A horizontal disc of moment of inertia 4.25 , | vedclass

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(D) Since there is no external torque on the system of discs,the angular momentum of the system remains constant.$I_1 \omega_1 + I_2 \omega_2 = (I_1 +

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$3 \,rps$ and anti-clockwise

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(D) Since there is no external torque on the system of discs,the angular momentum of the system remains constant.$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega \quad ...(i)$where $I_1$ and $I_2$ are the moments of inertia of the discs,and $\omega_1$ and $\omega_2$ are their angular speeds. $\omega$ is the angular speed of the combination of the discs.Given:$I_1 = 4.25 \,kg \cdot m^2, \omega_1 = 15 \,rps$ (counter-clockwise,taken as positive)$I_2 = 1.80 \,kg \cdot m^2, \omega_2 = -25 \,rps$ (clockwise,taken as negative)Substituting the values into Eq. $(i)$:$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega$$(4.25 	imes 15) + (1.80 	imes -25) = (4.25 + 1.80) \omega$$63.75 - 45 = 6.05 \omega$$18.75 = 6.05 \omega$$\omega = \frac{18.75}{6.05} \approx 3.099 \,rps$Since the result is positive,the direction is counter-clockwise (anti-clockwise).Thus,the new angular velocity is approximately $3 \,rps$ and anti-clockwise.

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