(18 points total) An expert gymnast performs a \"giant swing\" (a complete revol

Question

(18 points total) An expert gymnast performs a "giant swing" (a complete revolution around the high bar). At the top, he comes to a complete stop in the inverted position (a handstand on top of the bar), then lets himself fall under the influence of gravity, rotating around the bar. His body mass is 62 kg, and during the swing his center of mass (with arms extended to grasp the bar) is 1.2 m from the bar. As he rotates downwards, he releases one hand from the bar, holding on only with the other hand. Note: This really has to be seen. There are plenty of YouTube videos of this trick being performed. a. Describe the forces that are acting on him during the swing. Which external forces need to be 2. considered? Draw free-body diagrams for the forces acting on him at the top, at the bottom, and at two other points during the swing. Using the conversion of gravitational potential energy to kinetic energy, calculate his speed at the bottom of the swing. Calculate the maximum force on his hand at the bottom of the swing. Convert the force to pounds. Yes, the force is gigantic. Yet this trick is actually performed, although only at the highest level of gymnastics. Explain how it might be possible for a human being to perform this trick. Cite any sources you use. b. c. d.

Explanation / Answer

a)

force of gravity and normal force from bar acts on him during the swing.

At the top :

force of gravity downward

normal force Upward

At the bottom :

Force of gravity downward

normal force upward

At the horizontal right position :

Force of gravity downward

normal force rightward

At the horizontal left position :

Force of gravity downward

normal force leftward

b)

kinetic energy at the bottom + rotational kinetic energy = Potential energy at the top

(0.5) m v2 + (0.5) I w2 = mgh

(0.5) m v2 + (0.5) (1/3) (mr2) (v/r)2 = mgh

(0.5) m v2 + (0.5) (1/3) m v2 = mgh

(0.5) v2 + (0.5) (1/3) v2 = (9.8 x 1.2 x 2)

v = 6 m/s

c)

at the bottom of swing , force equation is given as

Fn - mg = m v2 /r

Fn - (62 x 9.8) = (62) (6)2 /(1.2)

Fn = 2467.6 N

d)

Fn = 2467.6 N = 554.74 pounds

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