A gymnast is swinging on a high bar. The distance between his waist and the bar is 1.2 m.At the top of the swing his speed is momentarily zero. Ignoring friction treating the gymnast as if all his mass is located at his waist,
There is an equation for calculating the time that it will take the gymnast to reach the other side which goes as follows: T = 2Pi X square root(L/G) where L is the distance of the gymnast from the rope which is 1.2m and G is acceleration due to gravity which is 9.8m/s^2
T = 2Pi X square root(1.2m/9.8m/s^2)
T = 2Pi X square root(0.12244897959184s^2)
T = 2pi(.349927106111887s)
T = 2.19865685171s
It is going to take half the time for the gymnast to reach the halfway point thus it will take him 1.09932842585s to reach the halfway point.
I'm going to make a little assumption here that the acceleration will still be 9.8m/s^2 even while falling in a circular pattern.
Thus to get the speed that the gymnast will reach once he is at the bottom, we take the time it will take him to reach it X acceleration since we can convert from acceleration = speed/time to speed = acceleration X time
His speed at the bottom is (9.8m/s^2)(1.09932842585s) = 10.77m/s
Oct 17th, 2014
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