Two fly wheels A and B are mounted side by side with frictionless bearings 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

Two fly wheels A and B are mounted side by side with frictionless bear

1.	Two fly wheels A and B are mounted side by side with frictionless bearings 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» `2 SQRT5` rps 0.5 rps 2 rps 1 rps SOLUTION :According to the law of CONSERVATION of angular MOMENTUM , `(5 xx 10) + (20 xx 0) = (5 + 20) omega or omega= (50)/(25) rps = 2rps`