AIPMT 2010 Physics and Chemistry Paper

As no external torque is applied to the system, the angular momentum of the system remains conserved.
∴‌‌L1=L1
According to given problem,
I1ω=(I1+Ib)ω1
or ωf=‌

Ifωi

(It+Ib)

......(i)
Initial energy, Ei=‌

1

2

Itωi2...(ii)
Final energy, Ef=‌

1

2

(It+Ib)ωf2 ...(iii)
Substituting the value of ωf from equation (i) in equation (iii) we get
Final energy, E1=‌

1

2

(I1+Ib)(‌

Itωi

I1+Ib

)2

=‌

1

2

‌

It2ωi2

(I1+Ib)

....(iv)
Loss of energy, ∆E=E−E1
=‌

1

2

Itωi2−‌

1

2

‌

I12ωi2

(I1+Ib)

(Using (ii) and (iv))
=‌

ωi2

2

(It−‌

It2

(I1+Ib)

)=‌

ωi2

2

(‌

I12+IbI1−I12

(It+Ib)

)
=‌

1

2

‌

IbIt

(It+Ib)

ωi2