How Does the Conservation of Energy Principle Apply to a Gymnast's Swing?

[tutojean]

Homework Statement

Example: A 50 kg gymnast does giant circles around a horizontal bar. At the top of her swing her center of mass is 1.0 m above the bar traveling 1.2 m/s. what is the speed of her center of mass at the bottom of her swing where her center at mass is 1.0 m below the bar? ( No given diagram)

Homework Equations

[K + Eg + Es] = [K + Eg + Es + Ef]
Initial Final

K=1/2mv^2
Eg=mgh
Es= 1/2 kx^2
Ef= uND

The Attempt at a Solution

What I did: [K] = [K + Eg]
or either [ Eg] = [K}. Please help me with setting this problem up. The first set up didn't work for me and I'm just needing help setting the problem up. No need for figuring out the problem. Thank You

 

[rock.freak667]

If we take the bar she is swinging from as having 0 potential and kinetic energy, that the top of her swing, what types of energy does he have and what the total energy she has due to these two types?

 

[tomcue]

I would first clarify the given information and assumptions. From the given information, we can assume that the gymnast's motion is in a vertical plane and that the horizontal bar is a pivot point for her circular motion. We can also assume that the gymnast's mass remains constant throughout her motion.

Next, we can apply the conservation of energy principle to this problem. This principle states that the total energy of a system (in this case, the gymnast) remains constant, and can only be transferred between different forms (kinetic, potential, elastic) but cannot be created or destroyed.

Using the given equations for kinetic energy (K), gravitational potential energy (Eg), elastic potential energy (Es), and non-conservative energy (Ef), we can set up the equation as follows:

Initial energy (at the top of the swing) = Final energy (at the bottom of the swing)

[Ki + Egi + Esi] = [Kf + Egf + Esf + Eff]

Since the center of mass remains at the same height (1.0 m) throughout the motion, we can eliminate the gravitational potential energy terms from the equation. Also, since no information is given about the elasticity of the horizontal bar, we can assume that there is no elastic potential energy involved, so we can eliminate that term as well.

This leaves us with the equation:

[Ki] = [Kf + Eff]

Using the equation for kinetic energy (K = 1/2mv^2), we can substitute the given values for the initial kinetic energy (Ki) and solve for the final kinetic energy (Kf):

[1/2 (50 kg) (1.2 m/s)^2] = [1/2 (50 kg) (vf)^2]

Solving for the final velocity (vf), we get:

vf = 1.2 m/s

Therefore, the speed of the gymnast's center of mass at the bottom of her swing is 1.2 m/s.