A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 30 rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20 s until it acquires the same final angular velocity as the turntable. What is the final angular momentum of the system? A) 0.93 kg∙m²/s B) 1.1 kg∙m²/s C) 1.3 kg∙m²/s D) 1.6 kg∙m²/s E) 1.9 kg∙m²/s

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Quizplus · Accepted Answer

The final angular momentum of the system is the sum of the angular momenta of the disk and the hollow cylinder. The disk has an angular momentum of:$$L_d = I_d \omega_d = \frac{1}{2}MR^2 \omega_d$$where:* $I_d$ is the moment of inertia of the disk* $M$ is the mass of the disk* $R$ is the radius of the disk* $\omega_d$ is the angular velocity of the diskThe hollow cylinder has an angular momentum of:$$L_c = I_c \omega_c = MR^2 \omega_c$$where:* $I_c$ is the moment of inertia of the hollow cylinder* $M$ is the mass of the hollow cylinder* $R$ is the radius of the hollow cylinder* $\omega_c$ is the angular velocity of the hollow cylinderThe final angular momentum of the system is:$$L = L_d + L_c = \frac{1}{2}MR^2 \omega_d + MR^2 \omega_c$$$$L = \frac{1}{2}(3.7 	ext{ kg})(0.40 	ext{ m})^2(30 	ext{ rpm}) + (3.7 	ext{ kg})(0.40 	ext{ m})^2(\frac{30 	ext{ rpm}}{2})$$$$L = 0.93 	ext{ kg}\cdot	ext{m}^2/	ext{s}$$

A Uniform Disk Has a Mass of 3