Question

A gymnast is swinging on a high bar. The distance between his waist and the bar is 1.37 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.

Answer 1

Let V be the speed of the swing at the bottom

Apply the law of conservation of the energy at the top of the swing and at the bottom of the swing assuming the line of waist of the gymnast as the datum line when gymnast at the bottom of the swing (so, at the bottom of the swing potential energy will be zero)

Since at the top of the swing his speed is zero, so Kinetic energy is zero

E_{at top of the swing} = E_{at the bottom of the swing}

(KE+PE)_{at the top of the swing} = (KE+PE)_{at the bottom of the swing}

0+mg*(1.37 + 1.37 ) = (1/2)mV²+0

9.81*2.74 = (1/2)*V²

V² = 53.7588

so, V = 7.332 m/s