A body of mass m = 1 kg on a rope of length l = 0.5 m…

A body of mass $m = 1\ \mathrm{kg}$ on a rope of length $l = 0.5\ \mathrm{m}$ rotates in a vertical plane with constant angular velocity $\omega = 10\ \mathrm{rad/s}$. Find the rope tension at the lowest point A, the highest point B, and the two points C and D where the rope is horizontal.

Explanation of Method

Rope tension in a vertical circle.

Step 1: Centripetal force needed

\[ m\omega^2 l = 1 \cdot 100 \cdot 0.5 = 50\ \mathrm{N}, \]

\[ mg = 9.81\ \mathrm{N}. \]

In each position, the rope tension plus the component of gravity toward the centre must give 50 N.

Step 2: Lowest point A

\[ T_A - mg = m\omega^2 l \;\Longrightarrow\; T_A = 50 + 9.81 \]

\[T_A = 59.81\ \mathrm{N}\]

Step 3: Highest point B

\[ T_B + mg = m\omega^2 l \;\Longrightarrow\; T_B = 50 - 9.81 \]

\[T_B = 40.19\ \mathrm{N}\]

Step 4: Horizontal positions C and D

Gravity is perpendicular to the rope, so the rope alone supplies the centripetal force: $T_C = T_D = 50\ \mathrm{N}$.

$A:\ 59.8\ \mathrm{N}$ $B:\ 40.2\ \mathrm{N}$ $C:\ 50\ \mathrm{N}$ $D:\ 50\ \mathrm{N}$ $m\omega^2 l = 50\ \mathrm{N}$ $mg = 9.81\ \mathrm{N}$

Tension arrows to scale, all pointing to the centre. Gravity adds to the needed tension at the bottom (A), helps at the top (B), and is perpendicular to the rope at C and D.

Result

\[\boxed{\,T_A = 59.81\ \mathrm{N}\,}\]

\[\boxed{\,T_B = 40.19\ \mathrm{N}\,}\]

\[\boxed{\,T_C = T_D = 50\ \mathrm{N}\,}\]

Remark

Note. The original results are correct. At C and D gravity is not "perpendicular to the motion", as the original says; it is along the motion, and it is perpendicular to the rope. Keeping $\omega$ constant then needs a tangential push from whatever drives the rotation, which a simple rope alone could not provide.