Calculate Maximum Velocity of Girl on Swing

In summary, the maximum velocity of the girl on the swing will be at the lowest point where she is 1m above the ground. This is related to her potential and kinetic energy. The period of the pendulum can be used to determine her maximum velocity. The equations for energy conservation and the period of the pendulum can be used to solve for her maximum velocity without needing to know her mass. The length of the swing is irrelevant for this problem.
  • #36
Rayquesto said:
oh! well then you'd think that technically there still is a potential energy at the lowest point, since it's 1 meter above ground, which means v=sqrt2ghfinal - sqrt2ghinitial, but that's incorrect, because it violates the law of conservation of energy. but hm...maybe h=2meters, because it's 2 meters from the ground. I'm really sorry if you are confused. I am too!

This problem has little to do with the actual Potential energy, and everything to do with the CHANGE in Potential energy.
In all problems, we get to set where zero height is, and invariably we choose either the lowest point, the highest point, the starting point or the finishing point.

For example

Lowest point - A swing or similar

Highest point - a satellite [rare]

Start point - projectile motion

Finish Point - roller coaster.
 
Physics news on Phys.org
  • #37
Rayquesto said:
the height is the y direction between the heighest point and the lowest point. does that make sense?

On in proper description: The h value [change in height] is the difference between highest point and lowest point.
 

Similar threads

  • Introductory Physics Homework Help
Replies
32
Views
10K
  • Introductory Physics Homework Help
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
814
  • Introductory Physics Homework Help
Replies
1
Views
790
  • Introductory Physics Homework Help
2
Replies
43
Views
2K
  • Mechanical Engineering
Replies
3
Views
252
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top