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`        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?`
7 years ago

147 Points
```										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
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
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