Small sphere in a circular surface

[Neolight]

Homework Statement

A small sphere of radius R held against the inner surface of a smooth spherical shell of radius 6R as shown in figure.

The masses of the shell and small sphere are 4m and m respectively. This arrangement is placed on a smooth horizontal table. The small sphere is now released. The x-coordinate of the center of the shell when the smaller sphere reaches the other extreme position is

Homework Equations

since the net force on x- axis is zero,

X_com for initial condition = X_com for final condition

The Attempt at a Solution

1. initial condition
when the small sphere is at rest

the distance X₁ for small sphere from origin is = 6R-R= 5R
the distance X₂ for shell from origin is = 0 , since the y-axis lies on the line of symmetry of the shell.

2. Final condition
when the small sphere reached the other extreme position
by this time i considered a small displacement L along -ve x-axis for the shell

( but i personally doubt that there should be any displacement because the only forces i see are the weight (mg) always directed downwards i.e. y-axis and the normal between the two surfaces of the small sphere and the shell , this normal forces will cancel out each other (Newton's third law) so there will be no forces acting along the X-axis . please correct me if I'm wrong)

so
the distance Z₁ for small sphere from origin is = 5R + L
the distance Z₂ for shell from origin is= L

X_com initial = X_com final

4m X 0 + m X 5R = 4m X L + m(5R+L)

please note that i haven't included the total mass since they will just cancel out each other

⇒ 5mR= 4mL + 5mR + mL
0= 4mL+mL
0= 5mL
L=0

so there is no displacement of the center of the shell along the x-axis so the X-coordinate is still zero

but somehow the anwser is 2R

i don't know what I'm doing wrong please help , is my concept wrong?

 

[DrClaude]

( but i personally doubt that there should be any displacement because the only forces i see are the weight (mg) always directed downwards i.e. y-axis and the normal between the two surfaces of the small sphere and the shell , this normal forces will cancel out each other (Newton's third law) so there will be no forces acting along the X-axis . please correct me if I'm wrong)

If there are no other forces, how does the small sphere follow the inside of the shell?

the distance Z₁ for small sphere from origin is = 5R + L

Be careful with the signs.

 

[Neolight]

So should it be 5R- L??

Because it is the -ve x-axis?

 

[Neolight]

If there are no other forces, how does the small sphere follow the inside of the shell?Be careful with the signs.

I took -ve signs

-5R -L

So
5mR = -4mL -5mR -mL
10mR= -5mL
L= 2R

Hahah I'm such an idiot
Thanks for your help

 

[haruspex]

I took -ve signs

-5R -L

So
5mR = -4mL -5mR -mL
10mR= -5mL
L= 2R

Hahah I'm such an idiot
Thanks for your help

Maybe a bit easier to find the common mass centre (the half-way arrangement, with the ball at the bottom) : (6R-R)m/(4m+m) and double it.

 

[rude man]

You have noted that there are zero net forces along x. More precisely, there is no net external force along the x-axis for the system comprising the shell and the ball.

That being the case, what can you infer about the c.g. of the system? From this you can readily determine the x displacement of the center of the shell, which is decidedly non-zero!
EDIT: looks like I got scooped.

 

[Charles Link]

It is a necessary assumption that the outer ring is stationary when the ball reaches the same vertical height on the left side of the ring, and that the ball actually reaches this full height. This is the result of energy conservation, but to prove this might be quite difficult=perhaps you can make a good argument to prove it... Once you make this assumption, it is a straightforward center of gravity problem.

 

[rude man]

It is a necessary assumption that the outer ring is stationary when the ball reaches the same vertical height on the left side of the ring, and that the ball actually reaches this full height. This is the result of energy conservation, but to prove this might be quite difficult=perhaps you can make a good argument to prove it... Once you make this assumption, it is a straightforward center of gravity problem.

Even if there was friction so the sphere would not reach the starting height, the shell would stop rotating at the high point of the sphere; the c.g is always at the same position along the x-axis so if the sphere stops moving, so does the shell. Of course, the answer would indeed be L < 2R.