Question

From a sphere of radius 2m, a sphere of radius 1 m is removed from the edge.The shift in the C.M. is​

Answer 1

Correct option is

D

14

1

m

Let us consider m1 as sphere of 0.5 radius and m2 as the remaining portion of sphere after removing m1

Let origin be at centre of sphere of radius 1m

x1 and x2 be the x components of centre of masses m1 and m2 respectively

x1 =0.5

x2=x

Let density of sphere be d

m1 = d

3

4

×π×0.5

3

m2=d

3

4

×π×(1

3

−0.5

3

)

centre of mass of m1 and m2 is at origin

0=d

3

4

×π×(0.5

3

)×0.5+d

3

4

×π×(1

3

−0.5

3

)×x

x=-

14

1

m

Answer 2

Given info : From a sphere of radius 2m, a sphere of radius 1 m is removed from edge.

To find : the shift in the centre of mass is ...

solution : let the centre of mass of sphere of radius 2 m is (0, 0). The mass of sphere is M.

now mass of sphere of radius 1 m = $\frac{M}{\text{volume of sphere of radius 2m}}\times\text{volume of sphere of radius 1m}$

= $\frac{M}{\frac{4}{3}\pi(2)^3}\times\frac{4}{3}\pi(1)^3$

= M/8

see diagram, centre of mass of sphere of radius 1 m = (1/2 , 0)

now it is removed from the sphere of radius 2m.

so new centre of mass , C.M' = $\left(\frac{M\times0-\frac{M}{8}\times\frac{1}{2}}{M-\frac{M}{8}},0\right)$

= (-1/6 , 0)

Therefore the shift in the centre of mass = 0 - (-1/6) = 1/6 m along negative x - axis when we removed the sphere from positive x - axis.