A homogeneous sphere starts from rest at the upper end of the track shown in the

Question

A homogeneous sphere starts from rest at the upper end of the track shown in the figure and rolls without slipping until it rolls off the right-hand end. If H = 53.1 m and h = 10.5 m and the track is horizontal at the right-hand end, how far horizontally from point A does the sphere land on the floor?

Explanation / Answer

m*g(H-h)= 0.5*m*v^2 +0.5*I*w^3 [ I is moment of inertia of sphere , w is angular speed] I=0.4*M*r^2 =>m*g(H-h)= 0.5*m*v^2 + 0.5*0.4*m*r^2 *v^2/r^2 [w=v/r] g(H-h)=0.5*v^2 + 0.2*v^2 g(H-h)=0.7*v^2 v=sqrt(14(H-h)) now h=0.5*g*t^2 =>time for which sphere flies = Sqrt(2*g*h) Since there is no acceleration in horizontal direction distance travelled by sphere = Sqrt(28gh(H-h)) =[28gh(H-h)]^0.5 =350.341m

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