The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is 'I'. It is rotating with angular velocity 'ω'. Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss in kinetic energy is

Question

The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is 'I'. It is rotating with angular velocity 'ω'. Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss in kinetic energy is (A) (I ω2/2) (B) (I ω2/4) (C) (I ω2/6) (D) (I ω2/8)

Tardigrade Admin · Accepted Answer

According to the law of conservation of angular momentum,

I w =

constant
Now, according to the question,

I_{1} ω_{1} = I_{2} ω_{2} or I w = (2 l) ω_{2}

ω_{2} = \frac{ω}{2}

New kinetic energy

= [\frac{1}{2} I_{2} ω_{2}^{2}] = \frac{1}{2} (2 I) \times (\frac{ω}{2})^{2} = \frac{l w ^{2}}{4}

Loss in kinetic energy

(K_{L}) = K_{i} - K_{j}

= \frac{1}{2} I w^{2} - \frac{I w ^{2}}{4} = \frac{I w ^{2}}{4}

$\frac{I ω ^{2}}{2}$

$\frac{I ω ^{2}}{4}$

$\frac{I ω ^{2}}{6}$

$\frac{I ω ^{2}}{8}$

2Iω2​

4Iω2​

6Iω2​

8Iω2​

According to the law of conservation of angular momentum, Iw= constant Now, according to the question, I1​ω1​=I2​ω2​ or Iw=(2l)ω2​ ω2​=2ω​ New kinetic energy =[21​I2​ω22​]=21​(2I)×(2ω​)2=4lw2​ Loss in kinetic energy (KL​)=Ki​−Kj​ =21​Iw2−4Iw2​=4Iw2​

$\frac{I ω ^{2}}{2}$

$\frac{I ω ^{2}}{4}$

$\frac{I ω ^{2}}{6}$

$\frac{I ω ^{2}}{8}$