Motion in a vertical circle

Motion in a vertical circle is a classic and important application of both Newton's second law and energy conservation together. Consider a mass $m$ tied to one end of a light, inextensible string of length $r$ (the radius of the circular path), the other end fixed, so the mass swings in a vertical circle.

At a general point where the position vector makes angle $\theta$ with the downward vertical, resolving the forces -- gravity $mg$ (always downward) and the string tension $T$ (always along the string, toward the centre) -- into tangential and radial components gives two equations:

Because the tangential acceleration $g\sin\theta$ is nonzero for every $\theta$ except $0^\circ$, this motion is never uniform circular motion -- both the speed $v$ and the tension $T$ change continuously around the loop.

At the lowest point (point 1), gravity points away from the centre while tension points toward it, giving $T_1 = mg + \dfrac{mv_1^2}{r}$. At the highest point (point 2), both gravity and tension point toward the centre, giving $T_2 = \dfrac{mv_2^2}{r} - mg$. Since gravity helps supply the centripetal force at the top but opposes it at the bottom, $T_1$ is always greater than $T_2$ for the same speed profile.

Applying energy conservation between the lowest point (taking $U_1=0$ there) and the highest point (a height $2r$ above it, so $U_2=mg(2r)$) gives $v_1^2 - v_2^2 = 4gr$. Combining this with the radial equations at both points (after eliminating $T_1,T_2$ or substituting directly) yields the difference in tension, $T_1-T_2 = 6mg$, independent of the speed.

The minimum speed to complete the loop is fixed by the requirement that the string can only pull, never push -- so the tension can never go negative. The critical point is the top, where tension is smallest; setting $T_2=0$ in the radial equation there gives the minimum speed at the top:

$v_2 = \sqrt{gr}$ …