What is moment of momentum called?

Angular momentum.
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What is the need of centre of mass?


Newton’s second law of motion is strictly applicable to point masses only. To apply the Newton's law of motion to rigid bodies, the concept of centre of mass is introduced.

The concept of centre of mass of a system enables us to discuss overall motion of the system by replacing the system by an equivalent single point object. 
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Define centre of mass.

Centre of mass of a body or a system of bodies is a point at which the entire mass of the body or system is supposed to be concentrated. 
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Is it necessary for centre of mass to lie within the body?

No, centre of mass needs not to lie within the body. It is not necessary that the total mass of the system be actually present at the centre.

The position of the centre of mass is calculated using the usual Newtonian type of equations of motion. 
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Center of mass of a system of three particles of masses 2kg, 4kg and 6kg is at P ↔ (-2,0,4). If 6kg mass is removed from system, the center of mass of system shifts to Q ↔ (1,2,4). Find the position co-ordinates of 6kg mass.

Consider, stack straight r subscript 1 with rightwards harpoon with barb upwards on top space comma space stack r subscript 2 with rightwards harpoon with barb upwards on top space a n d stack space r subscript 3 with rightwards arrow on top space be the position vectors of masses 2 kg, 4 kg and 6 kg respectively. The position coordinates of the centre of mass of 2 kg, 4 kg and 6 kg is (-2, 0, 4). 

Therefore position vector of the centre of mass of 2  kg, 4 kg and 6 kg is, 

R= -2straight i with hat on top + 0 straight j with hat on top + 4 straight k with hat on top 
Also, 

straight R with rightwards harpoon with barb upwards on top subscript 1 space equals space fraction numerator 2 space stack r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack r subscript 2 with rightwards harpoon with barb upwards on top space plus space 6 stack space r subscript 3 with rightwards arrow on top over denominator 2 space plus space 4 space plus space 6 end fraction 

Therefore, 

fraction numerator 2 space stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack straight r subscript 2 with rightwards harpoon with barb upwards on top space plus 6 stack space straight r subscript 3 with rightwards harpoon with barb upwards on top over denominator 2 space plus space 4 space plus space 6 end fraction space equals space minus space 2 space straight i with rightwards arrow on top space plus space 0 space straight j with hat on top space plus space 4 space straight k with hat on top space 
Therefore,

2 space stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack straight r subscript 2 with rightwards harpoon with barb upwards on top space plus space 6 space stack straight r subscript 3 with rightwards harpoon with barb upwards on top space equals space minus space 24 space straight i with hat on top space plus space 0 space straight j with hat on top space plus space 48 space straight k with hat on top   ... (1) 

The position coordinates of centre of mass of 2kg and 4kg is (1, 2, 4).

Therefore position vector of centre of mass of 2kg and 4 kg  is, 

stack straight R subscript 2 with rightwards harpoon with barb upwards on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 4 space straight k with hat on top space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space 
Also,

stack R subscript 2 space with rightwards harpoon with barb upwards on top space equals space fraction numerator 2 space stack r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack r subscript 2 with rightwards harpoon with barb upwards on top over denominator 2 space plus space 4 end fraction space 

therefore space fraction numerator 2 space stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack straight r subscript 2 with rightwards harpoon with barb upwards on top over denominator 2 space plus space 4 end fraction space equals space straight i with hat on top space plus space 2 straight j with hat on top space plus space 4 space straight k with hat on top space

2 space stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack straight r subscript 2 with rightwards harpoon with barb upwards on top space equals space 6 space space straight i with hat on top space plus space 12 space straight j with hat on top space plus space 24 space straight k with hat on top space space space space space space space space... space left parenthesis 2 right parenthesis 

From (2) from (1), we get

6 space stack straight r subscript 3 with rightwards harpoon with barb upwards on top space equals space minus space 30 space straight i with hat on top space minus space 12 space straight j with hat on top space plus space 24 space straight k with hat on top

stack straight r subscript 3 with rightwards harpoon with barb upwards on top space equals space minus 5 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 4 space straight k with hat on top
 
The position vector of 6kg mass is,

stack straight r subscript 3 with rightwards harpoon with barb upwards on top space equals space minus space 5 space i with hat on top space minus space 2 space j with hat on top space plus space 24 space k with hat on top

Therefore, position coordinate of 6 kg is ( -5, -2, 4).


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