P79* One end of a necklace of small pearls is attached to the outer surface of a fixed cylinder that has radius R and a horizontal axis; the attachment point P is at the same level as the axis. The necklace is wound once round the slippery surface of the cylinder, and the free end is left to dangle (see figure). How long, , does this free end need to be if the rest of the necklace is to touch the cylinder surface everywhere? eS | g |r

Key Concepts Needed

1. Tension variation on a heavy rope: For a rope of linear mass density $\lambda$ lying on a smooth curve, the tension changes because of the component of its weight along the curve.
2. Differential equilibrium on a cylinder: Take a tiny element of rope that subtends an angle $d\theta$ on the cylinder of radius $R$.
3. Tangential weight component: The component of the element’s weight along the tangent is $\lambda g R \sin\theta\,d\theta$.
4. Normal reaction test: For the rope to remain in contact, the normal force from the surface must stay non–negative. This leads to the condition

$T(\theta) \ge \lambda g R \cos\theta.$

5. Critical section: The largest right–hand side (most difficult place) occurs at $\theta = 0$ or $2\pi$ (where $\cos\theta = 1$). That gives the minimum tension needed at the exit point as $T_0 = \lambda g R$.
6. Relating to hanging length: The downward dangling bit is just a vertical string with tension at its top equal to its own weight: $T_0 = \lambda g L$. Setting them equal gives the required length.

Logical Student Chain-of-Thought

1. Label the glued point P as $\theta = 0$. Increase $\theta$ counter-clockwise.
2. Write the step-by-step balance of tangential forces for a small arc element to see how tension grows:
   $T(\theta + d\theta) = T(\theta) + \lambda g R \sin\theta\,d\theta.$
3. Integrate to get the tension distribution:
   $T(\theta) = T_0 + \lambda g R\bigl(1 - \cos\theta\bigr).$
4. Derive the normal force and impose no detachment:
   $N(\theta) = T(\theta) - \lambda g R \cos\theta \ge 0.$
5. Spot that the tight condition is at $\theta = 0$ giving $T_0 \ge \lambda g R$.
6. Recognise $T_0$ is produced by the weight of the free end: $T_0 = \lambda g L$.
7. Finally, equate and solve: $L = R$.

What’s happening?

Imagine you have a smooth cold-drink can and a beaded chain. You glue one end of the chain to the side of the can – exactly halfway up. Then you wrap the chain all the way around the can once (360°) and let whatever is left of the chain simply hang straight down.

What do we want?

We want every single bead that is resting on the can to stay pressed against the can’s surface – no little gap anywhere. The question is how long the part that hangs freely must be so that the wrapped part never lifts off.

Intuition like a 10-year-old

1. Chain on a curve needs a pull: A curve tries to throw the chain outward (just like when you swing a bucket of water in a circle). Something must pull it inward to keep it hugging the can. That something is the tension (stretching force) inside the chain.
2. Where does that inward pull come from? It comes from the weight of the hanging part – gravity pulls that part downward, which in turn pulls on (stretches) the wrapped part.
3. More hanging = more pull. If you make the hanging length longer, the pull (tension) gets stronger. If it’s too short, there isn’t enough pull and some section of the chain will lose contact.
4. Minimum trick: The most ‘demanding’ place is exactly where the chain leaves the cylinder (the same point where it’s glued). If there is enough pull there, it will be enough everywhere.
5. Magic number: The maths shows the pull needed equals the weight of a length of chain exactly equal to the cylinder’s radius.

So the hanging part must be as long as the radius of the cylinder – no more, no less!

Step-by-Step Calculation

Let the linear mass density of the necklace be $\lambda$ (mass per unit length).

1. Coordinate choice
   $\theta = 0$ at point $P$ (rightmost point, tangent is vertical downward). Positive $\theta$ increases counter-clockwise; the rope wraps to $\theta = 2\pi$ and returns to $P$.

2. Tangential equilibrium of a small element
   A small arc $R\,d\theta$ experiences a weight $\lambda g R\,d\theta$.
   Tangential component: $\lambda g R\sin\theta\,d\theta$.
   Hence
   $T(\theta + d\theta) - T(\theta) = \lambda g R \sin\theta\,d\theta.$

3. Integrate the differential relation
   $\frac{dT}{d\theta} = \lambda g R \sin\theta \;\Longrightarrow\; T(\theta) = T_0 + \lambda g R (1 - \cos\theta).$ Here $T_0 \equiv T(0)$ is the tension at point $P$ where the chain meets the hanging end.

4. Normal (radial) equilibrium and contact condition
   For the element to stay on the surface, the normal reaction $N$ must satisfy
   $N = T(\theta) - \lambda g R \cos\theta \ge 0.$ Substituting $T(\theta)$ gives

   = T_0 - \lambda g R \cos\theta.$$ The worst (largest $\cos\theta$) occurs at $\theta = 0$ (or $2\pi$): $$T_0 \ge \lambda g R.$$

5. Relate $T_0$ to the hanging length
   The hanging part is vertical; its tension at the top is just its own weight:
   $T_0 = \lambda g L.$

6. Minimum length for full contact
   Combining the two results:
   $\lambda g L = \lambda g R \;\Longrightarrow\; \boxed{L = R}.$

So the free end must be as long as the cylinder’s radius.