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Two disks are rotating about the same axis. Disk A has a moment of inertia of 3.2 kg · m2 and an angular velocity of +7.4 rad/s. Disk B is rotating with an angular velocity of -9.7 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 -2.3 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?

Answer :

MechEngineer

Answer:

[tex]4.2 kgm^2[/tex]

Step-by-step explanation:

As there's no external force when the disks are attached, the angular momentum of the system should be conserved.

[tex] I_A\omega_A + I_B\omega_B = (I_A + I_B)\omega[/tex]

[tex] 3.2*7.4 + I_B(-9.7) = (3.2 + I_B)(-2.3)[/tex]

[tex] 23.68 - 9.7I_B = -7.36 - 2.3I_B[/tex]

[tex]31.04 = 7.39I_B[/tex]

[tex]I_B = 31.04/7.39 = 4.2 kgm^2[/tex]

The moment of inertia of disk B when it rotates round the axis will be 4.2kgm².

How to calculate the moment of inertia?

Let the moment of inertia be represented by Ib.

Therefore, the moment of inertia will be calculated thus:

IaWa + IbWb = (Ia + Ib)w

(3.2 × 7.4) + Ib(-9.7) = (3.2 + In)(-2.3)

23.68 - 9.7Ib = -7.36 - 2.3Ib

Collect like terms

23.58 + 7.36 = -2.3Ib + 9.7Ib

31.04 = 7.39Ib

Divide through by 7.39

7.39Ib/7.39 = 31.04/7.39

= 4.2kgm²

Therefore, the moment of inertia of disk B when it rotates round the axis will be 4.2kgm².

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