The problem goes: Masses m and 2m are connected by a rod of negligible mass. The rod can be pivoted about several positions along the rod. to obtain the greatest angular acceleration for a fixed torque applied, the system should be pivoted about an:

-Axis through m

-Axis through 2m

-Axis through center of the rod

-Axis through the rod 1/3L from m

-Axis through the rod 1/3L from 2m

The rod is of length L. Don't think it should matter, but m is on the left, 2m is on the right.

My answer was the axis through the rod 1/3L from 2m. Since Torque = I * alpha, minimizing I leads to a higher alpha--the angular acceleration--because alpha = T/I

The moment of Inertia for an axis through the rod 1/3L from 2m is 2/3mL^2.

I've talked to many of my classmates, and they all said the first response was the correct one--axis through m--however, this has a moment of inertia of 2mL^2--the most out of the five choices. According to my reasoning, this would yield the least angular acceleration

Am I right? Or is my physics logic faulty somewhere?

## Physics(Mechanics) problem I'm having some trouble with...

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- lu6cifer
**Posts:**230**Joined:**Fri Mar 20, 2009 4:03 am UTC**Location:**That state with the all-important stone

### Physics(Mechanics) problem I'm having some trouble with...

lu6cifer wrote:"Derive" in place of "differentiate" is even worse.

doogly wrote:I'm partial to "throw some d's on that bitch."

- danpilon54
**Posts:**322**Joined:**Fri Jul 20, 2007 12:10 am UTC

### Re: Physics(Mechanics) problem I'm having some trouble with...

I believe you are correct but it's late so I may be wrong. You are correct that you want the lowest moment of inertia, and you're friends seem to be choosing the highest.

Mighty Jalapeno wrote:Well, I killed a homeless man. We can't all be good people.

- lu6cifer
**Posts:**230**Joined:**Fri Mar 20, 2009 4:03 am UTC**Location:**That state with the all-important stone

### Re: Physics(Mechanics) problem I'm having some trouble with...

Well, I'm still really unsure...The problem's from a test I took, and the presumably correct answer---axis through the rod 1/3L from m---was marked incorrect.

lu6cifer wrote:"Derive" in place of "differentiate" is even worse.

doogly wrote:I'm partial to "throw some d's on that bitch."

### Re: Physics(Mechanics) problem I'm having some trouble with...

Are you sure you have the question right then?

"I conclude that all is well," says Edipus, and that remark is sacred.

- Camus, The Myth of Sysiphus

Mental Health Break

- Camus, The Myth of Sysiphus

Mental Health Break

- MotorToad
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### Re: Physics(Mechanics) problem I'm having some trouble with...

By the common-sense method, the pivot point should be 1/3 from the larger mass. Pivoting about the larger mass means the smaller mass's r is too long,

I

I

r

You should be able to solve properly from this, or since it's multiple choice (and apparently already graded) we can bash through it manually.

I

I

I = 2/3

Pivoting around the large mass, the small mass has inertia of 1.

Pivoting around the midpoint gives .5 for the large mass and .25 for the small one, or .75 total.

Note that if we were being strict, analyzing the masses pivoting around the larger mass would also include the larger mass's inertia about itself, but I conveniently ignored that since it really didn't matter for this.

Oh, I forgot to mention... your classmates are idiots. Ignore them!

I

_{1}= m_{1}r_{1}^2I

_{2}= m_{2}r_{2}^2r

_{1}= r - r_{2}You should be able to solve properly from this, or since it's multiple choice (and apparently already graded) we can bash through it manually.

I

_{1}= 1*(^{2}/_{3})^{2}= .444...I

_{2}= 2*(^{1}/_{3})^{2}= .222...I = 2/3

Pivoting around the large mass, the small mass has inertia of 1.

Pivoting around the midpoint gives .5 for the large mass and .25 for the small one, or .75 total.

Note that if we were being strict, analyzing the masses pivoting around the larger mass would also include the larger mass's inertia about itself, but I conveniently ignored that since it really didn't matter for this.

Oh, I forgot to mention... your classmates are idiots. Ignore them!

What did you bring the book I didn't want read out of up for?

"MAN YOUR WAY TO ANAL!"

"MAN YOUR WAY TO ANAL!"

_{(An actual quote from another forum. Only four small errors from making sense.)}- danpilon54
**Posts:**322**Joined:**Fri Jul 20, 2007 12:10 am UTC

### Re: Physics(Mechanics) problem I'm having some trouble with...

Are you sure you aren't looking for the least angular acceleration?

Mighty Jalapeno wrote:Well, I killed a homeless man. We can't all be good people.

- lu6cifer
**Posts:**230**Joined:**Fri Mar 20, 2009 4:03 am UTC**Location:**That state with the all-important stone

### Re: Physics(Mechanics) problem I'm having some trouble with...

Well, verbatim from the problem set:

"To obtain the greatest angular acceleration for a fixed torque applied...."

"To obtain the greatest angular acceleration for a fixed torque applied...."

lu6cifer wrote:"Derive" in place of "differentiate" is even worse.

doogly wrote:I'm partial to "throw some d's on that bitch."

- MotorToad
- Really Repeatedly Redundantly Redundant
**Posts:**1114**Joined:**Sat Jul 07, 2007 10:09 pm UTC**Location:**Saint Joseph, CA-
**Contact:**

### Re: Physics(Mechanics) problem I'm having some trouble with...

That problem at least requires some thought, asking for the least acceleration is a bit obvious.

What did you bring the book I didn't want read out of up for?

"MAN YOUR WAY TO ANAL!"

"MAN YOUR WAY TO ANAL!"

_{(An actual quote from another forum. Only four small errors from making sense.)}- danpilon54
**Posts:**322**Joined:**Fri Jul 20, 2007 12:10 am UTC

### Re: Physics(Mechanics) problem I'm having some trouble with...

Well then, everyone is wrong except you. Congrats!

And to whoever marked the test, the beauty of physics is that you can prove them wrong by showing them what the moment of inertia is for your answer. They can't argue with math.

And to whoever marked the test, the beauty of physics is that you can prove them wrong by showing them what the moment of inertia is for your answer. They can't argue with math.

Mighty Jalapeno wrote:Well, I killed a homeless man. We can't all be good people.

- Yakk
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**Posts:**11120**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

### Re: Physics(Mechanics) problem I'm having some trouble with...

Let 1 be the length of the rod.

K be the distance you are from the heavier mass. 0 <= K <= 1

Then the moment of inertia I(K) of the rod is:

I(K) = K^2*2m + (1-K)^2*m = 2m K^2 + (1 - 2K + K^2)m

= 3m K^2 -2K + m

d I(K) /dK = 6m K -2 m

Solving for I'(K) = 0 we get:

6 m K - 2 m = 0

K = 1/3

Now, if you use the wrong moment of inertia formula (where it grows with R instead of R^2), you can get a different answer (axis through 2m).

So ya, write up your proof that your answer was correct, and go to the person who marked your test.

K be the distance you are from the heavier mass. 0 <= K <= 1

Then the moment of inertia I(K) of the rod is:

I(K) = K^2*2m + (1-K)^2*m = 2m K^2 + (1 - 2K + K^2)m

= 3m K^2 -2K + m

d I(K) /dK = 6m K -2 m

Solving for I'(K) = 0 we get:

6 m K - 2 m = 0

K = 1/3

Now, if you use the wrong moment of inertia formula (where it grows with R instead of R^2), you can get a different answer (axis through 2m).

So ya, write up your proof that your answer was correct, and go to the person who marked your test.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

### Re: Physics(Mechanics) problem I'm having some trouble with...

lu6cifer wrote:Well, I'm still really unsure...The problem's from a test I took, and the presumably correct answer---axis through the rod 1/3L from m---was marked incorrect.

...you mean from from 2m rigth?

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