Mini Physics

MENU
  • Forum
  • O Lvl
    • Topic List
    • Formula List
    • Definitions List
    • Practice MCQs
    • Mock Exam Paper 1
    • Close
  • A Lvl
    • Topic List
    • Definitions List
    • Close
  • H3
  • Uni
  • Chemistry
  • Misc
    • Interesting Stuffs
    • Short Q&A
    • Links
    • Close
  • Acc.
    • Mini Physics
      • About Mini Physics
      • Contact Mini Physics
      • Advertise Here
    • T&Cs
      • Acknowledgement
      • Disclaimer
      • Privacy Policy
    • Close
    • Close
MP University Year 1 Mechanics UY1: Calculation of moment of inertia of a hollow/solid cylinder

UY1: Calculation of moment of inertia of a hollow/solid cylinder

Derivation of the moment of inertia of a hollow/solid cylinder

A hollow cylinder has an inner radius R1, mass M, outer radius R2 and length L. Calculate/derive its moment of inertia about its central axis.

hollow cylinder

Guide:
– The cylinder is cut into infinitesimally thin rings centered at the middle. The thickness of each ring is dr, with length L.

We write our moment of inertia equation:

$$dI = r^{2} \: dm$$

Now, we have to find dm, (which is just density multiplied by the volume occupied by one ring)

$$dm = \rho \: dV$$

We’ve introduced dV in the above equation, so, we have to find out what dV is:

$$dV = dA \: L$$

The dA is just the area of the top of the ring, which is the area of the big (radius: r + dr) ring minus that of the smaller (radius: r) ring. We have:

$$dA = \pi (r + dr)^{2} – \pi r^{2}$$

$$dA = \pi (r^{2} + 2rdr + (dr)^{2}) – \pi r^{2}$$

Note: (dr)2 is equal to 0. An infinitesimally small number multiplied by another infinitesimally small number = 0.

$$dA = 2 \pi r \: dr$$

Note: Another way of obtaining dA is by differentiating.

$$A = \pi r^{2}$$

Differentiating wrt r, $$dA = 2 \pi r dr$$

Substituting dA into dV,

$$dV = 2 \pi r L \: dr$$

Using the above equation, substitute into dm,

$$dm = 2 \rho \pi r L \: dr$$

Finally, we have an expression for dm. We substitute that into the dI equation,

$$dI = 2 \rho \pi r^{3} L \: dr$$

Now, we can integrate to find the moment of inertia, (Note: I did not substitute in the expression for density because it is quite messy and it is not needed in the integration process – since the density is not dependent on r)

$$I = 2 \rho \pi L \: \int\limits_{R_{1}}^{R_{2}} r^{3} \: dr$$

I’m sure you are able to do this integration by yourself. Now, we can find the expression for density.

Recall:

$$\rho = \frac{M}{V}$$

Hence,

$$\rho = \frac{M}{\pi (R_{2}^{2} – R_{1}^{2}) L}$$

Substituting this back into the integrated solution, we have:

$$I = \frac{1}{2}M(R_{2}^{2}+R_{1}^{2})$$

Special Cases:

Hoop or thin cylindrical shell: (R1=R2=R)

$$I = M R^{2}$$

Disk or solid cylinder: (R1=0)

$$I = \frac{1}{2} M R^{2}$$

Sanity check: I is expected to be highest for hoop or cylindrical shell since all the mass are furthest away from the axis of rotation.

Back To Mechanics (UY1)

Derivation Of Moment Of Inertia Of Common Shapes:

  • Uniform Rigid Rod (“Beginners’ Lesson”)
  • Uniform Solid Sphere
  • Thin Spherical Shell

Share this:

  • Facebook
  • Twitter
  • WhatsApp
  • Email
  • More
  • Print
  • LinkedIn
  • Telegram
  • Skype
  • Reddit
  • Tumblr
  • Pinterest
  • Pocket

Filed Under: Mechanics

About Mini Physics

Administrator of Mini Physics. If you spot any errors or want to suggest improvements, please contact us.

Related Posts:


What Others Are Saying:

  1. ENRICO UVA scribbled

    July 27, 2018 at 9:15 AM

    Why not go from dV=dA* L directly to dV = 2*Pi*r*dr*L by considering a loop of thickness dr?

    Overall the moment of inertia derivations are very well done.

    Reply to ENRICO UVA
  2. Muhammad Mansoor scribbled

    June 24, 2017 at 11:37 AM

    In the beginning you have already supposed that for an infinitesimally thin cylindrical shell is I=mr^2 in the first equation I=r^2 *dm. No need to mention it again at the end.

    Reply to Muhammad Mansoor
  3. anon woker scribbled

    December 4, 2016 at 9:34 PM

    dude ur the plug thanks soo much

    Reply to anon woker
  4. RICARDO CECHELERO BAGATELLI scribbled

    June 13, 2016 at 5:24 PM

    Thanks a lot, it helped me substantially

    Just one question: In the picture on beginning, the variables should not be L, R1 and R2 instead of h, r1 and r2?

    thanks again

    Reply to RICARDO CECHELERO BAGATELLI
    • casid scribbled

      November 29, 2017 at 4:16 AM

      it doesn’t matter at all

      Reply to casid
  5. Thakur singh scribbled

    May 22, 2016 at 10:29 AM

    Thanx its help me easily

    Reply to Thakur singh
  6. sidhantha scribbled

    January 1, 2016 at 7:41 PM

    Thanks it helped me alot

    Reply to sidhantha
    • Thakur singh scribbled

      May 22, 2016 at 10:31 AM

      its help me very easily. Thanks

      Reply to Thakur singh
  7. Luiz scribbled

    May 22, 2015 at 1:39 AM

    Observe that in my comment below, something like a^2, means the variable ‘a’ to the exponent ‘2’ and R1, R2, the variables R index 2 and 1 respectively.

    I=(1/2)M(R2^2+R1^2) is still incorrect! Because the integration works out to be

    I = (1/2)M(R2^4 – R1^4),

    you could rewrite it as

    I = (1/2)M[R2^2 – R1^2 – 2*(R2^2)(R1^2)/(R2^2 – R1^2)]

    In the case R1 = 0, you’ll get

    I = (1/2)MR2^2. However, you CAN NOT use R1 = R2, for it would mean infinite material density. For that case, I is always ZERO, for only real objects (those with mass) can have I > 0.

    Please, let me know if you have trouble with the algebraic development. :0)

    Reply to Luiz
    • Muhammad Mansoor scribbled

      June 24, 2017 at 11:30 AM

      But refer to wikipedia article “List of moments of inertia” which says :-
      “This expression assumes that the shell thickness is negligible. It is a special case of the thick-walled cylindrical tube for r1 = r2.”

      Reply to Muhammad Mansoor
  8. John Adex scribbled

    May 19, 2015 at 5:27 AM

    Please can you simplify a little because it’s kinda confusing. If not for you I wouldn’t have been able to do my assignment. Thanks.

    Reply to John Adex
    • Mini Physics scribbled

      May 19, 2015 at 2:55 PM

      Could you point out the parts to be simplified?

      Reply to Mini Physics
  9. sonam scribbled

    August 26, 2014 at 11:21 AM

    While cosidering small element ring , thikness of ring is dr but what does it mean that length is L?

    Reply to sonam
    • Mini Physics scribbled

      August 27, 2014 at 8:33 PM

      The rings are actually very thin hollow cylinders, with height L.

      Reply to Mini Physics
  10. kaj1 scribbled

    December 7, 2012 at 9:46 PM

    Shouldn’t it be R2+R1 in the end?

    Reply to kaj1
    • Mini Physics scribbled

      December 8, 2012 at 12:28 AM

      Yes, you’re right. My apologies.

      Reply to Mini Physics
      • kaj1 scribbled

        December 8, 2012 at 1:34 AM

        So p (rho) = M/(pi*(R2^2+R1^2)*L)? Or what?

        Reply to kaj1
        • Mini Physics scribbled

          December 8, 2012 at 8:09 PM

          The density equation is correct in my post, which is $R_{2}^{2} – R_{1}^{2}$. The previous moment of inertia equation is wrong. I’ve corrected it.

          Reply to Mini Physics

Join In The Discussion: Cancel reply

Your email address will not be published. Required fields are marked *

ask-a-question

Top Posts & Pages

  • How To Read A Vernier Caliper
  • How To Read A Micrometer Screw Gauge
  • O Level Physics
  • Difference between boiling and evaporation
  • UY1: Calculation of moment of inertia of an uniform solid sphere
  • Practice On Reading A Vernier Caliper
  • Binding Energy Per Nucleon And Nuclear Stability

Recent Comments

  • Usman on Contact Us
  • zidan on Mock O Level Pure Physics Paper 1
  • Safi Ullah on Practice MCQs For Pressure
  • Abdul_wasay on O Level Physics
  • Abdul_Wasay on Work, Energy and Power (O Level)

Recent Posts

  • Looking to the Future During Aerospace Industry’s Transition Period
  • Contribute To Science With BOINC!
  • Mock O Level Pure Physics Paper 1
  • A.C. Generator (A Level)
  • Velocity Selector

With | 2010 - 2019 | Mini Physics |

loading Cancel
Post was not sent - check your email addresses!
Email check failed, please try again
Sorry, your blog cannot share posts by email.