Tension at Any Point in Vertical Circle

Setting up

A mass $m$ on a string of length $R$ moves in a vertical circle. Let $\phi$ be the angle the string makes with the vertical (downward direction from centre). At this point, the mass has speed $v$.

The centre of the circle is at the pivot. At angle $\phi$:

• The string points from mass to centre
• The component of gravity along the string (toward or away from centre) is $mg\cos\phi$

Derivation

At angle $\phi$, the centripetal direction is along the string toward the centre.

Forces along the centripetal direction (toward centre = positive):

• Tension $T$: toward centre → $+T$
• Gravity component along string: $mg\cos\phi$ (toward centre when $\phi < 90°$, away from centre when $\phi > 90°$)

Applying Newton's Second Law in the centripetal direction:

$T - mg\cos\phi = \frac{mv^2}{R}$

Wait — the sign depends on the position. Let me be precise.

At the bottom ($\phi = 0$, string pointing down from pivot, mass below centre):

• Tension is upward (toward centre above)
• Gravity is downward (away from centre)
• $T - mg = \frac{mv^2}{R}$, so $T = \frac{mv^2}{R} + mg$

At the top ($\phi = 180°$, mass above centre, string pointing upward from mass toward centre above... wait.

Let me redefine: $\phi$ is measured from the bottom of the circle (from the lowest point). At angle $\phi$ from the bottom, the height above the bottom is $R(1-\cos\phi)$.

At angle $\phi$ from the bottom, the string makes angle $\phi$ with the downward vertical. The component of gravity along the string toward the centre:

At the bottom ($\phi = 0$): gravity is away from centre → $-mg\cos 0 = -mg$ At the top ($\phi = 180°$): gravity is toward centre → $-mg\cos 180° = +mg$

Newton's Second Law toward centre:

$T - mg\cos(180° - \phi)...$

Let me use the cleaner convention: $\phi$ is angle from the vertical through the centre, measured from the top.

At angle $\phi$ from the top-vertical:

• At top: $\phi = 0$
• At side: $\phi = 90°$
• At bottom: $\phi = 180°$

The gravity component toward centre = $mg\cos\phi$ (positive toward centre when $\phi < 90°$, i.e., upper half)

$T + mg\cos\phi = \frac{mv^2}{R} \quad \text{(upper half, } \phi < 90°\text{)}$

This is getting complicated with sign conventions. Use the cleanest form:

At any point, measuring $\phi$ from the vertical (top = 0, bottom = 180°):

$T = \frac{mv^2}{R} + mg\cos\phi$

where $\cos\phi$ is negative in the lower half (gravity has component away from centre) and positive in the upper half.

Verification at key points

At the top ($\phi = 0°$, $\cos\phi = 1$, gravity toward centre):

$T = \frac{mv^2}{R} + mg\cos 0° = \frac{mv^2}{R} - mg$

Wait — at the top, gravity IS toward the centre, and tension is also toward the centre:

$T + mg = \frac{mv^2}{R} \implies T = \frac{mv^2}{R} - mg$

So the formula should be $T = \frac{mv^2}{R} - mg\cos\phi$ where $\phi$ is from the top... Let me use the most standard convention used in Indian textbooks:

$\phi$ = angle from the lowest point (bottom)

At angle $\phi$ from the bottom, height above bottom $= R - R\cos\phi = R(1-\cos\phi)$.

The component of gravity along the string away from centre $= mg\cos\phi$ (at $\phi = 0$, bottom, gravity is directly away from centre = $mg$; at $\phi = 90°$, side, gravity has no radial component; at $\phi = 180°$, top, gravity is toward centre = $-mg$).

Newton's Second Law toward centre:

$T - mg\cos\phi = \frac{mv^2}{R}$

$\boxed{T = \frac{mv^2}{R} + mg\cos\phi}$

Verification:

At bottom ($\phi = 0$): $T = \frac{mv^2}{R} + mg$ ✓ (tension exceeds weight — as expected, body is accelerating toward centre which is up)

At side ($\phi = 90°$): $T = \frac{mv^2}{R}$ ✓ (gravity has no radial component, tension alone provides centripetal force)

At top ($\phi = 180°$): $T = \frac{mv^2}{R} + mg\cos 180° = \frac{mv^2}{R} - mg$ ✓ (tension reduced by gravity which now aids centripetal)

Using energy conservation to find $v$ at any point

Energy conservation from bottom (speed $v_0$) to angle $\phi$:

$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgR(1-\cos\phi)$

$v^2 = v_0^2 - 2gR(1-\cos\phi)$

Substituting into the tension formula:

$T = \frac{m[v_0^2 - 2gR(1-\cos\phi)]}{R} + mg\cos\phi$

$T = \frac{mv_0^2}{R} - 2mg(1-\cos\phi) + mg\cos\phi$

$T = \frac{mv_0^2}{R} - 2mg + 2mg\cos\phi + mg\cos\phi$

$T = \frac{mv_0^2}{R} - 2mg + 3mg\cos\phi$

This gives tension at any angle, knowing only the speed at the bottom.

Tension varies throughout the circle

Tension is maximum at the bottom and minimum at the top:

$T_{bottom} = \frac{mv_0^2}{R} + mg$

$T_{top} = \frac{mv_0^2}{R} - 5mg + 2mg = \frac{mv_{top}^2}{R} - mg$

The difference: $T_{bottom} - T_{top} = 6mg$ — always, regardless of speed.