Problem Analysis

A ball is attached to a cord of length $l$ fixed to the top of a cylinder of radius $r$. As the ball moves, the cord wraps around the cylinder. We need to find the range of initial horizontal velocities $v$ such that the cord “slacks” (Tension becomes zero) during the motion.

Condition for Slacking

The cord will “certainly slack” if the velocity falls within a specific range: 1. Lower Limit: The velocity must be high enough to leave the region of simple oscillation (pendulum motion). This threshold is reached when the ball rises to the horizontal level of the center of curvature (similar to a pendulum reaching $90^\circ$). 2. Upper Limit: The velocity must not be so high that it completes the full loop (keeping tension $>0$ throughout).

Step 1: The Lower Limit (Entering the “Slackable” Zone)

For the cord to slack, the particle must rise at least to the horizontal level of the center of the cylinder’s curvature.
Initial State: Ball is at distance $l$ below the nail. Height $y_1 = -l$.
Critical Height: The level of the point of tangency when string is horizontal is $r$ below the nail. Height $y_{center} = -r$.
Elevation Change: $\Delta h = y_{center} – y_1 = -r – (-l) = l – r$.
Using conservation of energy: $$ \frac{1}{2}mv^2 = mg(l-r) \implies v = \sqrt{2g(l-r)} $$ If $v < \sqrt{2g(l-r)}$, the ball oscillates like a pendulum below this level and tension never vanishes. Thus, we need $v > \sqrt{2g(l-r)}$.

Step 2: The Upper Limit (Avoiding Looping)

We must find the condition for the ball to complete the loop (Tension $> 0$ at the highest point).
At the highest point, the string wraps halfway around the cylinder ($\pi$ radians).
Effective Length at Top: $R_{top} = l – \pi r$.
Height at Top: The ball is $R_{top}$ above the nail.
Total Height Rise: From bottom ($l$ below nail) to top ($l-\pi r$ above nail): $$ H = l + (l – \pi r) = 2l – \pi r $$

Critical Velocity at Top ($v_{top}$): For tension to be just zero at the top: $$ \frac{m v_{top}^2}{R_{top}} = mg \implies v_{top}^2 = g(l – \pi r) $$

Conservation of Energy: $$ \frac{1}{2} m v_{initial}^2 = \frac{1}{2} m v_{top}^2 + mgH $$ $$ v_{initial}^2 = g(l – \pi r) + 2g(2l – \pi r) $$ $$ v_{initial}^2 = gl – \pi gr + 4gl – 2\pi gr $$ $$ v_{initial}^2 = g(5l – 3\pi r) $$ If $v > \sqrt{g(5l – 3\pi r)}$, the ball completes the loop without slacking.