Expression for Centripetal Acceleration

To find the acceleration responsible for uniform circular motion, set up coordinates with the centre of the circular path at the origin O, and let the particle start (for convenience) at point $P_0$ on the positive x-axis. At a later instant $t$, the particle has moved anticlockwise to point P, and its radius vector makes an angle $\theta = \omega t$ with the x-axis (since $\dfrac{d\theta}{dt}=\omega$, a constant). The x and y components of the radius vector are therefore $r\cos\theta = r\cos(\omega t)$ and $r\sin\theta = r\sin(\omega t)$, so $\vec{r} = r\cos(\omega t)\,\hat{i} + r\sin(\omega t)\,\hat{j}$ (Eq. 3.50).

Differentiating the position vector with respect to time gives the instantaneous velocity, and differentiating the velocity gives the instantaneous acceleration — with $r$ and $\omega$ both constant throughout: $\vec{v} = \dfrac{d\vec{r}}{dt} = -r\omega\sin(\omega t)\,\hat{i} + r\omega\cos(\omega t)\,\hat{j}$ (Eq. 3.51), and differentiating once more, $\vec{a} = \dfrac{d\vec{v}}{dt} = -r\omega^2\cos(\omega t)\,\hat{i} - r\omega^2\sin(\omega t)\,\hat{j} = -\omega^2\vec{r}$ (Eq. 3.52). The minus sign shows the acceleration is opposite in direction to the radius vector $\vec{r}$ — that is, it points from the particle straight back toward the centre — confirming this is exactly the centripetal acceleration described qualitatively in section 3.4. Its magnitude is $a = \omega^2 r = \dfrac{v^2}{r}$ (Eq. 3.53), using $v=\omega r$ from section 3.4.1.

The force producing this acceleration is the centripetal force, $\vec{F} = m\vec{a} = -m\omega^2\vec{r}$ (Eq. 3.54), with magnitude $F = \dfrac{mv^2}{r} = m\omega^2 r$ (Eq. 3.55).

Conical pendulum. A conical pendulum is a mass $m$ (the bob) hung from a fixed point O by a string of length $l$, but instead of swinging in a vertical plane like an ordinary simple pendulum, it revolves steadily so that it traces a horizontal circle, while the string itself sweeps out the surface of a cone of constant half-angle $\theta$ (measured from the vertical) — hence the name. In the absence of friction, once started, such a system keeps revolving indefinitely at constant speed.

Only two forces act on the bob: its weight $mg$, straight down, and the string tension $T$, directed along the string toward O. Resolving the tension into a vertical component $T\cos\theta$ and a horizontal component $T\sin\theta$: for the bob to stay in its horizontal circle (no net vertical motion), the vertical component must exactly balance gravity, $T\cos\theta = mg$; the horizontal component, being the only unbalanced force, must supply the centripetal force needed for the circular motion, $T\sin\theta = \dfrac{mv^2}{r}$. …