A uniform disc of mass 100 kg and radius 2 | vedclass

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(A) Since there is no external torque acting on the system, the angular momentum of the system is conserved.$L_i = L_f$$I_i \omega_i = I_f \omega_f$In

Vedclass · Accepted Answer

$0.77$

Vedclass · Answer

(A) Since there is no external torque acting on the system, the angular momentum of the system is conserved.$L_i = L_f$$I_i \omega_i = I_f \omega_f$Initially, the boy is at the center, so his moment of inertia is zero. The moment of inertia of the disc is $I_{	ext{disc}} = \frac{1}{2} M R^2$.$I_i = \frac{1}{2} 	imes 100 	imes 2^2 = 200 \ kg \cdot m^2$.Finally, the boy is at a distance $r = 1 \ m$ from the center. His moment of inertia is $I_{	ext{boy}} = m r^2 = 60 	imes 1^2 = 60 \ kg \cdot m^2$.The final moment of inertia of the system is $I_f = I_{	ext{disc}} + I_{	ext{boy}} = 200 + 60 = 260 \ kg \cdot m^2$.Using conservation of angular momentum:$200 	imes 1 = 260 	imes \omega_f$$\omega_f = \frac{200}{260} = \frac{20}{26} \approx 0.77 \ rad/s$.

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