If the ball is released from rest at a height of 0.77 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track? Assume the ball is a solid sphere of radius 3.8 cm and mass 0.14 kg.
How high does the ball rise on the frictionless side? Please help me!
How high does the ball rise on the frictionless side? Please help me!
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w = v/r = (1/r)√[2gh/(1+k)] where k = 0.4 for a solid sphere.
w = 86.4 rad/sec; v = 3.283 m/s
H = v²/2g = 0.55 m
Note that mass does not matter.....
w = 86.4 rad/sec; v = 3.283 m/s
H = v²/2g = 0.55 m
Note that mass does not matter.....