A gymnast is swinging on a high bar. The distance between his waist and the bar is 1.1 m, as the drawing shows. At the top of the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.
To solve this problem, we will use the principle of conservation of mechanical energy. At the top of the swing, the gymnast's potential energy will be converted into kinetic energy at the bottom of the swing.
The potential energy at the top of the swing will be equal to the kinetic energy at the bottom of the swing.
At the top of the swing:
Potential energy = mgh = mgh_max
Kinetic energy = 0
At the bottom of the swing:
Potential energy = 0
Kinetic energy = 1/2mv^2
Therefore, mgh_max = 1/2mv^2
We can cancel out the mass "m" from both sides of the equation:
gh_max = 1/2v^2
Now, we need to find the height at the top of the swing. This can be done by using trigonometry:
h_max = L - Lcosθ
= L(1 -cosθ)
= 1.1(1 - cos180°)
= 1.1(1 - (-1))
= 2.2 m
Therefore, substituting this value into the equation:
9.81*2.2 = 1/2v^2
v^2 = 9.81*2.2*2
v = sqrt(9.81*2.2*2)
v ≈ 6.26 m/s
Therefore, the speed of the gymnast at the bottom of the swing is approximately 6.26 m/s.