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 a

1.	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.
Answer» 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 0.10x160+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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