A uniform disk (1=1/2 mR2) has a mass of 3 kg and a radius of 2m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 120 rpm (revolutions per minute, convert it to rad/s). A 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. (HINT: Note that some of the information given is not needed! Think which conservation law(s) apply.) The final angular momentum of the system is closest to: L= |Π| ΑΣΦ Submit Request Answer ? kg. m²/s 4

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Chapter1: Units, Trigonometry. And Vectors
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A uniform disk (I=1/2 mR²) has a mass of 3 kg and a radius of 2 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 120 rpm (revolutions per minute, convert it to rad/s). A 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.

(HINT: Note that some of the information given is not needed! Think which conservation law(s) apply.)

The final angular momentum of the system is closest to:

\[ L = \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \]

\[ \text{kg} \cdot \text{m}^2/\text{s} \]

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Transcribed Image Text:A uniform disk (I=1/2 mR²) has a mass of 3 kg and a radius of 2 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 120 rpm (revolutions per minute, convert it to rad/s). A 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. (HINT: Note that some of the information given is not needed! Think which conservation law(s) apply.) The final angular momentum of the system is closest to: \[ L = \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \] \[ \text{kg} \cdot \text{m}^2/\text{s} \] Submit Request Answer
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