Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 
Sep 27th 2017, 09:05 AM

#1  Member
Join Date: Jan 2015
Posts: 96
 Moment of inertia generic calculation
Hi everybody,
I am afraid I have come across another Moment of Inertia problem where I cannot see how to proceed to find the solution. It basically gives a cylinder, with generic length, radius and mass of values L, R and M and the fact that density of the cylinder varies linearly with the distance from axis of rotation, and asks for the generic calculation of the Moment of Inertia about the axis.
The problem details follow  Problem
A cylinder with radius 'R' and mass 'M' has density that increases linearly with distance 'r' from the cylinder axis given by the expression  p = a r .. where p = density; a = positive constant; r = distance from axis
(a) Calculate the Moment of Inertia of the cylinder about a longitudinal axis
through it's centre in terms of .. 'M' and 'R'
I attempted to solve this problem initially by assuming that the cylinder is a solid cylinder (although I realize there is nothing in the problem that specifically states this) I was under the impression that you would necessarily tackle this problem differently for a hollow cylinder, although I am not at all sure of this understanding of the problem.
Anyway that aside, I then used the general expression for the Moment of Inertia for a solid cylinder ... I = 1/2 M R(sqrd)
now from the density variation expression given in the problem .. p = a r
we can rearrange this to .. r = p/a
and substitute this into the first expression giving .. I = 1/2 M R(sqrd) = 1/2 M (p/a)(sqrd)
Now since we have a varying density we will have to integrate to find I
so .. I = integral [(1/2 M (p/a)(sqrd) dM]
and .. I = 1/2 (a(sqrd)/p(sqrd)) integral [M dM]
so .. I = 1/2 (a(sqrd)/p(sqrd)) (1/2 M(sqrd))
giving .. I = 1/4 (a(sqrd) M(sqrd)/p(sqrd))
This is nonsense .. the actual book solution is .. 3/5 M R(sqrd)
I did have other attempts equally unsuccessful, one involved using expression .. I = integral (1/2 M R(squared)dm) .. general expression for solid cylinder ..
giving .. I = 1/2 R(sqrd) integral [M dm] => I = 1/2 R(sqrd) (1/2 M(squared)) => I = 1/4 M(sqrd) R(sqrd)
Which again is nonsense .. I cannot see how to proceed from here and I think that this is because I do not have a proper understanding of the problem and hence how to go about solving it .. can anyone help ?
Jackthehat

 
Sep 27th 2017, 11:19 AM

#2  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,320

The problem is that you are starting with an equation that is based on a homogeneous density. Try starting with the definition of the moment of inertia:
$\displaystyle I = \int r^2 dm $
Here $\displaystyle dm = \rho (r) dV = a ( \frac r R ) L 2 \pi r dr$
From this you can get an expression for I in terms of a, L and R. Then determine the expression for mass, also in terms of a, L and R:
$\displaystyle M = \int _0 ^R \rho (r) L 2 \pi r dr $
Now combine this with your previous result and you get the same answer as the book.

 
Sep 27th 2017, 09:53 PM

#3  Member
Join Date: Jan 2015
Posts: 96

Hi ChipB,
Once again thank you for taking the time to try to help me understand a problem.
However, I am afraid I am still confused and still am finding difficulty in understanding things with this one.
Firstly, are you saying that from the above expression the final expression for 'I' would be .. I = Integral [r(sqrd) a ((r(sqt)/R) L 2 Pi) dr] .. which would end up (taking the constants outside of the integral as .. I = (a L 2 Pi) Integral [(r(sqrd)/R) dr] ?
Similarly taking the constants outside of the second integral does this mean that M = (a L 2 Pi) Definite Integral [Rho(r) dr] [from 0 to R]giving .. M = (a L 2 Pi) Definite Integral [a (r/R)(L 2 Pi r) dr][from 0 to R] M = (a L 2 Pi) (a L 2 Pi) Definite Integral [(r(sqrd)/R) dr][from 0 to R] ..and so M= (a L 2 Pi)(sqrd) Definite Integral [(r(sqrd)/R) dr][from 0 to R]
I am not sure that the above calculations are valid (My understanding is very cloudy about all this I'm afraid).
Finally when you say at the end of your explanation .. "Now combine this with your previous result and you get the same answer as the book." I am not sure what you mean by 'combine' the two results.
I am sorry for being a bit obtuse on this one but I have been very confused from the start I really don't understand fully the mechanics of the solution in detail.
Regards,
Jackthehat.

 
Sep 28th 2017, 05:17 AM

#4  Senior Member
Join Date: Aug 2008
Posts: 113

$\rho = \dfrac{dm}{dV} \implies dm = \rho \, dV = ar \cdot 2\pi r \, dr = 2\pi a r^2 \, dr$
$\displaystyle M = \int_0^R dm = 2\pi a \int_0^R r^2 \, dr = \dfrac{2\pi a \cdot R^3}{3}$
$\displaystyle I = \int_0^R r^2 dm = 2\pi a \int_0^R r^4 \, dr = \dfrac{2\pi a \cdot R^5}{5}$
$I = \dfrac{2\pi a \cdot R^3}{3} \cdot \dfrac{3R^2}{5} = \dfrac{3}{5}MR^2$

 
Sep 28th 2017, 06:17 AM

#5  Senior Member
Join Date: Aug 2010
Posts: 372

Originally Posted by skeeter $\rho = \dfrac{dm}{dV} \implies dm = \rho \, dV = ar \cdot 2\pi r \, dr = 2\pi a r^2 \, dr$
$\displaystyle M = \int_0^R dm = 2\pi a \int_0^R r^2 \, dr = \dfrac{2\pi a \cdot R^3}{3}$
$\displaystyle I = \int_0^R r^2 dm = 2\pi a \int_0^R r^4 \, dr = \dfrac{2\pi a \cdot R^5}{5}$
$I = \dfrac{2\pi a \cdot R^3}{3} \cdot \dfrac{3R^2}{5} = \dfrac{3}{5}MR^2$ 
Typo: that last fraction should be $\frac{2}{5}$, not $\frac{3}{5}$.

 
Sep 28th 2017, 06:58 AM

#6  Senior Member
Join Date: Aug 2008
Posts: 113

Originally Posted by HallsofIvy Typo: that last fraction should be $\frac{2}{5}$, not $\frac{3}{5}$. 
no typo ...
$\color{red}{M=\dfrac{2\pi a \cdot R^3}{3}}$
$I = \dfrac{2\pi a \cdot R^5}{5}= \color{red}{\dfrac{2\pi a \cdot R^3}{3}} \cdot \dfrac{3R^2}{5} = \color{red}{M} \cdot \dfrac{3R^2}{5}$

 
Sep 28th 2017, 07:20 AM

#7  Member
Join Date: Jan 2015
Posts: 96

Hi Skeeter,
Thank you for your input, it has helped me see things a bit clearer. However I am still not 100% about that final line. I followed the first three lines of your calculations but I am a bit confused with the final line I am afraid.
Where does the term .. 3 R(sqrd) / 5 ?
I can follow clearly each part of the process now up until that final line then I'm not sure about that final term multiplying the mass term ... sorry I am sure it must be obvious but I can't see it.
Regards,
Jackthehat.

 
Sep 28th 2017, 07:22 AM

#8  Member
Join Date: Jan 2015
Posts: 96

Hi HallsOfIvy,
Thank you for clearing that typo up .. I thought my arithmetic had gone haywire.
Thanks again for taking the time to help.
Regards,
Jackthehat

 
Sep 28th 2017, 07:32 AM

#9  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,320

Originally Posted by jackthehat Hi ChipB,
Firstly, are you saying that from the above expression the final expression for 'I' would be .. I = Integral [r(sqrd) a ((r(sqt)/R) L 2 Pi) dr] .. which would end up (taking the constants outside of the integral as .. I = (a L 2 Pi) Integral [(r(sqrd)/R) dr] ? 
That should be:
$\displaystyle I = a L 2 \pi \int_0 ^R \frac {r^4} R dr$
(it seems you dropped an r^2 term).
Originally Posted by jackthehat Similarly taking the constants outside of the second integral does this mean that M = (a L 2 Pi) Definite Integral [Rho(r) dr] [from 0 to R]giving .. M = (a L 2 Pi) Definite Integral [a (r/R)(L 2 Pi r) dr][from 0 to R] 
It seems when you pulled the constants out of the definite integral you accidentally left them in. This should be:
$\displaystyle M = \int \rho(r) dM = \int \frac {ar} R L 2 \pi r dr = a L 2 \pi \int _0 ^R \frac {r^2}R dr$
Originally Posted by jackthehat Finally when you say at the end of your explanation .. "Now combine this with your previous result and you get the same answer as the book." I am not sure what you mean by 'combine' the two results. 
Finishing up that first integral will give in terms of a, L , and and an R^4 term. The second integral fives you M in terms of a, L and and an R^2 term. Divide the first by the second and you get I in terms of M and R^2, which is what you're looking for.
Hope this helps.
Last edited by ChipB; Sep 28th 2017 at 07:35 AM.

 
Sep 28th 2017, 08:54 AM

#10  Senior Member
Join Date: Aug 2008
Posts: 113

Originally Posted by skeeter no typo ...
$\color{red}{M=\dfrac{2\pi a \cdot R^3}{3}}$
$I = \dfrac{2\pi a \cdot R^5}{5}= \color{red}{\dfrac{2\pi a \cdot R^3}{3}} \cdot \dfrac{3R^2}{5} = \color{red}{M} \cdot \dfrac{3R^2}{5}$ 
all I did was break the fraction representing $I$ into two factors, where one of those factors represents the mass of the cylinder. If you start with those two factors and multiply, you get back to $I$ ...
$\color{red}{\dfrac{2\pi a \cdot R^3}{3}} \cdot \dfrac{3R^2}{5} = \color{red}{\dfrac{2\pi a \cdot R^3}{\cancel{3}}} \cdot \dfrac{\cancel{3}R^2}{5} = \dfrac{2\pi a \cdot R^5}{5} = I$
another way to look at it ...
$I = kMR^2$, where $k$ is a constant to be determined ...
$k = I \div (MR^2) = \dfrac{2\pi a \cdot R^5}{5} \div \dfrac{2\pi a \cdot R^3 \cdot R^2}{3} = \dfrac{2\pi a \cdot R^5}{5} \cdot \dfrac{3}{2\pi a \cdot R^5} = \dfrac{3}{5}$
Last edited by skeeter; Sep 28th 2017 at 09:06 AM.

  Thread Tools   Display Modes  Linear Mode 
Similar Physics Forum Discussions  Thread  Thread Starter  Forum  Replies  Last Post  Moment of inertia  zinkid  Kinematics and Dynamics  1  Sep 12th 2011 07:52 AM  Moment of Inertia  pre pt marc  Equilibrium and Elasticity  3  Apr 15th 2011 01:30 PM  moment of inertia  dkumardeepthi  Kinematics and Dynamics  1  Feb 27th 2011 12:07 PM  Moment of Inertia?  Football Coach  Kinematics and Dynamics  3  Oct 17th 2010 07:06 PM  Moment of Inertia, lab  Undeadpaladin  Advanced Mechanics  1  Apr 23rd 2008 11:25 AM  