What Is the Tangential Speed of Your Foot When It Hits the Floor?

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In summary, the problem involves finding the tangential speed of a foot just before it lands after falling freely from a height while attached to a uniform rod that pivots freely about the hip. This can be solved using energy conservation with the expression for kinetic energy and by considering the center of mass of the system. The final answer is 3.2 m/s.
  • #1
will_lansing
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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


Homework Equations





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.
 
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  • #2
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.
 
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  • #3
I still don't understand what to do. Are you saying that I should use this formula [tex]E_{i}[/tex]=[tex]E_{f}[/tex]
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?
 
  • #4
will_lansing said:
I still don't understand what to do. Are you saying that I should use this formula [tex]E_{i}[/tex]=[tex]E_{f}[/tex]
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.
 
  • #5
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=[tex]\sqrt{(4/3)gh}[/tex]
v=[tex]\sqrt{(4/3)(9.81)(.79}[/tex]
v=3.2m/s

Is this how you are suppose to do it? I really don't know. Please help
 
  • #6
solved it. forgot about the center of mass bit.
 
  • #7
h=x/2 is what i mean about using the center of mass
 

Related to What Is the Tangential Speed of Your Foot When It Hits the Floor?

1. What is final tangential speed?

Final tangential speed refers to the velocity of an object at the end of its circular motion, tangent to its circular path. It is also known as tangential velocity.

2. How is final tangential speed calculated?

Final tangential speed can be calculated using the formula v = rω, where v is the tangential speed, r is the radius of the circular path, and ω is the angular velocity.

3. What is the relationship between final tangential speed and angular velocity?

The final tangential speed is directly proportional to the angular velocity. This means that as the angular velocity increases, the final tangential speed also increases, and vice versa.

4. Can final tangential speed be negative?

Yes, final tangential speed can be negative. A negative tangential speed indicates that the object is moving in the opposite direction of its circular path.

5. How does final tangential speed affect an object's centripetal acceleration?

Final tangential speed does not directly affect an object's centripetal acceleration. However, a change in tangential speed can result in a change in the object's centripetal acceleration, as they are both influenced by angular velocity and radius of the circular path.

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