Need help In Rotation Questions Please

[brainracked]

Homework Statement

A system consists of two particles, of masses m1 and m2, is connected by a light rigid rod of length L.
a) Find the rotational inertia I of system for rotations of tis object about an axis perpendicular to the rod and distance x from m1
b) Show that I is a minimum when x=x_cm

Homework Equations

I= I_com + mh²
I= 1/12 mr²

The Attempt at a Solution

The answer provided was given as: I₂=(m1+m2)x²+m2L²-2m2xL

I understand that it is juz applyin the formula I= I_com + mh² but wat i do not understand is that

The thing that I don't understand is the mh^2 part..why is only m2 used in the equation?
And why when h= (L-x)^2 is expanded..what happens to the m2x^2?

And wat about the part b? By juz replacing the value it wif x...it juz doesn't make any sense...

Thanks ya..any help will be very much appreciated..

 

[tiny-tim]

Welcome to PF!

Hi brainracked! Welcome to PF! smile:

The answer provided was given as: I₂=(m1+m2)x²+m2L²-2m2xL

I understand that it is juz applyin the formula I= I_com + mh² …

No, it isn't.

It's using the far simpler method of just using the basic formula for a point mass, I = mr².

Try that!

 

[brainracked]

Oh gosh.. -blush-
Dat simple?
Aiyo..I always complicate stuff...but..I still don't quite get it..haha..sorry ya..
So..okay..tryin out d simpler way..i get d 1st part which is I=(m1+m2)x^2 but wat about the rest of the answer where there is the +m2L^2 - 2m2xL?

And besides..i thought if it is not rotated about a central axis but at sum distance from it, the formula of I = Icom +mh^2 should b used as h comes into play as there is the mention of distance x from m1..

I can solve other problems similar to tis one but wif the numbers & figures given such as mass and distance and rev...but i juz can't seem to get my head round tis..

So sorry but I think I still need help in gettin tis rite.. =)

 

[tiny-tim]

Oh gosh.. -blush-
Dat simple?
Aiyo..I always complicate stuff...but..I still don't quite get it.

ok, let's do the easy one …

I for m₁ is momentum x distance, = m₁x².

So I for m₂ is … ?

 

[brainracked]

Haha..ok..so m2 should b...m2L^2? Correct?
So the -2m2xL?

 

[chislam]

No that is still incorrect. Do you even understand how tiny-tim gave you the answer for the inertia of m_1? If so, then do the same for m_2 and then add their inertia's together and you will indeed arrive at the answer provided.

 

[brainracked]

At first I don't think I fully understood..but now I get it..

I_m1=m₁x²

So..I_m2 should b..

I_m2=m₂ (L-x)²
= m₂ (L²+ x² - 2xL)

Then as u said..by adding the Inertia's together..tadaa..the answer..Haha..

Big thank you to tiny-tim & chislam!
Take care you ppl!