Question

Two discs of moments of inertia I₁ and I₂ about the 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 axis 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.

Answer 1

it is given, Moment of inertia of 1st disc = I1
Angular speed of 1st disc = ω1
So,  Angular momentum of 1st disc , L1 = I1ω1

Moment of inertia of 2nd disc = I2
Angular speed of 2nd disc = ω2
so,  Angular momentum of disc 2, L2 = I2ω2

⇒ Total initial angular momentum, Li = L1 + L2 = I1ω1 + I2ω2

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 + I2

Let ω is the angular speed of the system.
Let total final angular momentum, Lf = (I1 + I2) × ωf

Using the law of conservation of angular momentum, we have:
Li = Lf
⇒ I1ω1 + I2ω2 = (I1 + I2) × ωf
⇒ ωf = (I1ω1 + I2ω2) / (I1 + I2) ...…(i)

(b) Kinetic energy of 1st disc, K.E1  = 1/2 I1ω1²
Kinetic energy of 2nd disc , K.E2 = 1/2 I2ω2²
⇒ Total initial kinetic energy, K.Ei = 1/2 I1ω1² + 1/2 I2ω2² ...…(ii)


Angular speed of the system = ωf
Final kinetic energy, K.Ef = 1/2 (I1 + I2)ωf²

Substituting the value of ω from equation (i)
⇒ K.Ef = 1/2 (I1 + I2) {(I1ω1 + I2ω2) / (I1 + I2)}2 …(iii)

From equations (ii) and (iii),

K.Ei- K.Ef = [1/2 I1ω12 + 1/2 I2ω22] - 1/2 (I1 + I2) {(I1ω1 + I2ω2) / (I1 + I2)}2
By solving the equation we get,
K.Ei- K.Ef = I1I2(ω1-ω2)2/ 2(I1 + I2)

Since all Quantities on RHS are positive so we conclude that,
Ei-Ef> 0
i.e. Ei> Ef
The loss can be attributed to the frictional forces that arise when discs come into contact with each other.

Answer 2

Answer:

We have , L'=\Sigma r'_i\times P'_iL

′

=Σr

i

′

×P

i

′

Differentiating both sides with respect to time,

\begin{gathered}\frac{dL'}{dt}=\frac{d}{dt}\left(\Sigma r'_i\times p'_i\right)\\\\=p'_i\times\Sigma\frac{dr'_i}{dt}+\Sigma r'_i\times\frac{dp'_i}{dt}\\\\=\Sigma m_ir'_i\times v'_i+\Sigma r'_i\times\frac{dp'_i}{dt}\end{gathered}

dt

dL

′

=

dt

d

(Σr

i

′

×p

i

′

)

=p

i

′

×Σ

dt

dr

i

′

+Σr

i

′

×

dt

dp

i

′

=Σm

i

r

i

′

×v

i

′

+Σr

i

′

×

dt

dp

i

′

Where r'_ir

i

′

is the position vector with respect to centre of mass of system of particles.

But from definition of centre of mass,

\Sigma m_ir'_i=0Σm

i

r

i

′

=0

so, \frac{dL'}{dt}=\Sigma r'_i\times\frac{dp'_i}{dt}

dt

dL

′

=Σr

i

′

×

dt

dp

i

′

[ hence, proved]

We know, \bf{\tau=r\times\frac{dv}{dt}}τ=r×

dt

dv

So, \tau_{ext}=\Sigma r'_i\times\frac{dp'_i}{dt}τ

ext

=Σr

i

′

×

dt

dp

i

′

And hence, \frac{dL'}{dt}=\tau_{ext}

dt

dL

′

=τ

ext