Physics Help Forum Moment of inertia for a slender rod of varying mass

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 Jun 11th 2017, 09:08 AM #1 Member   Join Date: Jan 2015 Posts: 94 Moment of inertia for a slender rod of varying mass Hi, I have come across a problem where I eventually got the answers to two parts of the problem, but only after I had attempted each part first and got it wrong then, for each part sort of 'reverse-engineered' the solution from knowing what the correct solution was and although I came away with two correct solutions I really don't understand why the solutions are what they are, and of course before I go on to the next problem I would appreciate it if someone could explain the solutions to me. The problem is detailed below .. Problem A slender rod of length L has mass per unit length that varies with distance from the left end of the rod (where x=0), according to the formula .. dm/dx = Y x , where Y has units kg/m(sqrd) (a) Calculate the total mass of the rod in terms of 'Y' and 'L' (where L is length of the rod) (b) Use equation .. I = Integral [r (sqrd) dm] to calculate the Moment of inertia for an axis at the left end, perpendicular to the rod. Use the expression derived in part (a) to express I in terms of 'm' and 'L'. (c) Repeat part (b) for an axis at the right hand side of the rod. How do the results for part (b) and (c) compare ? Explain the results. As I stated I was able to solve part (a) easily enough and for parts (b) and (c) I sort of reverse-engineered the correct solutions once I knew what the correct solutions were, but I do not understand why the moments of inertia for the different ends of the rods should be different and why one is 1/3 the value of the other (that is part (b) I = m L(sqrd)/ 2 and for part (b) I = m L(sqrd) / 6). Can anyone explain these solutions to me as I am uncertain as to why they are what they are ? Thanks Jackthehat
 Jun 11th 2017, 11:09 AM #2 Senior Member     Join Date: Aug 2008 Posts: 113 I'm using $k$ instead of $Y$ if that's ok ... $\displaystyle \frac{dm}{dx} = kx \implies dm = kx \, dx$ $\displaystyle M = \int_0^L dm = \int_0^L kx \, dx = \frac{kL^2}{2} \implies k = \frac{2M}{L^2}$ from the end where the density is least to the end where the density is greatest ... $\displaystyle dI = r^2 \, dm \implies I = \int_0^L x^2 \cdot \frac{2M}{L^2} x \, dx = \frac{2M}{L^2} \int_0^L x^3 \, dx = \frac{2M}{L^2} \cdot \frac{L^4}{4} = \frac{ML^2}{2}$ from the end where the density is greatest to the end where the density is least ... $\displaystyle \frac{2M}{L^2} \int_0^L (L-x)^2 \cdot x \, dx = \frac{2M}{L^2} \int_0^L L^2x-2Lx^2+x^3 \,dx = \frac{2M}{L^2} \cdot \frac{L^4}{12} = \frac{ML^2}{6}$
 Jun 12th 2017, 07:56 AM #3 Member   Join Date: Jan 2015 Posts: 94 Moment of inertia for a slender rod of varying mass Hi Skeeter , Thank you for taking the time to look at the problem and give me an answer, much appreciated. However although I can see from the maths of it why the solutions turn out to have the final values they do, I still do not understand the reason behind it. For the first part you use .. 2ML2∫L0x3dx and for the second part you use .. 2ML2∫L0(L−x)2⋅xdx I am not sure of the reasons why you use different valued initial intergrals for the each part , that is axis at right hand side and axis at the left hand side of rod, after all the rod has the same total mass ? When you have the time, could you explain the thinking behind this ? Regards, Jackthehat
 Jun 12th 2017, 03:16 PM #4 Senior Member     Join Date: Aug 2008 Posts: 113 reference the attached sketch ... The rod has a linear mass density $\dfrac{dm}{dx} = kx$ based on the position of a representative "piece" of rod mass, $dm$, relative to $x = 0$. As $x$ increases from $0 \to L$, the rod becomes more dense, hence $\color{blue}{dm} < \color{red}{dm}$. Let's just consider the two marked pieces of $dm$ (blue & red) in the diagram ... ignore the remainder of the rod for now. For any point mass, its moment of inertia is $I = m \cdot r^2$, where $r$ is the distance from the axis of rotation. If the left end is the axis of rotation, the moment of inertia of just those two pieces would be $I = \color{blue}{dm \cdot x_1^2} + \color{red}{dm \cdot x_2^2}$ If the right end is the axis of rotation, the moment of inertia of just those two pieces would be $I = \color{red}{dm \cdot (L-x_2)^2} + \color{blue}{dm \cdot (L-x_1)^2}$ so, summing the infinitesimal moments for every $dm$ on the rod ... from left to right $\displaystyle \implies I = \int_0^L x^2 \, dm$ from right to left $\displaystyle \implies I = \int_0^L (L-x)^2 \, dm$ Attached Thumbnails
 Jun 12th 2017, 08:38 PM #5 Member   Join Date: Jan 2015 Posts: 94 Hi Skeeter, Thank you very much for that explanation of how the correct solution is achieved in this particular problem. The explanation is both clear and concise and I now get it. I just couldn't think of exactly how the varying mass (with position) could be tackled. I initially thought that you could possibly take an average of the mass variance (ie [greatest mass + least mass]/2) for the right hand axis and take the integral .. doing this I think I got the right answer (according to the book) but because I didn't have a full understanding of how the varying mass affects the problem I was unhappy with my solution .. a solution that came from knowing the answer and working backwards and trying to find a way to fit my calculations so as to achieve the desired result. Obviously this was no way tackle a problem, you really have to have an full understanding of the problem and this usually suggests how to proceed to find the solution and in this case I didn't have that understanding. Thank you again for your time and your help. Regards, Jackthehat

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