If the equation for the displacement of a particle moving on a circular path is given by (θ) =2t³+0.5, where θ is in radians and t in seconds, then the angular velocity of the particle after 2sec from its start is

To find the angular velocity of the particle, we need to understand the relationship between angular displacement ($\theta$) and angular velocity ($\omega$).

The angular velocity ($\omega$) is defined as the time rate of change of angular displacement. Mathematically, it is the first derivative of the angular displacement with respect to time ($t$):
$\omega =\frac{d\theta }{dt}$

The given equation for the angular displacement is:
$\theta \left(t\right)=2t^3+0.5$

Differentiating $\theta \left(t\right)$ with respect to $t$ using the power rule of differentiation ($\frac{d}{dt}\left(t^n\right)=n{t}^{n-1}$), we get:
$\omega =\frac{d}{dt}(2t^3+0.5)$
$\omega =2·3t^2+0$
$\omega =6t^2$

Now, we calculate the angular velocity at $t=2\text{seconds}$:
$\omega =6·(2{)}^2$
$\omega =6·4$
$\omega =24\text{rad /sec}$

Therefore, the angular velocity of the particle after 2 seconds is 24 rad /sec.