Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω₁ and ω₂. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is (a) 1/4I(ω₁-ω₂)² (b) I(ω₁-ω₂)² (c) 1/8I(ω₁-ω₂)² (d) 1/2I(ω₁+ω₂)²

Answer: (a). Conservation of L: Iω₁ + Iω₂ = 2Iω ⇒ ω = (ω₁+ω₂)/(2). Ki = 1/2I(ω₁²+ω₂²), Kf = 1/2(2I)ω² = 1/4I(ω₁+ω₂)². Loss = Ki - Kf = 1/4I(ω₁-ω₂)² So the answer is (a) 1/4I(ω₁-ω₂)².

Two discs of same moment of inertia rotating about their regular axis… | NEET 2017

Q68NEET 2017Medium

Angular Momentum & its Conservation

Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $\omega_1$ and $\omega_2$ . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is

Show answer and solution

Answer: (a)

Conservation of $L$ : $I\omega_1 + I\omega_2 = 2I\omega \Rightarrow \omega = \frac{\omega_1+\omega_2}{2}$ .
$K_i = \frac{1}{2}I(\omega_1^2+\omega_2^2)$ , $K_f = \frac{1}{2}(2I)\omega^2 = \frac{1}{4}I(\omega_1+\omega_2)^2$ .
Loss $= K_i - K_f = \frac{1}{4}I(\omega_1-\omega_2)^2$
So the answer is (a) $\frac{1}{4}I(\omega_1-\omega_2)^2$ .

Report a mistake

Angular Momentum & its Conservation question from NEET 2017

More questions on Angular Momentum & its Conservation