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A small sphere of radius R is held again...

A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R. The masses of large and small spheres 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 centre of the larger sphere when the smaller sphere reaches the other extreme position.

Text Solution

Verified by Experts

The correct Answer is:
B
KEY CONCEPT : The concept of centre of mass can be
applied in this problem.
When small sphere M change its position to other extreme
position, there is no external force in the horizontal direction.
Therefore the x-coordinate of c.m. will not change.
`[x_(c.m.)]_(initial) = [x_(c.m.)]_(final)`

Thin line of sphere represents intial state, dotted line of
sphere represents final state.
frm(i)
`(x_(c.m.)_(initial) = (x_(c.m.)_(final)`
`rArr(M_1x_1 + M_2x_2)/(M_1+M_2) = (M_1x'_1+M_2x'_2)/(M_1+M_2)`
`rArr (4mxxL+mxx(5R+L)/(4m + m) = (4mxxL'+mxx(L'-5R)/(4m+m)`
`rArr 5L + 5R = 5L' - 5R`
`rArr 5L + 10R = 5L rArr L+2R = L'`
Since, the individual center of mass of the two spheres
has a y co-rodinate zero in its initial state and its final state
therefore the y-coordinate of c.m of the two sphere system
will remain zero.
Therefore the coordinate of c.m. of bigger sphere is (L
+ 2R, 0).
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