Question

A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R. The masses of the large and small sphere are 4M and M respectively. This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the coordinates of the center of the larger sphere when the smaller sphere reached the other extreme position.

Answer 1

Hint: We have to calculate this problem with the concept of centre of mass. Since the system is freely moving on the surface, we can assume that there are no external forces. That is, even if the system is changing its position, the centre of mass will be the same.
The formula used: ${{x}_{i}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$, where $({{m}_{1}}+{{m}_{2}})$ is the total mass of the system and $({{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}})$ is the sum of the moments of individual masses in the system.

Complete step by step answer:
We are going to do this problem with the help of the centre of mass. Since we are dealing with the spheres of smooth surfaces. No external forces are acting on the sphere. Thus, the centre of mass of the system remains stationary. Like the figure above, the small sphere movement will make a change in the position of the large sphere. Initially, the centre of the large sphere will be at (x, 0), and the small sphere will be at (x+5R, 0). Since our small sphere of radius R and a large sphere of radius 6R. So, the difference will be the distance between the centre of the large sphere and the centre of the small sphere.
From this, we can calculate the centre of mass of the system.

We can write down the given information below.

Mass of the large sphere is 4M.

Mass of the small sphere is M.

Centre of mass can be written as,

${{x}_{i}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$, where $({{m}_{1}}+{{m}_{2}})$ is the total mass of the system and $({{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}})$ is the sum of the moments of individual masses in the system.

For the initial case, the values will be,

& {{m}_{1}}=4M \\\ & {{m}_{2}}=M \\\ & {{x}_{1}}=x \\\ & {{x}_{2}}=x+5R \\\ \end{aligned}$$ $${{x}_{i}}=\dfrac{4Mx+M(x+5R)}{4M+M}$$………….(1) For the final case, the values will be, $$\begin{aligned} & {{m}_{1}}=4M \\\ & {{m}_{2}}=M \\\ & {{x}_{1}}=L \\\ & {{x}_{2}}=L-5R \\\ \end{aligned}$$ $${{x}_{f}}=\dfrac{4ML+M(L-5R)}{4M+M}$$…………….(2) Since there is no change in the centre of the mass of the system, we can equate both equations. $${{x}_{i}}={{x}_{f}}$$ $$\dfrac{4Mx+M(x+5R)}{4M+M}=\dfrac{4ML+M(L-5R)}{4M+M}$$ $$4x-4L=(L-5R)-(x+5R)$$ $$5x-5L=-10R$$ $$-5L=-5x-10R$$ We can divide both sides by -5. $$L=x+2R$$ This is the X coordinate of the big sphere. Since the sphere is moving along the X-axis, the Y component will be zero. Hence, the coordinates of the large sphere, when the smaller sphere reaches the other extreme point are (x+2R,0). Note: We can do this problem in many ways. Since our question is not providing the options. But the change in position will be 2R. This change in position will be the same for any methods that can be done here. Do not forget that, this centre of the mass problem is arising since we are not applying any external forces. If we are applying any forces such as friction, air resistance etc. we have to consider that also. So we can’t equate the centre of masses.