A ring of mass 2.40 kg, inner radius 6.00 cm, and outer radius 8.00 cm rolls (without slipping) up an inclined plane that makes an angle of 36.9°. At the moment the ring is at position x  2.00 m up the plane, its speed is 2.80 m/s. The ring continues up the plane for some additional distance and then rolls back down. It does not roll off the top end. How far up the plane does it go?

Dear Amrit pal

at the given instant total energy of ring is E = 1/2 I w²  + 1/2 mv²

                                                            =1/2 I (V/R)² + 1/2 mV²

here I =1/2 m (R² -r²)

E = 1/4 m[1-(r/R)²]V² + 1/2 mV²

let the ring move L distance in the plane

 increse in potential energy = mgLsin36.9

so from energy balance

1/4 m[1-(r/R)²]V² + 1/2 mV²   = mgLsin36.9

find L from these equation

so total distance move by ring =L+2

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