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

AI Thread Summary
The discussion focuses on calculating the speed of a gymnast's center of mass at the bottom of a swing using energy conservation principles. The initial conditions include a 50 kg gymnast with a center of mass 1.0 m above the bar, traveling at 1.2 m/s. The correct approach involves equating initial and final mechanical energy, accounting for the 2 m height difference between the top and bottom of the swing. A calculation error was identified, with the correct total mechanical energy at the top being 526 instead of 520. The final speed of the gymnast's center of mass at the bottom of the swing is determined to be approximately 4.56 m/s after correcting the errors.
pharmguy
Messages
8
Reaction score
0

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.
 
Physics news on Phys.org
Hi farmguy

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

if E0=Ef ...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.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top