Finding Angular Velocity of a Torsion Spring Assembly

[qweazy]

Homework Statement

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.

Homework Equations

T1+∑U=T2

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
ω=3.59rad/s
actual answer: 4.9rad/s
What am I doing wrong?

 

[paisiello2]

I think a diagram would help.

 

[qweazy]

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

 

[tms]

Homework Equations

T1+∑U=T2

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

 

[paisiello2]

Diagram works.

What is I for a disk?

 

[qweazy]

I think its (3/2)mr^2

 

[paisiello2]

I don't think that is correct.

 

[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

 

[qweazy]

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

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

 

[paisiello2]

What is the z'z' axis?

 

[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

 

[paisiello2]

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

 

[qweazy]

I'm guessing its the pin?

 

[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.

 

[qweazy]

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

 

[paisiello2]

Looks better. Does it give the right answer now?

 

[qweazy]

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

 

[paisiello2]

Ok, no surprise I guess.

Can you show your calc for mgh?

 

[qweazy]

15(9.81)(.405)= 59.6

 

[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.

 

[qweazy]

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