Conservation of Energy and Final Tangential Speed

Yes.In summary, the conversation discusses a problem involving a person lying on their back with one leg pointing straight up. The person is asked to find the tangential speed of their foot just before it hits the floor when allowed to fall freely. Assumptions are made that the leg can be treated as a uniform rod pivoting freely around the hip. The conversation then discusses different attempts at solving the problem, with the final solution involving calculating the change in gravitational potential energy and rotational kinetic energy using the center of mass as a reference point.
  • #1
Nghi
18
0

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 0.83 m long that pivots freely about the hip.


Homework Equations



KE = 0.5mv^2(1 + I/(mr^2))
PE = mgh
I = (1/3)mr^2

The Attempt at a Solution



This problem is driving me crazy! I've tried so many different ways, and I can't seem to get it. Since the problem gives us that it's a uniform rod that pivots around the hip, I equated that to the rod being rotated on an axis at an endpoint. Hence me getting I = (1/3)mr^2.

From there, I basically plugged and chugged. Since it's starting at a height, the initial PE = mgh, and initial KE = 0. The final KE = the long equation in 2, and the final PE = 0 since it lands on the floor.

initial PE + initial KE = final PE + final KE
mgh + 0 = 0 + 0.5mv^2(1 + I/(mr^2))
mgh = 0.5mv^2(1 + I/(mr^2))

The two masses then canceled out to give the following:

gh = 0.5v^2(1 + I/(mr^2))

Since I = (constant)mr^2, the two terms cancel out, so all that's left is:

gh = 0.5v^2(1 + (constant))
(9.81)(0.83) = 0.5v^2(1 + (1/3))
v = [2(9.81)(0.83)/(4/3)]^0.5

And when I solved for v, I got 3.49 m/s. But this isn't the right answer. I'm not sure why it isn't, and I'm not too sure what to do now. I thought maybe that the leg was only doing rotational kinetic energy since it wasn't really doing any linear movement, but that sort of got me confused, so I abandoned that trail of thought.

Some help would be very appreciated. Thanks!
 
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  • #2
Nghi said:
KE = 0.5mv^2(1 + I/(mr^2))
I have no idea where you got this equation from.
PE = mgh
I = (1/3)mr^2
These are OK.

(When calculating the change in PE as the leg falls, remember that it is a uniform rod.)

The Attempt at a Solution



This problem is driving me crazy! I've tried so many different ways, and I can't seem to get it. Since the problem gives us that it's a uniform rod that pivots around the hip, I equated that to the rod being rotated on an axis at an endpoint. Hence me getting I = (1/3)mr^2.
So far, so good.

...
And when I solved for v, I got 3.49 m/s. But this isn't the right answer. I'm not sure why it isn't, and I'm not too sure what to do now. I thought maybe that the leg was only doing rotational kinetic energy since it wasn't really doing any linear movement, but that sort of got me confused, so I abandoned that trail of thought.
Time to get back on that trail! Since the leg is in pure rotation (it pivots) just compute the increase in rotational KE. Use that to find the angular speed and then the tangential speed of the foot.
 
  • #3
The KE I got is the equation used for rolling motion without slipping. But I guess that wouldn't make sense, since the leg isn't really doing any translational kinetic energy. But about the rotational energy thing: I also did that, I believe, and didn't get the right answer. :(

initial PE + initial KE = final PE + final KE
mgh + 0 = 0.5Iw^2 + 0
mgh = 0.5Iw^2
mgh = 0.5((1/3)mr^2)w^2
gh = (1/6)(r^2)w^2

I substituted in x = 0.83 m for both height 'h' and radius 'r' to get the following:

(9.81)(0.83) = (1/6)(0.83^2)w^2
w = 8.421
v/r = 8.421
v/(0.83) = 8.421
v = 6.9895 m/s

This answer wasn't also the right one, either. Gahhhf;sdlfisf
 
  • #4
This time your method is correct but you're using the wrong value for 'h'. While the foot falls a distance equal to 0.83 m, other parts of the leg do not. Hint: Follow the center of mass.
 
  • #5
So the foot falls in a quarter of a circle instead. So... that mean I have to use C = 0.5(pi)r to find height h?
 
  • #6
Wait, I just reread what you wrote. If I'm suppoesd to follow its center of mass, and it's a uniform rod, does that mean the center of mass is located at the geometric center? And would that geometric center be located at the knees? Because if so, then that would make the height h = 0.83/2, or 0.415. Would that be the actual height the leg falls?
 
  • #7
Yes, the center of mass of a uniform rod is in the middle. In calculating the change in gravitational PE of an extended object, what counts is the change in height of the center of mass. Obviously, being a rotating stick, some parts move more that others. But the average change in height (and the change in height of the c.o.m.) is 0.83/2 m. That's what you need to use.
 
  • #8
Ahh, that makes sense. So a few small after-questions, then:

1. does this center of mass come into play for objects that rotate? Because all of the other problems I've done concerning PE and KE didn't include center of mass as a factor.

2. If it was a rotating sphere instead of a stick, would you still have to use center of mass?

Sorry if I'm clogging up your inbox. I have a test on Thursday, and the whole 'center of mass' being introduced is creating a little bit of an obstacle for me. :/
 
  • #9
Nghi said:
1. does this center of mass come into play for objects that rotate? Because all of the other problems I've done concerning PE and KE didn't include center of mass as a factor.
If you need to find the change in gravitational PE associated with a rotating or moving object, then sure you'd use the center of mass. Depending on the exact problem, you might not need it.

Realize that this is equivalent to asking for the work done by gravity. And where does gravity act? Effectively at the center of mass.

2. If it was a rotating sphere instead of a stick, would you still have to use center of mass?
I don't understand the situation you have in mind.

Sorry if I'm clogging up your inbox. I have a test on Thursday, and the whole 'center of mass' being introduced is creating a little bit of an obstacle for me.
Feel free to keep asking until it's clear.

Perhaps a specific example might help clarify things. Find one that bugs you and we can discuss it.
 
  • #10
I guess what I don't understand is WHEN to use center of mass. Is it used exclusively for gravitational PE? It just seems really random for the problem to pull out center of mass. There really isn't any specific problem in the book that I've encountered (sorry! :(), but it's just in general, I guess. How would I know when the problem calls for COM?

Oh, and as for the rotating sphere/rod thing. You know how in the example you have the uniform rod rotating at an endpoint? Well, you had to go and find the COM in order to be able to find the "h" height that the rod really fell through. Well, if it was a uniform sphere that was rotating around an endpoint instead, would you still have to find the COM to find the height "h"?
 

Related to Conservation of Energy and Final Tangential Speed

1. What is the conservation of energy?

The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but can only be transferred or transformed from one form to another.

2. How does the conservation of energy apply to final tangential speed?

In the context of final tangential speed, the conservation of energy applies in the sense that the total energy of a system remains constant. This means that the sum of the kinetic and potential energies at any point in time will equal the sum of the kinetic and potential energies at any other point in time.

3. Can the final tangential speed of an object be greater than its initial tangential speed?

Yes, the final tangential speed of an object can be greater than its initial tangential speed. This is because energy can be transferred into or out of the system, resulting in a change in the object's speed.

4. How does friction affect the conservation of energy in determining final tangential speed?

Friction can affect the conservation of energy when determining final tangential speed by converting some of the object's kinetic energy into heat energy. This means that the total energy of the system may not remain constant, resulting in a lower final tangential speed than what is calculated based on the conservation of energy principles alone.

5. What factors can influence the final tangential speed of an object?

The final tangential speed of an object can be influenced by factors such as the object's mass, the force applied to it, the distance it travels, and any external forces acting on it (such as friction). The conservation of energy principle can also be used to determine the final tangential speed of an object.

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