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? ET / N s / % \ / \ \ ( \ B : ( oreo] Le | 3 a8 / \ RO] $8/ A /

1. Setting up the picture

1. The cylinder has radius $R$ and its axis is horizontal.
2. Point $P$ is on the right-hand side at the same height as the axis.
3. The chain (mass per unit length $\lambda$) is perfectly flexible & the surface is perfectly smooth (no friction).
4. Angle $\phi$ is measured from $P$ while you move clockwise around the cylinder (so the chain touches $\phi = 0 \rightarrow 2\pi$).

2. Two equilibrium equations for each tiny piece

For a small arc element of length $ds = R\,d\phi$:

1. Tangential direction (along the chain):
   $\frac{dT}{ds} = \lambda g \cos\phi$ This comes from resolving the weight $\lambda g$ along the tangent.

2. Normal (radial) direction:
   Net inward force per unit length must be zero because the chain does not accelerate. Curvature gives an inward pull $\frac{T}{R}$, gravity contributes $\lambda g \sin\phi$ inward, and the cylinder supplies an outward normal reaction $N$.
   $N = \frac{T}{R} + \lambda g \sin\phi$ For the chain to keep contact we need $N \ge 0$ everywhere.

3. Integrating the tangential equation

Take $T_0$ as the tension at $\phi = 0$ (the point $P$ where the free part hangs). Integrating,

$T(\phi) = T_0 + \lambda g R \sin\phi$

4. Find where the normal reaction is the smallest

Insert $T(\phi)$ into the $N$ expression:

$N(\phi) = \frac{T_0}{R} + 2\lambda g \sin\phi$

$\sin\phi$ is minimum (-1) at the lowest point $\phi = 3\pi/2$. Thus the smallest normal force is

$N_{\text{min}} = \frac{T_0}{R} - 2\lambda g$

For just-in-contact condition we set $N_{\text{min}} = 0$:

$\frac{T_0}{R} - 2\lambda g = 0 \quad\Longrightarrow\quad T_0 = 2\lambda g R$

5. Relate $T_0$ to the hanging length $\ell$

The free vertical part simply hangs, so its tension at the top is its own weight:

$T_0 = \lambda g \ell$

Combine with the previous result:

$\lambda g \ell = 2\lambda g R \;\;\Rightarrow\;\; \boxed{\ell = 2R}$

Hence the dangling segment must be twice the cylinder’s radius.

Imagine this

• You have a smooth round pipe (the cylinder).
• You fix one end of a bead-string on the side of the pipe, exactly half–way up.
• You wrap the string once all the way around the pipe.
• Whatever is left of the string is allowed to simply hang straight down by gravity.

The question is: how long must that hanging part be so that the wrapped part always hugs the pipe and never lifts off anywhere?

Answer (we will prove it): the hanging bit must be twice the radius of the cylinder.

Step-by-Step Mathematical Solution

Let $\lambda$ = mass per unit length, $g$ = acceleration due to gravity.

1. Coordinate choice:
   Angle $\phi$ measured clockwise from the attachment point $P$.

2. Tangential equilibrium on element $ds = R\,d\phi$:
   $\frac{dT}{ds} = \lambda g \cos\phi \;\;\Rightarrow\;\; dT = \lambda g R \cos\phi\,d\phi$
   Integrate with $T(0)=T_0$:
   $T(\phi) = T_0 + \lambda g R \sin\phi$

3. Normal equilibrium:
   Curvature provides inward pull $\frac{T}{R}$; gravitational inward component is $\lambda g \sin\phi$. Balancing with the outward normal reaction $N$: $N = \frac{T}{R} + \lambda g \sin\phi$ Substitute $T(\phi)$:
   $N(\phi) = \frac{T_0}{R} + 2\lambda g \sin\phi$

4. Just-contact criterion:
   Minimum $N$ occurs at $\sin\phi = -1$. Set $N_{\min}=0$:
   $0 = \frac{T_0}{R} - 2\lambda g \quad\Longrightarrow\quad T_0 = 2\lambda g R$

5. Relate $T_0$ to free length $\ell$:
   $T_0 = \lambda g \ell$
   Hence $\ell = 2R$

Final Answer

$\boxed{\ell = 2R}$