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System of Particles and Rotational Motion

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Physics Part I

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Physics

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω1 and ω2 are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω1 ≠ ω2.

(a) Moment of inertia of disc I= I

Angular speed of disc I = ω

Moment of inertia of disc II = I

Angular speed of disc II = ω

Angular momentum of disc I, L1 = I1ω

Angular momentum of disc II, L2 = I2ω

Total initial angular momentum Li = I1ω1 + I2ω

When the two discs are joined together, their moments of inertia get added up. 

Moment of inertia of the system of two discs, I = I1 + I

Let ω be the angular speed of the system. 

Total final angular momentum, LT = (I1 + I2ω 

Using the law of conservation of angular momentum, we have

               Li = L

I1ω1 + I2ω2 = (I1 + I2)ω 

∴              ω =fraction numerator straight I subscript 1 straight omega subscript 1 plus space straight I subscript 2 straight omega subscript 2 over denominator straight I subscript 1 space plus space straight I subscript 2 end fraction

(b) Kinetic energy of disc I, E1 = open parentheses 1 half close parentheses I1ω1
Kinetic energy of disc II, E2= open parentheses 1 half close parentheses I2ω2
Total initial kinetic energy, Ei =open parentheses 1 half close parenthesesI1ω12 + I2ω22

When the discs are joined, their moments of inertia get added up. 

Moment of inertia of the system, I = I1 + I

Angular speed of the system = ω 

Final kinetic energy is given by,

      
E= open parentheses 1 half close parenthesesI1 + I2) ω2

          = open parentheses 1 half close parentheses ( I1 + I2open square brackets fraction numerator open parentheses straight I subscript 1 space straight omega subscript 1 space plus space straight I subscript 2 straight omega subscript 2 close parentheses over denominator open parentheses straight I subscript 1 space plus space straight I subscript 2 close parentheses end fraction close square brackets squared 

          =open parentheses 1 half close parentheses  fraction numerator open parentheses straight I subscript 1 space straight omega subscript 1 space plus space straight I subscript 2 straight omega subscript 2 close parentheses squared over denominator open parentheses straight I subscript 1 space plus space straight I subscript 2 close parentheses end fraction 

∴  Ei - Ef 1 half open square brackets fraction numerator straight I subscript 1 straight omega subscript 1 squared space plus straight I subscript 2 straight omega subscript 2 squared space space minus space open parentheses begin display style bevelled 1 half end style close parentheses space open parentheses straight I subscript 1 straight omega subscript 1 space plus space straight I subscript 2 straight omega subscript 2 close parentheses space squared over denominator straight I subscript 1 space plus space straight I subscript 2 end fraction close square brackets
Solving the equation, we get 

          = fraction numerator straight I subscript 1 space straight I subscript 2 space open parentheses straight omega subscript 1 space minus space straight omega subscript 2 close parentheses squared over denominator 2 space left parenthesis straight I subscript 1 space plus space straight I subscript 2 right parenthesis end fraction 

All the quantities on RHS are positive 

Therefore,

 
Ei - Ef > 0 

i.e.,       Ei > E

When the two discs come in contact with each other, there is a frictional force between the two. Hence there would be a loss of Kinetic Energy.
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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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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. 
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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.
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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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