A uniform ring rolls down an inclined plane without slipping. If it re – InterviewSolution

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1.	A uniform ring rolls down an inclined plane without slipping. If it reaches the bottom of a speed of `2m//s`, then calculate the height of the inclined plane (use `g=10m//s^(2)`)
Answer» `h=(v^(2))/(2g)(1+(k^(2))/(R^(2)))` `M.I.` of ring `=MR^(2)`, `K=R` `h=(v^(2))/(2g)(1+(k^(2))/(R^(2)))=(v^(2))/(2g)xx2=(v^(2))/(g)=(2xx2)/(10)=0.4m` Hint : `h=(v^(2))/(2g)(1+(k^(2))/(R^(2)))`

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