A very small solid sphere is set rolling on a rough horizontal track which becomes circular after certain distance. Radius of the circular portion of the track is R which is large compared to radius r of the sphere. Ifaim is that sphere should complete the circle successfully then minimum required speed at the bottom is found to be sqrt((27gR)/(n)) Calculate .

A very small solid sphere is set rolling on a rough horizontal track w

1.	A very small solid sphere is set rolling on a rough horizontal track which becomes circular after certain distance. Radius of the circular portion of the track is R which is large compared to radius r of the sphere. Ifaim is that sphere should complete the circle successfully then minimum required speed at the bottom is found to be sqrt((27gR)/(n)) Calculate .
Answer» SOLUTION : Situation described in the question is as shown in the figure. When sphere reaches the top point of circle then normal reaction and weight P of the body acta vertically downwards ASSUMING v to be the speed at the highest point we can write the following equation: `N+Mg=(mv^(2))/(R)` Here note that radius of circular motion corresponding to centre of sphere is taken to be R because it is large in comparison to radius of sphere. To calculate minimum speed at the bottom we can assume critical most situation at the top point when normal reaction becomes zero (N=0) `rArr 0+mg+(mv^(2))/(R) rArr v= sqrt(gR) ""...(i)` Let us assume that corresponding speed at the bottom is w. We can apply conservation of mechanical energy because sphere performs pure rolling Loss in K.E.Gain in gravitational potential energy `(7)/(10) "mu"^(2)-(7)/(10) mv^(2)= mg (2R)` Substituting from equation (1) we get the following `(7)/(10) "mu"^(2)-(7)/(10) m ( sqrt(gR))^(2)= mg (2R)` `u= sqrt((27 gR)/(7))` On comparing with the GIVEN RESULT we get n=7.