Two discs of moment of inertia i1 and i2

Asked by admin @ 13/08/2021 in Physics viewed by 298 People

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. Question 7.25 System Of Particles And Rotational Motion

Answered by admin @ 13/08/2021

Let L1 and L2 are the angular momentum of two discs.

(A)Total angular momentum of two discs =L1 + L2
= I1w1 + I2w2
=
Total moment of inertia when discs are brought in contact= I1 + I2
Let w is angular speed of two discs system ,
Then, total angular momentum = (I1 + I2)×w

Here , external torque = 0
So, angular momentum is conserved .
I1w1 + I2w2 = (I1 + I2)×w
w = (I1w1 + I2w2)/(I1 + I2)
(B)
Initial K.E of two discs = 1/2×I1w1² + 1/2×I2w2²
Final KE of two discs system = 1/2(I1 + I2)×w²
= 1/2(I1 + I2)×(I1w1+I2w2)²/(I1 + I2)²
= (I1w1 + I2w2)²/2(I1 + I2)
Now, E1 - E2 = 1/2×I1w1² + 1/2×I2w2² - (I1w1 + I2w2)²/2(I1 + I2)
= 1/2 { I1I2/(I1 + I2) } (w1 - w2)²
Which is positive quantity , hence, kinetic energy losses .as torque produced by fractional forces is only inernal ,therefore angular momentum remains conseved .

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