jackthehat

Hi everyone,

I wonder if anyone can help me with a problem which has had me stuck for some time. The problem is as follows -

PROBLEM

A thin, flat, uniform disk has mass 'M' and radius 'R'. A circular hole of radius 'R/4' centred at the point 'R/2' from the disk's centre is then punched in the disk.

(A) Find the Moment of Inertia of the disk with the hole about an axis through
the original centre of the disk, perpendicular to the plane of the disk.

[ Hint - Find the Moment of Inertia of the piece punched from the disk ].

There is a part (B) to this question, but it is part (A) above that has me stuck.

I attempted this problem by using the 'Parallel Axis Theorem',which states
I = I(center of mass) + M d (sqrd)
where I is the Moment of Inertia we are trying to find, I(Centre of mass)
is Moment of Inertia about centre of mass axis, M is mass of object and
d is distance of axis from centre of mass.
Calculating Moment of Inertia of piece punched out I first looked the mass of the object. Now Mass = Density x Pi x Radius(sqrd) x height
Since the disk is uniform then Density, and height do not change and Pi is a constant , only radius changes and changes by the term squared.
So we have ..
I = I(com) + M d(sqrd)
I(com) = 1/2 M R(sqrd) .. uniform disk
M = mass of punched piece = (1/4)(sqrd) .. changes by order squared
d = (R/4) .. thus giving ..

I = 1/2 M R(sqrd) + 1/2 [(1/16M) R(sqrd) R(sqrd)/16]
=> I = 1/2 M R(sqrd) +1/2 (M R(sqrd)/256]
=> I = M R(sqrd)/2 + M R(sqrd)/512
=> I =[(256 +1)/512] M R(sqrd)
=> I = 257/512 M R(sqrd)

However the book gives a different answer ...
I = (247/512) M R(sqrd) ??

Can anyone point out where I have gone wrong because I cannot see it ?

Jackthehat

ChipB

The moment of inertia of the disc with the hole punched out is equal to the moment of inertia of a complete disc minus the moment of inertia of the part that's missing. So start by determining the moment of inertia of the punched out piece, taken about the center of the large disc. I'll use m and r for the mass and radius of the small disc, and M and R for the mass and radius of the large disc.

$\displaystyle I_{small \ disc} = \frac 1 2 m r^2 + m d^2$

$\displaystyle m = \frac M {16}$

$\displaystyle r = \frac R 4$

$\displaystyle d = \frac R 2$

$\displaystyle I_{small} = \frac 1 2 (\frac M {16}) (\frac R 4)^2 + \frac M {16} (\frac R 2 )^2 = MR^2[\frac 1 {512} + \frac 1 {64}] = MR^2 (\frac 9 {512})$

So:

$\displaystyle I_{total} = I_{large\ disc} - I _{small \ disc} = \frac 1 2 MR^2 - MR^2 ( \frac 9 {512} ) = (\frac {247}{512}) MR^2$

skeeter

Moment of inertia of the hole about the large disk center if it were filled ...

$I_{hole}=\dfrac{1}{2}\cdot \dfrac{M}{16} \cdot \left(\dfrac{R}{4}\right)^2 + \dfrac{M}{16} \cdot \left(\dfrac{R}{2}\right)^2$

$I_{hole}= \dfrac{MR^2}{512} + \dfrac{8MR^2}{512} = \dfrac{9MR^2}{512}$


$I_{disk \, with \, hole} + I_{hole} = I_{disk}$

$I_{disk \, with \, hole} = I_{disk} - I_{hole}$

$I_{disk \, with \, hole} = \dfrac{MR^2}{2} - \dfrac{9MR^2}{512}$

$I_{disk \, with \, hole} = \dfrac{256MR^2-9MR^2}{512} = \dfrac{247MR^2}{512}$

... the Chipster beat me to it.

jackthehat

Hi ChipB,
Thank you for taking the time to look at my problem and give me some help, much appreciated. When I look at the analysis of the problem as you have set out above
it looks obvious that the correct solution should be the Moment of Inertia of the whole disk less the Moment of Inertia of the missing section. I am at a loss as to why I didn't see that straight off.
I think the reason that I seem to miss these things often is that I do not sit back and try to work the solution out on a top-down basis I frequently just jump straight in without this initial analysis. I will just have to try to train myself to take the time to do the initial analysis of the problem first before jumping in, I may have better luck in forming a viable solution more often. Something for the future I think.
Anyway thanks again for your time and clear helpful explanations they are really helpful to me in my efforts with these problems.

Jackthehat

jackthehat

Hi Skeeter,
Thank you for taking the time to look at my problem and give me some help, much appreciated. When I look at the analysis of the problem as you have set out above
it looks obvious that the correct solution should be the Moment of Inertia of the whole disk less the Moment of Inertia of the missing section. I am at a loss as to why I didn't see that straight off.
I think the reason that I seem to miss these things often is that I do not sit back and try to work the solution out on a top-down basis I frequently just jump straight in without this initial analysis. I will just have to try to train myself to take the time to do the initial analysis of the problem first before jumping in, I may have better luck in forming a viable solution more often. Something for the future I think.
Anyway thanks again for your time and clear helpful explanations they are really helpful to me in my efforts with these problems.

Jackthehat

PS - How do you manage to get the special characters within your replies .. that is
square and square-root characters .. etc ?

ChipB

You can write math equations using LaTeX. You start a LaTeX string with the word "math" in square brackets and terminate it with [\math]. Within the math and \math delimiters you write out the equation using special LaTeX formatting commands, which generally take the form of a backslash followed by a command like \sqrt for square root or \frac for fraction, followed by argument(s) for that command in squiggly brackets. The squiggly brackets can be nested. Other common characters include ^ for superscript (usually used for "raised to the power of") and _ for subscript. Thus you can do things like:

\frac {a^2} {b_i} yields $\displaystyle \frac {a^2} {b_i}$

\frac { -b \pm \sqrt{b^2-4ac}}{2a} yields: $\displaystyle \frac { -b \pm \sqrt{b^2-4ac}}{2a}$

\ln x = \int _ 1 ^x \frac {dw} w gives $\displaystyle \ln x = \int _ 1 ^x \frac {dw} w $

You can a lot more complicated things such as matrices. Do some Google searches to find more LaTeX formatting options. A word of caution though: not all LaTeX compilers support all possible LaTeX commands.