What is the speed of a gymnast's center of mass at the bottom of a swing?

[pharmguy]

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

[K + Eg] = [K]

1/2mv^(2) + mgh = 1/2mv^(2)
1/2(50 kg)(1.2m/s)^(2) + (50kg)(9.8 Nt/kg)(1.0m) = 1/2(50 kg)(v)^(2)
520 = 25(v)^(2)
v=4.56 m/s

is this right? If its not can someone please walk me through what I did wrong. Thank You so much. I'm just getting ready for a test soon.

 

[renaldocoetz]

Hi farmguy

youre using the right approach but you've left out information and made a calculation error:

if E₀=E_f ...Initial total mechanical energy = final tot mech enrgy

then...

remember that if u make final height 0, like u did, u have to make initial height 2m. From reading the problem statement u can conclude that there's 2m difference between her top height and bottom height, right? That has to reflect in your equation as well. Also u made a slight calculation error on the left. Should get 526, not 520.

Tip: In calculating problems like these the values used for the initial and final heights doesn't matter as long as the difference is correct.

 

[tomcue]

Your attempt at a solution is correct! You have correctly applied the conservation of energy principle, stating that the total energy at the beginning of the swing (kinetic energy + gravitational potential energy) is equal to the total energy at the bottom of the swing (kinetic energy). Good job!