Moment of inertia of a rigid body

[doktorwho]

Homework Statement

Determine the moment of inertia of a rigid body on the picture:

The radius of the inner cylinder is R and the outer is 2R.

Homework Equations

3. The Attempt at a Solution [/B]
I thought of subtracting the big cylinder inertia from the small and adding the hanging body and i get:
$I=\frac{m}{2}(R^2 + 4R^2) + m(2R)^2$
but for some reason the book states its:
$I=\frac{m}{2}(R^2 + 4R^2) + \frac{m}{2}R^2$
Why is it like this? Isn't the body hanging from the the outher radius and is like a point mass?
Could you explain?

 

[BvU]

Try inverting Kinetic energy = ${1\over 2} I\omega^2$ for m

and you'll find you are correct. As you could expect from such a dimensional mismatch, the book is wrong

 

[BvU]

PS

I thought of subtracting the big cylinder inertia from the small

the other way around ?

the drawing is confusing: does the inner cylinder have mass m ? Or is the cylinder hollow and does the axle with radius 0 have mass m ?

 

[doktorwho]

PS
the other way around ?

the drawing is confusing: does the inner cylinder have mass m ? Or is the cylinder hollow and does the axle with radius 0 have mass m ?

Sorry i made numeorus errors, what i meant was that i got
$I=\frac{m}{2}(-R^2 + 4R^2) + m(2R)^2$
Since i subtracted the smaller one from the bigger one. and they got:
$I=\frac{m}{2}(R^2 + 4R^2) + \frac{m}{2}R^2$

 

[PeroK]

Sorry i made numeorus errors, what i meant was that i got
$I=\frac{m}{2}(-R^2 + 4R^2) + m(2R)^2$
Since i subtracted the smaller one from the bigger one. and they got:

Why would you subtract like that?

PS The answer given is correct for a dual-density cylinder with an inner cylinder of mass M and radius R and an outer hollow cylinder also of mass M and outer radius 2R.

The hanging block is not part of the "rigid body", so perhaps should be excluded for now. That might become clear in the next part of the question.

 

[BvU]

Sorry i made numeorus errors, what i meant was that i got
$I=\frac{m}{2}(-R^2 + 4R^2) + m(2R)^2$
Since i subtracted the smaller one from the bigger one. and they got:
$I=\frac{m}{2}(R^2 + 4R^2) + \frac{m}{2}R^2$

Yes, and I look like a fool now, referring to dimensional errors that have disappearedited . Never mind. PeroK is correct on $\forall$ counts.

 

[doktorwho]

Why would you subtract like that?

PS The answer given is correct for a dual-density cylinder with an inner cylinder of mass M and radius R and an outer hollow cylinder also of mass M and outer radius 2R.

The hanging block is not part of the "rigid body", so perhaps should be excluded for now. That might become clear in the next part of the question.

Oh i see. So the moment of inertia is the sum of two inertias, the sum of body on the inside plus the one on the outside.
$I=m/2*(R^2+4R^2) + X$ where x is the part i don't get. Why is the thrid term there, isn't this all ther is? I combined the total system?

 

[PeroK]

Oh i see. So the moment of inertia is the sum of two inertias, the sum of body on the inside plus the one on the outside.
$I=m/2*(R^2+4R^2) + X$ where x is the part i don't get. Why is the thrid term there, isn't this all ther is? I combined the total system?

That's not quite it. You have three components to the system. I suggest that the hanging block is excluded from your calculation, as it isn't part of a rigid body.

The first two terms $I=m/2*(R^2+4R^2)$ together are the MoI of the outer hollow cylinder. This may be something you have to show. In any case, it would be a good exercise to derive this.

The third term $m/2*(R^2)$ is the MoI of the inner cylinder of mass $M$ and radius $R$. It's not the MOI of the external block.

 

[BvU]

Agree with PerOK; he's giving away a lot -- but then again: the link I gave in #3 tells the story in full also. You subtract the smaller one cylinder from the bigger one and end up with a plus sign anyway because of the m, Do the exercise !

 

[doktorwho]

Got it, thanks :)