1.

Two fly wheels A and B are mounted side by side with frictionless bearing on a common shaft. Their moments of inertia about the shaft are 5.0 kg m^(2) and 20.0 kg m^(2) respectively. Wheel A is made to rotate at 10 revolution per second. Wheel B, initially stationary is now coupled to A with the help of a clutch. The rotation speed of the wheels will become

Answer»

`2sqrt(5)` rps
0.5 rps
2 rps
None of these

Solution :Here applying law of CONSERVATION of angular momentum ,
`5xx10+20xx0=(5+20)omega`
`THEREFORE omega=(50)/(25)=2r.p.s`


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