Question

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 spheres is now released. Find the co-ordinates of the centre of the larger sphere when the smaller sphere reaches the other extreme position.

Answer 1

As no external force is applied in the x-direction so position of center of mass will be constant.

For finding center of mass of complex system we use the following formulae

$X_0=\frac{X_1{M}_1+X_2{M}_2}{M_1+M_2}$

when the smaller sphere reaches the extreme left position x co-ordinates of center of mass is compared initial position of center of mass.

$\frac{\left(4M\right)(L+x)+\left(M\right)(L+x-5R)}{4M+M}=\frac{\left(4M\right)\left(L\right)+\left(M\right)(L+5R)}{4M+M}$

$4ML+4Mx+ML+Mx-5MR=4ML+ML+5MR$

$5Mx=10MR$

$x=2R$

their is no change in position of center of mass in y direction so final co-ordinates of center of large sphere is (L+2R , 0).