A body on a rope is whirled in a vertical circle, moving…

A body on a rope is whirled in a vertical circle, moving only under gravity and the rope (no driving force along the path). Prove that the greatest rope tension is at least 6 times the body's weight.

Explanation of Method

Rope tension in a vertical circle is at least 6mg.

Step 1: Tensions at the top B and bottom A

\[ T_B + mg = \frac{mv_B^2}{R}, \]

\[ T_A - mg = \frac{mv_A^2}{R}. \]

Step 2: Energy

Going down $2R$ from B to A: $v_A^2 = v_B^2 + 4gR$.

Step 3: Difference

\[ T_A - T_B = \frac{m\,(v_A^2 - v_B^2)}{R} + 2mg, \]

\[ T_A - T_B = 4mg + 2mg. \]

\[T_A = T_B + 6mg \ge 6mg \ \ (\text{since } T_B \ge 0)\]

The limit $T_A = 6mg$ is reached when the body only just gets over the top ($v_B^2 = gR$).

$v_B \ge \sqrt{gR}$ $v_A^2 = v_B^2 + 4gR$ $T_A$ $T_B \ge 0$ $T_A - T_B = 6mg$ $\Rightarrow T_A \ge 6mg$

For a body whirled freely (speed changing only by gravity), the bottom tension always exceeds the top tension by exactly $6mg$; since the top tension cannot be negative, the maximum is at least $6mg$.

Result

\[\boxed{\,T_A = T_B + 6mg \ge 6mg \ \ (\text{since } T_B \ge 0)\,}\]

Remark

About the statement. The original says the body rotates "with constant angular velocity", but the proof uses energy conservation, i.e. a speed that changes under gravity. If the angular velocity really were constant, then $T_A - T_B = 2mg$, and the maximum tension would only have to be at least $2mg$. The $6mg$ result holds for a body whirled freely. The original's proof is otherwise correct, though its sign in the top-point equation is garbled.