Question

A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a horizontal frictionless surface initially. When the ball is, again magically, released inside the shell, it sloshes around in the shell and eventually stops at the bottom of the shell. How far does the shell move from its initial contact point with the surface

Answer 1

Answer:

$x =\frac{-R}{2}$

Explanation:

From the question we are told that mass

Thin layer radius $= 2R$

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state  

Considering the center of mass of both bodies

$xcm=\frac{m*x+m*x)}{2m} =x$

$x =\frac{-R}{2}$

Therefore the enclosing layer moves $x =\frac{-R}{2}$