A uniform flexible chain on a smooth sphere

Just released, the whole chain slides at the same acceleration rate against the sphere. Consider a slip of a small displacement of the whole chain

The kinetic energy corresponding to this velocity should be converted from part of the gravitational potential energy during the chain's descent. When calculating the change of gravitational potential energy, it is equivalent.

It is considered that the length of the top of the chain is *dl * is tiny length of chain A small section is transferred directly to the end. Note that the linear density of chains is

The quality of this section is Gravitational potential

Can change to

Every point on the chain moves in a circle. Since the velocity is zero at the time of release, the centripetal acceleration is zero, and only the size of the phase is different. Equal tangential acceleration When discussing the velocity change in a small process, it can always be regarded as uniformly accelerated linear motion.

Measure the angle $\phi$ from the vertical direction. A segment of the chain between $\phi$ and $\phi+d \phi$ has a mass of $\rho d \phi$ , where $\rho$ is a constant. The tangential tension force pulling this segment backwards is $F$ , the tension pulling forward is $F+d F$ and the tangential component of the gravitational force is $\rho g \sin\phi d \phi$ . Newton's law for this segment is

$(\rho g \sin\phi )d \phi - F+ F+dF= (a \rho) d \phi$

where $a$ is the magnitude of the acceleration that is independent of the angle $\phi$ . This leads to

$\frac{1}{\rho g}\frac{dF}{d \phi}=\frac{a}{g}- \sin \phi$

We solve this differential equation with the boundary condition that the tension is zero at $\phi=0$ and at $\phi=\pi/3$ . The solution is

$\frac{F}{\rho g}=\frac{a}{g}\phi +cos \phi -1$

where $\frac{a}{g}=\frac{3}{2\pi}$ , otherwise the boundary condition is not satisfied. The force is maximum when $\frac{dF}{d \phi}=0$ , or $\frac{1}{\rho g}\frac{dF}{d \rho}=\frac{a}{g}- \sin \phi =0$ . This yields

$\phi= \sin^{-1} \frac{3}{2\pi} = 28.52^o$