# Homework Help: Principle of work energy

1. May 11, 2014

### qweazy

1. The problem statement, all variables and given/known data
When Θ=0, the assembly is held at rest, and the torsional spring is untwisted. If the assembly is released and falls downward, determine its angular velocity at the instant Θ=90°. Rod AB has a mass of 6kg, and disk has a mass of 9kg.

Rod is 450mm and disk has a radius of 75mm
So there is a pin holding the assembly upwards which is when Θ=0 and at the pin there is a torsional spring with constant of k=20N m/rad. One end of the rod is attached to the pin and the other is attached to the disk.

2. Relevant equations
T1+∑U=T2

3. The attempt at a solution
So first I found the center of mass of the combined mass
I called mass of the rod M and mass of disk m.
Center of mass = (.5(.45)6+(.45+.075)9)/(6+9)= .405m
Then I found the moment of inertia
I=(1/3)(6(.45)^2)+(3/2)(9)(.075)^2+9(.525)^2= 2.96

T1=0 since the assembly started from rest.
Then I found T2
.5mv^2+.5Iω^2
.5(M+m)(.405ω)^2+.5(2.96)ω^2=2.71ω^2

Solved for the potential energy of the spring
.5(20)(π/2)^2=24.674
solved for ∑U= mgh-24.674= 34.861
solved for ω
What am I doing wrong?

2. May 11, 2014

### paisiello2

I think a diagram would help.

3. May 11, 2014

### qweazy

Dont have a camera but I drew a picture :)
and the numbers are in mm

#### Attached Files:

• ###### hw.png
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4. May 11, 2014

### tms

You are assuming that $\sum U$ is 0 at the end. Is that true?

5. May 11, 2014

### paisiello2

Diagram works.

What is I for a disk?

6. May 11, 2014

### qweazy

I think its (3/2)mr^2

7. May 12, 2014

### paisiello2

I don't think that is correct.

8. May 12, 2014

### qweazy

Ixx=Iyy=(1/4)mr^2
Izz=(1/2)mr^2
Iz'z'=(3/2)mr^2
This is what the book gives me

#### Attached Files:

• ###### mmi.png
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9. May 12, 2014

### qweazy

No, I'm assuming it equals mgh-1/2ks^2

10. May 12, 2014

### paisiello2

What is the z'z' axis?

11. May 12, 2014

### qweazy

I think its just saying that its rotating about that axis. Izz is rotating about the z axis and Ixx is rotating about the x axis and so on

12. May 12, 2014

### paisiello2

Yeah, I get that part. My question is what is the z'z' axis?

13. May 12, 2014

### qweazy

I'm guessing its the pin?

14. May 12, 2014

### paisiello2

Not sure where the pin is but, based on your second diagram, you are not being consistent when applying the parallel axis theorem. I would take I about the z-z axis instead.

15. May 12, 2014

### qweazy

ok so it would be (1/2)9(.075)^2+9(.525)^2 for the disk?

16. May 12, 2014

### paisiello2

Looks better. Does it give the right answer now?

17. May 12, 2014

### qweazy

No, for I I got 2.91 and solved it like I how I did before and got 3.60rad/s

18. May 12, 2014

### paisiello2

Ok, no surprise I guess.

Can you show your calc for mgh?

19. May 12, 2014

### qweazy

15(9.81)(.405)= 59.6

20. May 12, 2014

### paisiello2

OK, I see your mistake now. Your kinetic energy included a translational term but all the kinetic energy here is rotational only. Try and see if this gives the right answer.

21. May 12, 2014

### qweazy

Yea I got it. Thanks a lot for helping me. I appreciate it.