A disc of moment of inertia I₁ is rotating in horizontal plane about an axis passing through centre and perpendicular to its plane with constant angular speed ω₁. Another disc of moment of inertia I₂ having zero angular speed is placed co-axially on a rotating disc. Now, both the discs are rotating with constant angular speed ω₂. The energy lost by the entire system is (xω₁²)/(2) where the value of x is (a) [((I₁+I₂))/(I₁I₂)] (b) (I₁I₂)/((I₁-I₂)) (c) ((I₁-I₂))/(I₁I₂) (d) (I₁I₂)/((I₁+I₂))

Answer: (d). I₁ω₁=(I₁+I₂)ω₂⇒ω₂=(I₁ω₁)/(I₁+I₂) Loss =12I₁ω₁²-12(I₁²ω₁²)/(I₁+I₂)=(ω₁²)/(2)·(I₁I₂)/(I₁+I₂) So x=(I₁I₂)/(I₁+I₂), answer (d).

A disc of moment of inertia I₁ is rotating in horizontal plane about… | MHT CET 2026 (14 May, Shift 1)

Q13MHT CET 2026 · 14 MAY · SHIFT 1Medium

Angular Momentum & its Conservation

A disc of moment of inertia $I_1$ is rotating in horizontal plane about an axis passing through centre and perpendicular to its plane with constant angular speed $\omega_1$ . Another disc of moment of inertia $I_2$ having zero angular speed is placed co-axially on a rotating disc. Now, both the discs are rotating with constant angular speed $\omega_2$ . The energy lost by the entire system is $\frac{x\omega_1^2}{2}$ where the value of $x$ is

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Answer: (d)

$I_1\omega_1=(I_1+I_2)\omega_2\Rightarrow\omega_2=\frac{I_1\omega_1}{I_1+I_2}$
Loss $=\frac12I_1\omega_1^2-\frac12\frac{I_1^2\omega_1^2}{I_1+I_2}=\frac{\omega_1^2}{2}\cdot\frac{I_1I_2}{I_1+I_2}$
So $x=\frac{I_1I_2}{I_1+I_2}$ , answer (d).

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Angular Momentum & its Conservation question from MHT CET 2026 (14 May, Shift 1)

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