Q.Define angular velocity. Show that the centripetal force on a particle undergoing uniform circular motion is $-m\omega^2\vec{r}$.

[!TLDR] Angular velocity is the rate of sweep of the radius vector, with direction along the rotation axis; differentiating the position vector twice for UCM gives acceleration = -ω²r. [!ANSWER] $\omega=\dfrac{d\theta}{dt}$ (direct …

Angular velocity ω is a vector quantity equal to the rate at which the radius vector (from the centre of rotation to the particle) sweeps out angle, $\omega = \dfrac{d\theta}{dt}$, with its direction along the axis of rotation (given by the right-hand rule) — although for motion confined to one plane, as in this chapter, only its magnitude (angular speed) is generally used. For a particle in UCM at radius r, write the position vector as $\vec{r}=r\cos(\omega t)\hat{i}+r\sin(\omega t)\hat{j}$ (starting on the x-axis at t=0). Differentiating once gives velocity $\vec{v}=-r\omega\sin(\omega t)\hat{i}+r\omega\cos(\omega t)\hat{j}$, and differentiating again gives acceleration $\vec{a}=-r\omega^2\cos(\omega t)\hat{i}-r\omega^2\sin(\omega t)\hat{j}=-\omega^2\left[r\cos(\omega t)\hat{i}+r\sin(\omega t)\hat{j}\right]=-\omega^2\vec{r}$. The …

Write the position vector of a UCM particle using ω and differentiate it twice with respect to time; multiply the resulti …

Stopping after computing only the velocity (one derivative) and forgetting a second differentiation is n …