# Final tangential Speed

## Homework Statement

After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.79 m long that pivots freely about the hip.
here's the link to the image
http://img229.imageshack.us/my.php?image=20071019105225633283879gs4.gif

## The Attempt at a Solution

I have no idea what to do am i suppose to use Conservation of Energy to solve this problem or use this formula K=(1/2)mv^2[1+(I/mr^2)] I have no idea where to start.

Gokul43201
Staff Emeritus
Gold Member
You should use energy conservation with the kinetic energy given by the expression above. You should also understand why it is possible to use energy conservation, when there is clearly an external force acting on the system.

Last edited:
I still don't understand what to do. Are you saying that I should use this formula $$E_{i}$$=$$E_{f}$$
so than
mgh+(1/2)mv^2[1+mr^2/mr^2]=0
v=sqrt(gh)
but i still get the wrong answer. What should i do? Where did i go wrong?

Gokul43201
Staff Emeritus
Gold Member
I still don't understand what to do. Are you saying that I should use this formula $$E_{i}$$=$$E_{f}$$
Yes.

so than
mgh+(1/2)mv^2[1+mr^2/mr^2]=0
I can not make any sense of that equation. Could you please start from Ei=Ef, and substitute step by step and show what you get? Also, explain all the terms in the expressions you use. This is not so hard if you are just a little careful.

Ei=mgh
Ef=(1/2)mv^2[1+(1/2)mr^2/mr^2]
mgh=(1/2)mv^2[1+(1/2)]
the mass cancels out
gh=(3/4)mv^2
v=$$\sqrt{(4/3)gh}$$
v=$$\sqrt{(4/3)(9.81)(.79}$$
v=3.2m/s