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WITCHER [35]
3 years ago
5

A uniform disk has a mass of 3.5 kg and a radius of 0.5 m. The disk is mounted on frictionless bearings and is used as a turntab

le. The turntable is initially rotating at 50 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?
Physics
1 answer:
Alexandra [31]3 years ago
4 0
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Answer:

<h3>The area of second coil is ≅ 0.025 m^{2}</h3>

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According to the law of electromagnetic induction,

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Since given in question emf of both coil is same so we compare above equation.

    -\frac{N_{1} d\phi _{1}   }{dt_{1} }  = -\frac{N_{2} d\phi _{2}   }{dt_{2} }

   \frac{N_{1} A_{1}   dB_{1}  }{dt_{1} }  = \frac{N_{2} A_{2} dB_{2}     }{dt_{2} }

        A_{2} = \frac{N_{1} A_{1}  }{N _{2}  }

        A_{2} = \frac{41 \times 0.074 }{123  }

        A_{2} = 0.0246 = 0.025 m^{2}

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