Inertia Question, simple but yet confused? Help greatly appreciated

[jcfor3ver]

Homework Statement

QUESTION 1: Calculate the rotational moment of inertia of a meter stick, with mass 0.56 kg, about an axis perpendicular to the stick and located at the 20 cm mark. Treat the stick as a "thin rod."

Homework Equations

I=m(l)^2
BUT SINCE IT ASKED TO consider the stick as a "thin rod" you would have I=(ml^2)/12

Then Iaxis=Icm+ml^2.

I=inertia
m=mass
l=length

The Attempt at a Solution

Took the inertia through the center of mass, which I split the meter stick in half to get .5m, multiplied it by .56 kg and divided by 12. Then added that to the mass multiplied by the distance between the axis of rotation and the center of mass squared.

got .62 kg*m^2. What am I doing wrong? The computer will tell me when I'm right.

 

[collinsmark]

Took the inertia through the center of mass, which I split the meter stick in half to get .5m, multiplied it by .56 kg and divided by 12.

Don't forget to square the 0.5 m.

Then added that to the mass multiplied by the distance between the axis of rotation and the center of mass squared.

I agree that using the parallel axis theorem should give you the correct answer.

got .62 kg*m^2. What am I doing wrong? The computer will tell me when I'm right.

Your result is off by a factor of 10 from what I am getting. Maybe you are dropping a 0 somewhere?

[Edit: not exactly a factor of 10, but roughly an order of magnitude. (I probably spoke too soon about simply dropping a 0.)]

 

[jcfor3ver]

I changed my answer considering this factor of 10, and still is wrong? Maybe my equation for a "thin rod" should be changed to I=ml^2? I don't know, this is frustrating because it is a simple problem.

 

[collinsmark]

I changed my answer considering this factor of 10, and still is wrong? Maybe my equation for a "thin rod" should be changed to I=ml^2? I don't know, this is frustrating because it is a simple problem.

Don't just change your answer by a factor of 10 all willy nilly.

Your thin rod equation of I = mr²/12 is fine for the rod spinning around its center. And your use of the parallel axis theorem (adding the mass multiplied by the distance between the axis of rotation and the center of mass squared), should work fine too.

Show us your work on how you obtained 0.62 [kg m²], and we we can help figure out what went wrong.

 

[collinsmark]

Oh, wait, I think I see what's going on.

The formula for moment of inertia of a thin rod is I = mL²/12 when rotated about its center. But here L is the entire length of the rod, not the distance from the center to the end of the rod. You were using 0.5 meters for this before, but it should be 1 m.

 

[jcfor3ver]

Ive tried a bunch of things. But here is what I did:

Inertiacentermass=.56kg(either 1m or .5 m)^2+.56kg(.3m)^2= .61

I am still wrong. Seriously, what is going on.

 

[jcfor3ver]

could someone look at my original question and try it for themself to compare answers? That would help a lot.

 

[collinsmark]

Ive tried a bunch of things. But here is what I did:

Inertiacentermass=.56kg(either 1m or .5 m)^2+.56kg(.3m)^2= .61

What happened to dividing by 12?

Like I said before, the moment of inertia of a thin rod rotating around its center is

I = mL²/12

Where L is the length of the entire rod (not just the length from the center to one of its ends, but the entire rod).

After that, apply the parallel axis theorem as you've done above, with the pivot point 0.3 m away from the center of the rod.