Answer to Question #109852 in Mechanics | Relativity for Chinmoy Kumar Bera

Question

A particle of mass m can move on a smooth horizontal table.it is attached to a string which pass through a smooth hole in the table goes under a small smooth pulley of mass M and is attached to a point in the under side of the table so that the parts of string hang vertically.if the motion be slightly disturbed when the mass m is describing a circle uniformly so that the angular momentum unchanged find the apsidal angle?

Answer 1

The length of hanging string is

$$y=l-r,$$

where $l$ - length of the string, $r$ - distance between the particle and the point where the string is attached.

The equation of motion of the particle with mass m will be

$$m\bigg(\frac{d^2u}{d\theta^2}+u\bigg)=\frac{T}{h^2u^2};\\ r^2\frac{d\theta}{dt}=h,\space d\theta/dt=h/r^2=hu^2.$$

The pulley with mass M moves downward, the equation of motion will be

$$M\frac{d^2}{dt^2}\bigg(\frac{l-r}{2}\bigg)=Mg-2T.$$

We see that

$$\frac{d^2r}{dt^2}=-h\frac{d^2u}{d\theta^2};\frac{d\theta}{dt}=-h^2u^2\frac{d^2u}{d\theta^2};$$

Therefore, the equation of motion of M is

$$\frac{M}{4}\frac{d^2u}{d\theta^2}=\frac{Mg}{2h^2u^2}-\frac{T}{h^2u^2}.$$

Add this to the first equation, eliminate T:

$$\frac{4m+M}{4m}\frac{d^2u}{d\theta^2}+u=\frac{Mg}{2mh^2u^2}.$$

If the path is a circle of radius $1/c$ , u=c, i.e. constant, and $d^2u/d\theta^2=0.$

Therefore, from the last equation with masses:

$$c=\frac{Mg}{2mh^2c^2},\space h^2=\frac{Mg}{2mc^3}.$$

Hence

$$\frac{4m+M}{4m}\frac{d^2u}{d\theta^2}+u=\frac{c^3}{u^2}.$$

This is the circular path of m.

If we displace particle so that its angular momentum remains unaltered, in the last equation we can substitute $u=x+c.$ Therefore:

$$\frac{4m+M}{4m}\frac{d^2x}{d\theta^2}+c+x=c\bigg(1+\frac{x}{c}\bigg)^{-2}.$$

If we neglect the higher powers of x other than first, we get

$$\frac{4m+M}{4m}\frac{d^2x}{d\theta^2}=-3x,\\ \space\\ \frac{d^2x}{d\theta^2}=-\frac{12m}{M+4m}x,\\$$

the apsidal angle becomes

$$\pi/\sqrt{\frac{12m}{M+4m}}=\pi\sqrt{\frac{M+4m}{12m}}.$$