Q: Find out the moment of inertia of a uniform circular disc of radius $r$ & mass $M$ & the axis passes through a point on the circumference.

My attempt: Let the axis passes through $O$ on the circumference. Then I took a ring at $x$ unit from the axis and width $dx$. Now, area of the disc is$$dA =2\pi(r - x).dx$$ . Since, the disc is uniform, density $$ D = \dfrac{M}{\pi .r^2} .$$ Therefore, mass of the ring $$ dM = \dfrac{2M(r - x)dx}{r^2}$$ . Now, moment of inertia w.r.t. the axis $$dI = \dfrac{2M(r - x)x^2 .dx}{r^2}$$ . Therefore, moment of inertia of the disc $$ \int_0^{2r} dI = \int_0^{2r} \dfrac{2M(r - x)x^2 .dx}{r^2} \implies I = \dfrac{2M}{r^2} [ \int_0^{2r} (r - x)x^2 .dx \implies I = \dfrac{2M}{r^2} [ r\int_0^{2r} x^2 .dx - \int_0^{2r} x^3 .dx] \implies I = \dfrac{2M}{r^2} [\frac{8r^4}{3} - 4r^4] \implies I = - \dfrac{8Mr^2}{3}.$$ A hard work! But all in vain!! Moment of inertia is negative??? Where did I mistake?? Please help.


closed as off-topic by David Z Nov 29 '14 at 7:11

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  • $\begingroup$ Are the rings you are integrating concentric? Where are they centered? There are two axes involved; which one do you mean in your attempted solution? $\endgroup$ – DJohnM Nov 29 '14 at 5:44
  • $\begingroup$ @David Z: Sir, what do want to mean by "effort" ? Is thisn't an effort?? Sorry, I can't agree with you! Frustrating!! $\endgroup$ – user36790 Nov 29 '14 at 8:06
  • $\begingroup$ ...and is it off-topic? Then something is wrong really. My attempt is not a minimal to you, right?? Can you define for me, what is actually an effort ? Thanks, sir! $\endgroup$ – user36790 Nov 29 '14 at 8:16
  • $\begingroup$ That is one great bit of effort! I have flagged for reopening, but I am not exactly all that 'valued' (not a theoretical physicist) $\endgroup$ – user60063 Nov 29 '14 at 15:35
  • $\begingroup$ See the associated meta post. @SabreTooth I'd suggest taking a look at the arguments made there against reopening this question. $\endgroup$ – David Z Nov 29 '14 at 16:15

with your definition of $dA$ you must integrate between r and 0, because you start at the center, and the rings grow in radius as you go for (r-x) from x=r to x=0

  • $\begingroup$ Sir, then I have taken wrong limits, right? $\endgroup$ – user36790 Nov 29 '14 at 5:49
  • $\begingroup$ yes, the rest seems ok, but you can always check if the result is correct by computing it in a different way as suggested in the other answer. But I agree with you, you must be able to calculate it any way you you want. $\endgroup$ – Wolphram jonny Nov 29 '14 at 5:51
  • $\begingroup$ ...nevertheless, the limits should be $\int_{r}^{0}$ , right? Thanks again:) You saved me more than once! $\endgroup$ – user36790 Nov 29 '14 at 8:21
  • $\begingroup$ @wolprhramjonny the meta post explains exactly why this is happening. So now you don't need to believe that we don't read the questions anymore. :-) $\endgroup$ – David Z Nov 30 '14 at 7:13

There is a general and simple formula to calculate the moment of inertia with respect to some axis if the moment of inertia with respect to another axis is known. I am pretty sure you'll be able to find it in your textbook. The theorem is parallel axis theorem.

  • $\begingroup$ Would that be "another parallel axis"? $\endgroup$ – DJohnM Nov 29 '14 at 5:23
  • $\begingroup$ Yes, we know that moment of inertia of a disc about the com of the disc is mr^2/2. Using parallel axis theorem we can find moment of inertia about circumference $\endgroup$ – Sashurocks Nov 29 '14 at 5:25
  • $\begingroup$ The parallel axis theorem is the right way to solve this problem and avoids a lot of the mathematical manipulation (with the potential for mistakes). You would immediately get $I=\frac12mr^2 + mr^2 = \frac32mr^2$ (assuming the axis is perpendicular to the disk - this is not specified in the question). $\endgroup$ – Floris Dec 11 '14 at 14:44