1.

The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3kgm^(2), find torque required to bring it to rest in 20s.

Answer»

Solution :Torque `TAU=Ialpha=I((omega_(2)-omega_(1)))/(t)`
`:.tau=(0.3xx(0-10pi))/(20)`
`=-0.15pi=-0.471Nm`


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