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 ω 1 . Another disc of moment of iner

Question

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 ω 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 ω 2 ​ . The energy lost by the entire system is 2 x ω 1 2 ​ ​ where the value of x is

TestHub · Accepted Answer

By the principle of conservation of angular momentum:

$I_{1} ω_{1} + I_{2} (0) = (I_{1} + I_{2}) ω_{2}$

$ω_{2} = \frac{I _{1} ω _{1}}{I _{1} + I _{2}}$

Initial kinetic energy of the system:

$K_{i} = \frac{1}{2} I_{1} ω_{1}^{2}$

Final kinetic energy of the system:

$K_{f} = \frac{1}{2} (I_{1} + I_{2}) ω_{2}^{2}$

Substituting the value of $ω_{2}$ :

$K_{f} = \frac{1}{2} (I_{1} + I_{2}) (\frac{I _{1} ω _{1}}{I _{1} + I _{2}})^{2} = \frac{1}{2} \frac{I _{1}^{2} ω _{1}^{2}}{I _{1} + I _{2}}$

Loss in kinetic energy of the system:

$Δ K = K_{i} - K_{f} = \frac{1}{2} I_{1} ω_{1}^{2} - \frac{1}{2} \frac{I _{1}^{2} ω _{1}^{2}}{I _{1} + I _{2}}$

$Δ K = \frac{1}{2} ω_{1}^{2} (I_{1} - \frac{I _{1}^{2}}{I _{1} + I _{2}})$

$Δ K = \frac{1}{2} ω_{1}^{2} (\frac{I _{1}^{2} + I _{1} I _{2} - I _{1}^{2}}{I _{1} + I _{2}})$

$Δ K = \frac{1}{2} ω_{1}^{2} (\frac{I _{1} I _{2}}{I _{1} + I _{2}})$

Comparing this with the given loss $\frac{x ω _{1}^{2}}{2}$ , we get:

$x = \frac{I _{1} I _{2}}{I _{1} + I _{2}}$

Answer: $\frac{I _{1} I _{2}}{( I _{1} + I _{2} )}$

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