Consider a disc of mass 5 ,kg and radius 2 ,m | vedclass

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(D) The moment of inertia of the disc is $I = \frac{1}{2} M R^2 = \frac{1}{2} 	imes 5 	imes (2)^2 = 10 \,kg \cdot m^2$.The initial angular momentum

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(D) The moment of inertia of the disc is $I = \frac{1}{2} M R^2 = \frac{1}{2} 	imes 5 	imes (2)^2 = 10 \,kg \cdot m^2$.The initial angular momentum is $L_i = I \omega_i = 10 	imes 10 = 100 \,kg \cdot m^2/s$.The initial kinetic energy is $E_i = \frac{1}{2} I \omega_i^2 = \frac{1}{2} 	imes 10 	imes (10)^2 = 500 \,J$.When an identical disc is placed on top, the final moment of inertia becomes $I_f = I + I = 2I = 20 \,kg \cdot m^2$.By the conservation of angular momentum, $L_i = L_f$, so $100 = I_f \omega_f = 20 \omega_f$.Thus, the final angular velocity is $\omega_f = \frac{100}{20} = 5 \,rad/s$.The final kinetic energy is $E_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} 	imes 20 	imes (5)^2 = 10 	imes 25 = 250 \,J$.The energy dissipated is $\Delta E = E_i - E_f = 500 \,J - 250 \,J = 250 \,J$.

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