# Work-energy problem (units!)

1. Apr 25, 2013

### drewta

1. The problem statement, all variables and given/known data
The 8 lb uniform rod AB shown in the figure rotates in the vertical plane about a pin at C. The spring attached to the rod at D has a spring constant of 1.25 lb/in., and its unstretched length is 24in. If the rod is released from rest in the position shown, determine its angular velocity when it is in the vertical position.

2. Relevant equations
U = mgh
U = .5*k(x^2)
U = .5*Iω^2
I = I_cm + mr^2

3. The attempt at a solution
Initial length = sqrt(36^2+20^2) = 41.183 in
Initial stretch = 41.183 - 24 = 17.183 in

Since the movement will tend towards lower energy (i.e. the spring will compress), the spring will become less stretched as the rod goes towards its vertical position (as opposed to stretching further into the opposite vertical position).

Final length = sqrt(26^2+10^2) = 27.857 in
Final stretch = 27.857 - 24 = 3.857 in

As the spring becomes less stretched, elastic potential energy is converted into other forms of energy (rotational or gravitational potential).

Initial elastic potential energy = (.5)(k)(x)^2
= .5(1.25 lb/in)(17.183^2)
= 184.53 lb*in

Final elastic potential energy
= .5(1.25 lb/in)(3.857^2)
= 9.298 lb*in

Gravitational potential energy is gained as the rod goes into its vertical position.
Final U = mgh
= (8 lb)(386.4 in/s^2)(10 in)
= 30912 lb*in

Rotational energy is gained = .5*Iω^2
Final I = ml^2/12 + mr^2
= (8 lb)(40 in)^2/12 + (8 lb)(10 in)^2
= 13600 lb*in^2

Initial energy = final energy
184.53 lb*in = 9.298 lb*in + 30912 lb*in + .5(13600)ω^2

Solve for ω
ω = imaginary...?

So, obviously, I'm screwing up somewhere. Probably with the units, but I don't know which is wrong. Is there anything I'm not accounting for? What value should I be using for gravity?

Thank you so much.

2. Apr 25, 2013

### drewta

3. Apr 25, 2013

### Staff: Mentor

Take a closer look at your gravitational PE. Not all of the rod is rising...

Also, your moment of inertia for the rod looks a tad high.

4. Apr 25, 2013

### drewta

I revised my calculations in the other thread and came out to a sensible answer (not sure if it's correct though.

I did not take into account your note about the gravitational PE. Like I asked in the other thread, shouldn't I only consider the mass center, which moves up 10 inches, like I did in my revised calculations in the other thread?
If I were to take into account your advice, how I would I change my calculations accordingly?

Last edited: Apr 25, 2013
5. Apr 25, 2013

### Staff: Mentor

Because the rod pivots about a point that is not at its end, a portion of the mass falls.

edit: Although, upon reflection, this doesn't change the net result; the change in height of the entire rod will still yield the net change in PE. Never mind

Last edited: Apr 25, 2013
6. Apr 25, 2013

### drewta

So would that detract from the gravitational potential energy?
Here are calculations, which I'm not sure are correct.

Portion that rose: mgh = wh = (8*.75)*(30/2)
Portion that fell: -mgh = -wh = -(8*.25)*(10/2)

Sum = (8*.75)*(30/2)-(8*.25)*(10/2) = 80 lb*in
That is the same as my previous calculation: wh = 8*10 = 80 lb*in.

It doesn't seem to me that the gravitational potential energy is changing, even if you attempt to account for portion of the mass that falls by treating that portion and the portion that rose as separate masses. Am I doing something wrong?

7. Apr 25, 2013

### Staff: Mentor

See the edit of my previous post. Considering just the motion of the center of mass of the entire rod is fine.

8. Apr 25, 2013

### drewta

All right, sounds good.
Do you mind taking a look at the other thread and seeing what's going wrong?

I have a variation of the same problem with different numbers and the corresponding answer for that variation, but using the same method, I can't get that correct answer.

9. Apr 25, 2013

### Staff: Mentor

Your calculations there look fine to me. Could be the provided answer is not correct.