S.T. v^2=(2gh)/(1+(K^2)/(R^2)) for the rolling object on an inclined plane of height (h) using dynamical Consideration.

S.T. v^2=(2gh)/(1+(K^2)/(R^2)) for the rolling object on an inclined p

1.	S.T. v^2=(2gh)/(1+(K^2)/(R^2)) for the rolling object on an inclined plane of height (h) using dynamical Consideration.
Answer» Solution :Since energy is conserved `mgh=1/2mv^2+1/2Iomega^2` `mgh1/2mv^2+1/2mK^2(v^2)/(R^2)` `2gh=v^2+(K^2v^2)/(R^2)` or `2gh=v^2(1+(K^2)/(R^2))` where K=radius of GYRATION `v^2=(2gh)/((1+(K^2)/(R^2))` Note : for a CIRCULAR ring, k=R, For a circular disk K`=R/sqrt2`, for a solid cylinder `K=R/sqrt2`, for a hollow cylinder, `K=R`, For a solid sphere K=, and for a hollow sphere `K=(sqrt(2/3))R`