A wheel of moment of inertia `0.10 kg-m^2` is rotating about a shaft at an angular speed of `160 (rev)/(min)`. 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 of the second wheel.

Answer

B, D

Explanation

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`