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Hanuman Aadil Maharaj
Asked: 2022-11-10T08:34:45+05:30 In:

A wheel of moment of inertia `0.10 kg-m^2` is rotating about a shaft at an angular speed o 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with as common angular speed of 200 rev/minute. Find the moment of inertia o the second wheel.

A wheel of moment of inertia `0.10 kg-m^2` is rotating about a shaft at an angular speed o 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with as common angular speed of 200 rev/minute. Find the moment of inertia o the second wheel.
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  1. Correct Answer – B::D
    Wheel 1 has : `I_1=10kg-m^2`
    `omega_1=160 rev/min
    Wheel 2 has `I_2=?`
    `omega_2=300 rev/min`
    Given that after they are coupled
    `=omega =200 rev/min `
    Theredore if we take the two wheels to be an isolated system.
    Totla exterN/Al torque =0
    Therefore `I_1omega_1+I_2omega_2=(I_1+I_2)omega`
    `gt 0.10×160+I_2xx300=(0.10+I_2)xx200`
    `rarr 16+300I-2=20+200I_2` ltbr. `rarr 100I_2=4`
    `rarr I_2=4/100=0.04kg-m^2`

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