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

Torque Calculation

  1. Jan 12, 2009 #1
    Hi,

    I am stuck on what should be a fairly simple problem. I need to calculate the torque required to rotate a pipe weighing 30 lbs resting on bearings at constant speed. How should my approach be? Do I need to actually measure the torque required using a sensor?

    Thanks,
    Mike
     
  2. jcsd
  3. Jan 12, 2009 #2

    minger

    User Avatar
    Science Advisor

    The problem that you're more than likely running into is that torque or moments cause angular acceleration. The relation is analagous to F=ma, where rather than a force, we have a moment; a mass moment of inertia rather than mass, and angular acceleration rather than linear acceleration.

    The force needed to maintain constant speed will be the force needed to overcome friction. Your main frictional forces will be in the bearings, which calculating decent numbers may be tricky.
     
  4. Jan 14, 2009 #3
    My problem specification is as follows:

    mass, m = 300 lbs
    Hollow tube with OD = 6" & ID = 5"
    Initial angular velocity W1= 0 rpm
    Final angular velocity W2= 20rpm
    Change in time, Delta t = 3sec

    So using,
    I=(m(OD^2 + ID^2))/2 = 9160 lb-in^2
    Angular acceleration, alpha = (W2-W1)/delta t = 6.67 rad/s^2

    Torque, T = I*alpha = 61031 lb-in

    I think the torque value looks too high....Do you see any problem in my procedure?
     
  5. Jan 14, 2009 #4
    Look at your units:

    You have (lb-in^2)*(rad/s^2) [=] lb-in^2/s^2
    which is not the in-lb you claimed for the resulting torque.

    The problem is that you did not compute the mass moment of inertia. Instead you computed a quantity usually called Wk^2, the weight * radius of gyration^2. The relation you need is
    I = Wk^2/g = Mk^2

    This is a result of confusing weight and mass. If you have a quantity expressed in pounds (lb), it is a weight or a force. If it is a mass, in the US Customary system it will either be in slugs, or if you are working in inches, the units for mass will be (lb-s^2/in). (Yes, I know it seems awkward, but that is just how it is!)
     
  6. Jan 15, 2009 #5
    Thanks for the reply! Yes you are right.....Here is my updated calculation,

    mass, m = 300 lbs = 9.324 slug
    Hollow tube with OD = 6" & ID = 5"
    Initial angular velocity W1= 0 rpm
    Final angular velocity W2= 20rpm = 2.094 r/s
    Change in time, Delta t = 3sec

    So using,
    I=(m(OD^2 + ID^2))/2 = 284.38 lb-in^2
    Angular acceleration, alpha = (W2-W1)/delta t = 0.7 rad/s^2

    Torque, T = I*alpha = 200 lb-in

    This torque value looks sensible. Does this look right?

    Thanks,
    Mike
     
  7. Jan 15, 2009 #6
    You still have an error because you used the wrong value for g. Look at your units for I.

    You have slug-in^2 [=] (lb-s^2/ft)*in^2

    Your units don't work.

    You needed to get the mass in units of lb-s^s/in, and to do that you have to express g in units of in/s^2, not ft/s^2. Your torque value is still considerably too high.

    You really do need to check your units, really check them, carefully, if you want to get correct results. You could have saved yourself a lot of time on this problem if you had paid attention to the units.
     
  8. Jan 16, 2009 #7
    Thanks for the input! I realy appreciate it!

    I redid my calculations and verified them by working in metric. here they are:

    In English units:
    mass, m = 300 lbs = 300/386.4 = .7764 slug (lb-s^2/in)
    Hollow tube with OD = 6" & ID = 5"
    Initial angular velocity W1= 0 rpm
    Final angular velocity W2= 20rpm = 2.094 r/s
    Change in time, Delta t = 3sec

    So using,
    I=(m(OD^2 + ID^2))/2 = 23.68 lb-s^2-in
    Angular acceleration, alpha = (W2-W1)/delta t = 0.7 rad/s^2

    Torque, T = I*alpha = 16.7 lb-in

    In Metric Units:
    mass, m = 300 lbs = 136.07 kg
    Hollow tube with OD = .1524m & ID = .127m
    Initial angular velocity W1= 0 rpm
    Final angular velocity W2= 20rpm = 2.094 r/s
    Change in time, Delta t = 3sec

    So using,
    I=(m(OD^2 + ID^2))/2 = 2.67 kg-m^2
    Angular acceleration, alpha = (W2-W1)/delta t = 0.7 rad/s^2

    Torque, T = I*alpha = 1.87 N-m = 16.6 lb-in

    Significant difference and now I'm not sure if it is too low. But the calculations seems ok now

    Thanks Again,
    Mike
     
  9. Jan 16, 2009 #8
    You really should have considerable confidence in these calcs by now. You have done them in two different unit systems and gotten essentially the same result (the difference is probably in the slight difference in your value for g). If it is any consolation, this is the result I got also.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Torque Calculation
  1. Torque calculations (Replies: 3)

  2. Torque calculations (Replies: 3)

  3. Torque calculation (Replies: 4)

  4. Torque calculation (Replies: 3)

Loading...