Circular Motion — Non-Uniform

Let a particle of mass $m$ move on a vertical circle of radius $r$, attached to a light inextensible string. Let $\theta$ be the angle from the downward vertical (so $\theta = 0$ at the bottom, $\theta = \pi$ at the top). At a general point the string tension is $T$ and the weight is $mg$ downward.

Radial direction (towards centre): The component of gravity along the inward radial direction is $-mg\cos\theta$ at the bottom-ish region (away from centre) but flips sign as you climb. Resolving carefully gives $T - mg\cos\theta = \tfrac{mv^2}{r}$, i.e.

So at the bottom ($\cos\theta = 1$): $T_{\text{bot}} = m v_{\text{bot}}^2/r + mg$. At the top ($\cos\theta = -1$): $T_{\text{top}} = m v_{\text{top}}^2/r - mg$.

Energy conservation links the two speeds. Taking the bottom as the reference height, the top is at height $2r$:

Substituting into the two tension formulas and subtracting gives the classic result $T_{\text{bot}} - T_{\text{top}} = 6mg$ — independent of speed.

General point on vertical circle: $T = mv^2/r + mg\cos\theta$ (taking $\theta$ from downward vertical) · Bottom: $T = mv^2/r + mg$; Top: $T = mv^2/r - mg$ · Energy: $v_{\text{bot}}^2 - v_{\text{top}}^2 = 4gr$ · Key identity: $T_{\text{bot}} - T_{\text{top}} = 6mg$ (independent of speed) · Minimum at top (string): $v^2_{\text{top}} = gr$, equivalently $v^2_{\text{bot}} \geq 5gr$

Pause — copy the vertical-circle tension formulas at bottom and top, the energy relation $v_{\text{bot}}^2-v_{\text{top}}^2=4gr$, the identity $T_{\text{bot}}-T_{\text{top}}=6mg$, and the minimum-speed condition $v_{\text{top}}^2 = gr$ into your book.

We just saw the vertical-circle analysis: radial Newton's law at angle $\theta$ gives tension $T = mv^2/r + mg\cos\theta$ (bottom) or $T = mv^2/r - mg$ (top), energy gives $v_{\text{bot}}^2 - v_{\text{top}}^2 = 4gr$, and the identity $T_{\text{bot}}-T_{\text{top}} = 6mg$ holds independently of speed. That raises a question: what are the equilibrium equations for a conical pendulum, and how do they connect to banked turns? This card answers it → $T\cos\theta = mg$, $T\sin\theta = mr\omega^2$; divide to get $\tan\theta = r\omega^2/g$; period $T_{\text{period}} = 2\pi\sqrt{\ell\cos\theta/g}$.

A mass $m$ swings on a light string of length $\ell$, making constant angle $\theta$ with the vertical, tracing a horizontal circle of radius $r = \ell \sin\theta$ at constant speed $v$. Two equations only:

• Vertical: $T\cos\theta = mg$ (no vertical acceleration).
• Horizontal/radial: $T\sin\theta = \dfrac{mv^2}{r} = m r \omega^2$.

Dividing the second by the first gives the canonical relation:

The period is $\mathbf{T_{\text{period}} = 2\pi\sqrt{\ell\cos\theta/g}}$ — independent of mass. A banked frictionless track of bank angle $\theta$ satisfies the same equation: at the "designed speed" $v^2 = rg\tan\theta$ no friction is needed to hold the car in the circle.

Conical pendulum vertical: $T\cos\theta = mg$ · Conical pendulum radial: $T\sin\theta = mv^2/r = mr\omega^2$ · $\tan\theta = v^2/(rg) = r\omega^2/g$ · Period $T_{\text{period}} = 2\pi\sqrt{\ell\cos\theta/g}$ — independent of $m$ · Banked turn (frictionless designed speed): $v^2 = rg\tan\theta$

Pause — copy the conical-pendulum equations ($T\cos\theta=mg$, $T\sin\theta=mr\omega^2$, $\tan\theta=r\omega^2/g$), period $T_{\text{period}}=2\pi\sqrt{\ell\cos\theta/g}$, and the banked-turn design speed $v^2=rg\tan\theta$ into your book.