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

Question #109852
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?
1
2020-04-22T10:07:48-0400

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};\\\\\nr^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,\\\\\n\\space\\\\\n\\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}}."

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