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प्रश्न
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 speed ω2 and ω2 are brought into contact face to face with their axes of rotation coincident.
- Does the law of conservation of angular momentum apply to the situation? why?
- Find the angular speed of the two-disc system.
- Calculate the loss in kinetic energy of the system in the process.
- Account for this loss.
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उत्तर
a. As there is no net external torque on the system, the law of conservation of angular momentum may be applied.
Gravitational forces and their normal reactions are external forces, but their net torque is zero, therefore they have no impact.
b. By applying the Law of conversation of angular momentum,
`L_j = L_i`
⇒ `I_ω = I_1ω_1 + I_2ω_2`
I = moment of inertia
ω = angular speed
∴ ω = `(I_1ω_1 + ω_2ω_2)/I` ∴ I = I1 + I2
∴ ω = `(I_1ω_1 + ω_2ω_2)/I`
c. As `KE_f = KE_R + KE_T`
Translational Energy = 0
∴ KET = 0
∴ `KE_f = KE_R = 1/2 Iω^2 = 1/2 (I_1 + I_2) [(I_1ω_1 + I_2ω_2)/(I_1 + I_2)]^2`
`KE_f = 1/2 (I_1ω_1 + I_2ω_2)^2/((I_1 + I_2))`
`KE_i = KE_(1R) + KE_(2R) + KE_(1T) + KE_(2T)`
Because there is no translational motion which in turn results,
KE1T = 0
KE2T = 0
∴ `KE_i = 1/2 I_1ω_1^2 + 1/2 I_2ω_2^2 = 1/2 (I_1ω_1^2 + I_2ω_2^2)`
∴ `ΔKE = KE_f - KE_i = 1/2 (I_1ω_1 + I_2ω_2)^2/((I_1 + I_2)) - 1/2 (I_1ω_1^2 + I_2ω_2^2)`
= `1/2 [(I_1^2ω_1^2 + I_2^2ω_2^2 + 2I_1I_2ω_1ω_2 - [(I_1 + I_2)(I_1ω_1^2 + I_2w_2^2)]]/(I_1 + I_2)]`
= `[([I_1^2ω_1^2 + I_2^2ω_2^2 + 2I_1I_2ω_1ω_2] - [I_1^2ω_1^2 + I_2^2ω_2^2 + I_1I_2ω_1^2 + I_2^2ω_2^2])/(2(I_1 + I_2))]`
= `(-I_1I_2)/(2(I_1 + I_2)) (-2ω_1ω_2 + ω_2^2 + ω_1^2)`
ΔKE = `(-I_1I_2)/(2(I_1 + I_2)) (ω_1 - ω_2)^2 < 0`
d. `K_f < K_i` since energy is wasted due to friction between discs' moving surfaces.
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