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  • A circular disc of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed <img alt="\omega_{\text{i}}" src="https://entrancecorner

A circular disc of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed \omega_{\text{i}}. Another disc of moment of inertia Ib is dropped coaxially onto the rotating disc. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed \omega_{\text{f}}. The energy lost by the initially rotating disc to friction is

Option: 1

\frac{1}{2}\frac{\text{I}_{\text{b}}\text{I}_{\text{t}}}{(\text{I}_{\text{t}}+\text{I}_{\text{b}})}\omega_{\text{i}}^{2}


Option: 2

\frac{1}{2}\frac{\text{I}_{\text{b}}^{2}}{(\text{I}_{\text{t}}+\text{I}_{\text{b}})}\omega_{\text{i}}^{2}


Option: 3

\frac{1}{2}\frac{\text{I}_{\text{t}}^{2}}{(\text{I}_{\text{t}}+\text{I}_{\text{b}})}\omega_{\text{i}}^{2}


Option: 4

\frac{\text{I}_{\text{b}}-\text{I}_{\text{t}}}{(\text{I}_{\text{t}}+\text{I}_{\text{b}})}\omega_{\text{i}}^{2}


Answers (1)

best_answer

Clearly, there is no external force or torque acting on the system, then we can say that, then angular moment of the first and second cases are equal and conserved.

\begin{aligned} & \Longrightarrow L_i=L_f \\ & \Longrightarrow I_t \omega_i=\left(I_t+I_b\right) \omega_f \\ & \Longrightarrow \omega_f=\frac{I_t \omega_i}{I_t+I_b} \end{aligned}

Then, we know that the energies of the system is given as K . E=\frac{1}{2} I \omega^2

Then, here the loss in energy is given as \Delta E=E_1-E_2

\begin{aligned} & \Longrightarrow \Delta E=\frac{1}{2} I_t \omega_i^2-\frac{1}{2}\left(I_t+I_b\right) \omega_f^2 \\ & \Longrightarrow \Delta E=\frac{1}{2} I_t \omega_i^2-\frac{1}{2}\left(I_t+I_b\right)\left(\frac{I_t \omega_i}{I_t+I_b}\right)^2 \\ & \Longrightarrow \Delta E=\frac{1}{2} I_t \omega_i^2\left[1-\frac{I_t}{I_t+I_b}\right] \\ & \therefore \Delta E=\frac{1}{2} I_t \omega_i^2\left[\frac{I_b}{I_t+I_b}\right] \end{aligned}

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Ritika Harsh

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