }}{{\text{s}}^{ - 1}}$ i.e. SI unit of angular velocity along with answer as 17. Also we must know how to do derivatives using the concept of differentiation $\left( {n{x^{n - 1}}} \right)$.