Question 14 of 14 Map Sapling Learning macmillan learning A solid 0.6750-kg ball

Question

Question 14 of 14 Map Sapling Learning macmillan learning A solid 0.6750-kg ball rolls without slipping down a track toward a loop-the-loop of radius R 0.6550 m. What minimum translational speed Vmin must the ball have when it is a height H= 0.9727 m above the bottom of the loop, in order to complete the loop without falling off the track? Number m/ s min figure not to scale Previous e Check Answer Next Exit Y Hint This problem can be solved by applying the principle of energy conservation. The total energy of the system-including gravitational potential energy, translational kinetic energy, and rotational kinetic energy- must remain constant throughout the process Assume that the radius of the ball itself is much smaller than the loop radius R Use g = 9.810 m/s2 for the acceleration due to gravity

Explanation / Answer

The kinetic energy of a sphere = 0.7mV^2
At the top of the loop the sphere must have a tangential speed

V^2/r = g in order to prevent it from falling down
V = g*r = 9.8*0.655 = 2.534 m/sec
In addition a further potential energy U = m*g*2r must be given to hold the sphere

Thus the total energy E the sphere must posses at the bottom of the loop is
E = 0.7mV^2+m*g*2r = 0.675*(0.7*2.534^2+9.8*1.31) = 12.53 joule
At height H = 0.9727 m, the enegy will be
E = 12.53 = m*g*H+0.7mVo^2
12.53 = 0.675*9.8*0.9727+0.675*0.7*Vo^2
Vo = (12.53 -(0.675*9.8*0.9727))/(0.675*0.7) = 3.59 m/sec

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