Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Moment of inertia: off-central diameter

  1. Oct 15, 2012 #1
    Question for the brilliant minds around here:

    I'm trying to figure the amount of torque (I*a) needed to rotate a cylinder of a given mass, diameter and length around a pivot point that is off-center. Typically I'm assuming I would find it out by calculating both I for central diameter and end diameter, and then multiply the off-center percentage by the difference between Icentral and Iend... Right?

    Anyhow, here's the catch:
    The cylinder CoG is at the pivot point (the cylinder is heavier on one end, and the pivot point is positioned so that it's perfectly balanced). I've been working off the assumption that I could simply use the central diameter equation to find the moment of inertia, since the weight is equally balanced, but I'm not sure...

    So if the cylinder is 120cm long, and the pivot point is at 90cm, knowing that both end (long and short) are exactly 2.5kg, can I calculate the MoI using the central diameter equation with 120cm of length, or should I use the length of the longer end*2 (180cm) ?

    That changes the torque requirements quite significantly, so any help would be immensely appreciated!

    PS: I'm referring to the formulas to calculate the Moment of Inertia from Hyperphysics - Moment of Inertia: Cylinder

    Thanks much!

  2. jcsd
  3. Oct 16, 2012 #2
    You can't make that assumption because the local contribution to the moment of inertia depends on the square of the distance from the axis of rotation. You need to calculate the moment of inertia from the general formula. (Integrate dI = density*distance^2*dV over the volume)

    Can you be more specific about where precisely the axis of rotation is, and confirm that your cylinder is three-dimensional? (you didn't specify a radius).
  4. Oct 16, 2012 #3
    Thank you for your reply!

    Yes, the cylinder is three-dimensional; with roughly 7.5cm radius.
    It is 1 meter long (I decided to shorten it, to ease the choice of motor), and weight 5kg - but most of the weight is on one end, so that when it is supported on a central pivot point located at 75cm from one end (3/4), it is perfectly balanced.

    I've used:
    I = (Mr2)/4 + (ML2)/12 to get the MoI - which would be correct if the cylinder mass was consistent across its length and the pivot point was central, but since both are offset equally, I was hoping it would get me in the same ballpark?

    To put it in context, it is a small, lightweight DIY motorised camera arm that I'm putting together. The camera is on a mini pan/tilt head at one end, the pivot point is offset to allow as much travel distance as possible and counter-weight is added at the other end of the arm to keep it balanced. I'm going through these calculation to find out the torque needed for the motor on the arm axis to get decent speed...

    Thanks a lot!
  5. Oct 16, 2012 #4
    Ah. I've found your formula and the derivation. I'm fairly sure the result is different for your specific cylinder. What matters is not only the lengths and the weights, but the positions of the weights along the rod. For example, if you have a rod with a camera on one end plus a counter-weight on the other end, it's a lot easier to calculate the moment of inertia by adding the three individual mass contributions than to model the rod as a solid cylinder with variable density.


    Part 1: calculate the moment of inertia of the uniform density rod first. You can use your formula to calculate the MOI of a 1.5m rod from the centre, and also a 50cm rod from the centre. So if you work out the MOI of a 1.5m version (with central pivot), subtract the MOI of a 0.5m version (with central pivot), and divide by 2, you will end up with the contribution to the MOI of the rod from the 50cm section which is most distant from your pivot. Add this to the 0.5m calculation and you've got your answer, for a uniform density rod!

    Part 2: simply add the counterweight/camera masses together. Eg. if your camera weighs 1kg and it is 75cm from the pivot, its contribution is 1*0.75^2. Same with the counterweight.

    So the thing as a whole can be calculated using a series of symmetry tricks, due to the additive nature of the moment of inertia.
  6. Oct 16, 2012 #5

    Thanks much Mikey. Just to sum it up - and to make sure I understood correctly - here's my process:

    The arm is 1m long, has a radius of 7.5cm and a mass of 1Kg.
    A camera weighting 1Kg is attached on one end, and a 3Kg counter-weight is attached on the other end.
    The arm can pivot around an axis located 75cm from the camera - at the CoG. A motor must be able to move it at up to 25RPM.

    First, the MoI of the arm alone; using cylinder equation for the long and short ends:

    I=(Mr2)/4 + (ML2)/12;

    Long end: (1*0.0752)/4 + (1*1.52)/12 = 0.1889
    Short end: (1*0.0752)/4 + (1*0.52)/12 = 0.0222
    Substract short end from long end, and divide by 2: (0.1889-0.0222)/2 = 0.0834
    Add short end contribution: 0.0834 + 0.0222 = 0.1056

    Camera & counterweight contributions:
    Camera (1Kg @ 75cm): 1*0.752 = 0.5625
    Counterweight (3Kg @ 25cm): 3*0.252 = 0.1875

    Total I for the arm with camera & counterweight: 0.1056 + 0.5625 + 0.1875 = 0.8556

    If the above is correct, then in order to move it at 25RPM:
    25 RPM == 2.618rad/s

    T = I*a; so 0.8556*2.618 = 2.24

    2.24Nm == 22.842 Kg-cm

    All the provided measures are approximate, and this does not take into account the bearing friction and other forces - but is it safe to assume that a motor with a rated torque of 50kg-cm (doubling for comfortable margin) would have no problems moving the arm around at speeds up to 25RPM?

    Thanks again for your help...

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Moment of inertia: off-central diameter
  1. Moment of Inertia (Replies: 3)