A solid 0.7150-kg ball rolls without slipping down a track toward a loop-the-loop of radius r = 0.9150 m. what minimum translational speed vmin must the ball have when it is a height h = 1.433 m above the bottom of the loop, in order to complete the loop without falling off the track?

Answer: kinetic energy of a sphere is worth 0.7mV^2 at the top of the loop the sphere must have a tangential speed such that V^2/r = g in order to prevent it from falling dawn ... so : V = âšg*r = âš9.8*0.915 = 3.00 m/sec in addition a further potential energy U = m*g*2r must be given to hold the sphere ... this means that the total energy E the sphere must posses at the bottom of the loop is equal to : E = 0.7mV^2+m*g*2r = 0.475 * (0.7*2.455^2+9.8*1.23) = 7.73 joule @ height H = 0.9133 m : E = 7.73 = m*g*H+0.7mVo^2 7.73 = 0.475*9.8*0.9133+0.475*0.7*Vo^2 Vo = âš (7.73 - (0.475*9.8*0.9133)) / (0.475*0.7) = 3.23 m/sec