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System of Particles and Rotational Motion

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Physics Part I

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Class 10 Class 12
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Mass per unit area of the original disc = σ 

Radius of the original disc = 

Mass of the original disc, M = πR2σ 

The disc with the cut portion is shown in the following figure,


Radius of the smaller disc = R/2 

Mass of the smaller disc, M = π (R/2)2σ

                                          = π R2σ / 4 

                                           =  M / 4 

Let O and O′ be the respective centres of the original disc and the disc cut off from the original.

As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O′.

It is given that, 

OO′= begin inline style straight R over 2 end style

After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses.

The two masses are, 

M (concentrated at O), and –Mequals straight M over 4concentrated at O′.

(The negative sign indicates that this portion has been removed from the original disc.)

Let x be the distance through which the centre of mass of the remaining portion shifts from point O. 

The relation between the centres of masses of two masses is given as, 

x =open parentheses fraction numerator straight m subscript 1 straight r subscript 1 space plus space straight m subscript 2 straight r subscript 2 over denominator straight m subscript 1 space plus straight m subscript 2 end fraction close parentheses

For the given system,

x =fraction numerator left square bracket straight M space straight x space 0 space minus space straight M apostrophe space straight x space open parentheses begin display style straight R over 2 end style close parentheses over denominator straight M space plus space left parenthesis negative straight M apostrophe right parenthesis end fraction
   = negative straight R over 6

(The negative sign indicates that the centre of mass gets shifted toward the left of point O.) 

The centre of gravity of the resulting flat body is R/6; from the original centre of the body and opposite to the centre of the cut portion.

What is the significance of defining the center of mass of a system?

The motion of n particle system can be reduced to one particle motion.

An equivalent single point object would enable us to discuss the overall motion of the system. 

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. 

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. 

Is it necessary that there should be matter at the centre of mass of system?

No, it is not necessary that there be matter at the centre of mass of the system.

For e.g., if two equal point masses are separated by certain distance, the centre of mass lies at the mid point of two point masses and there is no mass at that point.

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.