# Homework Help: Newtown 2nd law of rotation problem

1. Jul 3, 2009

### Puchinita5

1. The problem statement, all variables and given/known data

10-42 shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 0.80 m and L2 = 6.2 m. The rod is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of (a) particle 1 and(b) particle 2?

see http://www.webassign.net/hrw/hrw7_10-43.gif for image.

2. Relevant equations

Tau=I alpha

3. The attempt at a solution
I am not sure if the center of mass is important, since it will be rotating about a fulcrum. So i'm not sure if I should use I=mr^2 or I=mr^2 +mh^2. Also, would I set Torque equal to mg? Since there are no other forces identified I would assume this. I have no other clues as to how to go about this.

2. Jul 3, 2009

### tiny-tim

Welcome to PF!

Hi Puchinita5! Welcome to PF!
Since the rod is massless, it'll be much simpler just to treat the two masses separately, and just use the fundamental r x v and r x F formulas for anglar momentum and torque (about the fulcrum), and then torque = rate of change of angular momentum.

3. Jul 3, 2009

### Puchinita5

okay, so we actually haven't discussed angular momentum yet, I am supposed to use NEwton's second law of rotation, but if you say it is easier I will try to figure it out using ur method. I skipped to the next chapter to see the forumlas you talked about.

So if torque=r x F...would I then say torque=rFsintheta? and in which case, would r be the distance from the particle to the fulcrum? and the force would be gravity (mg), and the angle between would be 90?

and then how would i relate this to momentum?

4. Jul 4, 2009

### tiny-tim

oh ok … that case, use mr2 as the moment of inertia of a point mass m about an axis at a distance r from the mass (but treat the two masses separately).
better leave that chapter until your professor does it.

5. Jul 5, 2009

### Puchinita5

ok i still have NO idea....

so what i am thinking so far is that initially, at mass two there is a force mg...which I would think is equal to Torque since there are no other forces happening. So Torque=mg= Inertia x angular acceleration.... so for say, mass two, the inertia would be mr^2=m(6.2)^2 since the mass is 6.2m away from the fulcrum. So then I would say:

mg=m(6.2^2)alpha....and solve for angular acceleration. So 9.8/(6.2^2)....this angular acceleration gives me the wrong answer. So then I thought maybe i needed to find Tangential acceleration. so I multiplied the angular acceleration by 6.2 since Tangential acceleration=radius times angular acceleration. This also gave me the wrong answer....So where am I totally not getting this?

6. Jul 5, 2009

### Puchinita5

any help would be HUGE...i have had this post up for a few days now and i only have two hours to figure out how to do this problem....it's not life or death but this will be the first problem all summer i will get wrong on a homework assignment. I have another version of the question (exactly the same but with different numbers) as well as the answer for it, and i still cannot figure out how to get the answer even when i know what it is! I also have an exam tomorrow and I literally have no one i can go to for help with this. PLEASE HELP!

7. Jul 5, 2009

### Puchinita5

okay, new tactic, that STILL doesn't work!!! i even took this method out of an example in my book, and it worked for one question but not this one...

and this i definitely don't get why, because it worked on another problem....

so I have that Tnet= T1 and T2=-mg(6.2)+mg(.8)=-5.4mg

I=m(.8^2)+m(6.2^2)+2m(2.7^2)

so then alpha equals .986209467

and to change that to tangential acceleration i multiplied that by .8 and 6.2 to figure out the initial accelerations of each particicle respectively.....this still didn't work....

WHYYY!!!?????

8. Jul 6, 2009

### tiny-tim

Sorry, Puchinita5, you've completely lost me

this is a massless rod … is that 2.72 supposed to be for a mass of 2m in the middle of the rod?

9. Jul 6, 2009

### Puchinita5

well i figured the center of mass between the two particles would be at 3.5m but the fulcrum(rotation axis) is at .8m, 2.7m away...

10. Jul 6, 2009

### tiny-tim

But you don't need the centre of mass …

you must only include things once

you can't calculate moment of inertia on the basis of the masses being where they are, and being all at the centre.

11. Jul 6, 2009

### Puchinita5

hmmm, i thought the formula for inertia though takes it from the central axis, but then you have to account for how far off away the fulcrum is from the center?

12. Jul 6, 2009

### tiny-tim

ah, you mean the parallel axis theorem …

yes, I = Ic + 2md2, with d = 2.7

but Ic is the moment of inertia about the centre of mass, which is 2m(3.5)2

13. Jul 6, 2009

### Puchinita5

ahhhh i already used up all my attempts in my homework so now it won't tell me if this gives me the right answer....but it makes sense! i kept getting so confused with stratagies that I was reading in my book and for other questions and then me being the queen of dumb mistakes doesn't ever help...thanks so much for your input!