Question

A disc of moment of inertia 'I₁' is rotating in horizontal plane about an axis passing through a 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 initial rotating disc is ______.

विकल्प

• `1/2[("I"_1+"I"_2)/("I"_1"I"_2)]omega_1^2`

• `1/2[("I"_1"I"_2)/("I"_1-"I"_2)]omega_1^2`

• `1/2[("I"_1-"I"_2)/("I"_1"I"_2)]omega_1^2`

• `1/2[("I"_1"I"_2)/("I"_1+"I"_2)]omega_1^2`

Answer 1

A disc of moment of inertia 'I₁' is rotating in horizontal plane about an axis passing through a 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 initial rotating disc is `underlinebb(1/2[("I"_1"I"_2)/("I"_1+"I"_2)]omega_1^2)`.

Explanation:

From conservation of angular momentum, as net torque on the system is zero

I₁ω₁ = (I₁ + I₂)ω₂

⇒ `omega_2/omega_1="I"/("I"_1+"I"_2)` 

Energy lost ΔE = E₁ - E₂

= `1/2"I"_1omega_1^2-1/2("I"_1+"I"_2)omega_2^2`

= `1/2omega_1^2["I"_1-("I"_1+"I"_2)omega_2^2/omega_1^2]`

= `1/2omega_1^2["I"_1-("I"_1+"I"_2)"I"_1^2/("I"_1"I"_2)^2]`  ` [∵ omega_2/omega_1="I"_1/("I"_1+"I"_2)]`

= `1/2omega_1^2[("I"_1^2+"I"_1"I"_2-"I"_1^2)/("I"_1+"I"_2)]`

or ΔE = `1/2[("I"_1"I"_2)/("I"_1+"I"_2)]omega_1^2`