a small sphere of radius a is cut from homogeneous sphere of radius r. find the position of centre of mass of the remaining part with the respect to the centre of mass of the original sphere.
.
.
.
riya13128
Answers
Explanation:
Let mass per unit volume =\sigmaσ
Mass of bigger sphere=\sigma \left( \dfrac { 4 }{ 3 } \pi { R }^{ 3 } \right) σ(
3
4
πR
3
) at O.
Mass of smaller sphere=\sigma \left( \dfrac { 4 }{ 3 } \pi { a }^{ 3 } \right) σ(
3
4
πa
3
) at center
X_{com}X
com
= \dfrac { \sigma \left( \dfrac { 4 }{ 3 } \pi { R }^{ 3 } \right) \times 0-\sigma \left( \dfrac { 4 }{ 3 } \pi { a }^{ 3 } \right) b }{ \sigma \left( \dfrac { 4 }{ 3 } \pi { R }^{ 3 } \right) -\sigma \left( \dfrac { 4 }{ 3 } \pi { a }^{ 3 } \right) \\ }
σ(
3
4
πR
3
)−σ(
3
4
πa
3
)
σ(
3
4
πR
3
)×0−σ(
3
4
πa
3
)b
=\dfrac { -{ a }^{ 3 }b }{ { R }^{ 3 }-{ a }^{ 3 } } =
R
3
−a
3
−a
3
b
directed back away from O to center of small sphere.
Answer:
The centre of mass of the remaining part w.r.t. the centre of mass of the original sphere is x = r/14 units.
Explanation:
Let us assume a sphere of radius 'r'. Then, its mass will be 'M'. Now, let us cut out a small sphere of radius 'a' which equals r/2 from the original sphere.
We know that, volume of sphere = 4/3 πr³. So, the mass of small sphere will be 'M/8'. Also, the remaining mass of the original sphere after subtracting the small sphere will be 7M/8. So, let the new centre of mass for the remaining sphere be at a distance of 'x' from the original centre of mass.
[Refer to the attached image for the systematic representation of the given case]
Now assuming the two spheres to be in a two particle system. Then, from moment of mass we have;-
Mass₁ × Distance₁ = Mass₂ × Distance₂
On putting the given values, we get;-
7M/ 8 × x = M/8 × r/2
x = r / 2 × 7
x = r / 14 units.
Hence, the new centre of mass of the remaining part w.r.t. the centre of mass of the original sphere is x = r/14 units.
Hope it helps! ;-))
