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Two disks are rotating about the same axis. Disk A has a moment of inertia of 3.4 kg m^2 and an angular velocity of 7.2 rad/s. Disk B is rotating with an angular velocity of -9.8 rad/s. The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of -2.4 rad/s. The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B

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cjrodriguez89

Answer:

I = 4.4 kg*m^2

Explanation:

  • As no external torques are present, total angular momentum must be conserved, as follows:

       [tex]L_{o} = L_{f} (1)[/tex]

  • The initial angular momentum of the two disks rotating separately, can be written as follows:

       [tex]L_{o} =I_{A} * \omega_{oA} + I_{B} * \omega_{oB} (2)[/tex]

  • Replacing by the givens, we get:

       [tex]L_{o} = 3.4kg*m2 * 7.2rad/s + I_{B} * (-9.8 rad/s) (3)[/tex]

  • The final angular momentum Lf, as the axis of rotation remains the same, is the product of the moment of inertia of both disks rotating as one, and the common angular velocity ωf, as follows:

       [tex]L_{f} = (I_{A} + I_{B}) *\omega_{f} (4)[/tex]

  • Replacing by the givens, we get:

       [tex]L_{f} = (3.4 kg*m2 + I_{B} ) * (-2.4 rad/s) (5)[/tex]

  • From (3) and (5), we can solve for IB, as follows:
  • IB = 4.4 kg*m2

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