MD5 - Conical pendulums

One and two strings

Two strings join a rotating particle to fixed points above and below it. Can spinning too slowly make the lower string go slack? A string can pull along its own length, but cannot push. That simple restriction will decide which algebraic answers describe possible motion.

Picture a small particle moving steadily around a horizontal circle beneath a fixed point. A taut string joining them sweeps out a cone. The particle stays at the same height, so its vertical acceleration is zero, even though its horizontal velocity is continually turning.

With one string of length $l$, making angle $\theta$ with the downward vertical, the circle's radius is $R=l\sin\theta$. Its centre is the point on the vertical axis at the particle's height, not the suspension point. The vertical distance below the suspension point is $l\cos\theta$. For a genuine one-string horizontal circle under gravity, $0<\theta<\pi/2$. The endpoint $\theta=0$ has zero radius and is not a nondegenerate circular path; a horizontal string alone cannot balance the weight.

[DIAGRAM: asset_name: Forces in one-string and two-string conical systems; asset_slug: aqa7367-conical-forces; caption: Vertical components balance while horizontal components provide the inward resultant; description: A one-string conical pendulum beside a two-string system with equal string lengths and fixed attachments A and B on a vertical axis. In the two-string system the particle is midway in height between A and B, each a distance h above or below the plane of its circular path. Both tensions point towards their respective attachments. Upper tension has an upward component; lower tension has a downward component. Weight points down. The horizontal circle radius R is marked separately from either string length.]

In a two-string system with the particle between the attachment heights, upper tension pulls upwards and inwards, while lower tension pulls downwards and inwards. The vertical components oppose each other, but the horizontal inward components add. The lower string does not help to support the weight; it makes the upper string work harder.

For equal lengths $l$ and attachments separated vertically by $2h$, both strings taut place the particle in the horizontal plane midway between them. Right-angled triangles give $R=\sqrt{l^2-h^2}$, requiring $0<h<l$. If either string goes slack, that two-taut-string geometry can no longer be assumed.

Vertical equilibrium and horizontal radial force

Draw the actual forces before using a circular-motion formula. With one string, the forces are tension $T$ along the string and weight $mg$ downwards. There is no extra “centripetal force”. For angle $\theta$ measured from the vertical, the adjacent component $T\cos\theta$ is vertical and the opposite component $T\sin\theta$ is horizontal.

Mass cancels from the speed and angular-speed relationships. A heavier particle at the same angle needs a larger tension, but not a larger angular speed. For a given string length, the nondegenerate one-string model requires $\omega^2>g/l$, since $0<\cos\theta<1$. Equality is the zero-radius limiting configuration.

Use $g=9.8\,\mathrm{m\,s^{-2}}$ in the following examples.

Keep the mass and string length the same, but reduce the angle so that $\cos\theta=4/5$. Both the horizontal radius and the required tension change.

For two taut strings, use the same two directions. If upper and lower strings make acute angles $\alpha$ and $\beta$ with their respective vertical directions, the equations are