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Thread: Calculating Acceleration and Torque Required in an Output Shaft

  1. #1
    Associate Engineer
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    Calculating Acceleration and Torque Required in an Output Shaft

    Hello All,
    I am learning Engineering Science at a college in England. One of my assignment questions asks me to calculate the torque required to accelerate a hollow shaft from 500rpm to 3600rpm in 2.4 secs. The shaft is 1.8kg and 0.65m in length. 45mm diameter and 4mm wall thickness.
    I worked the above figures into T=mar2, where a is the angular acceleration (don't know how to do the Greek symbol) and got a worryingly small answer of 0.12 Nm. It might be my lack of experience with these calculations or I might be missing something, but it doesn't feel right. I'm just hoping for a few helpful comments

  2. #2
    Lead Engineer
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    Assuming you used consistent units, quick combing of the basic equations for Jm (Polar Moment of inertia) for a tube and T (torque, Nm) indicates that your equation is generally correct except for one issue. For a tube about its central axis: Jm = 1/2M(R2+ r2) which would revise your formula to T = 1/2M(R2+ r2)a

    For your reassurance on your equation calculation, by calculating the two rotational velocities used to determine the rotational acceleration value in radians/sec gives me (377-52.4)/2.4 = 135.25 radians/sec^2 and using that in your equation my solution is also: T = 1.8 x 135.25 x (.045/2)^2 = .123 Nm, so all other quantities kept equal, using the above revised equation should obtain a correct value for your solution.

  3. #3
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    Fantastic, thanks very much for taking the time to help me

  4. #4
    Administrator Kelly Bramble's Avatar
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    Awesome JAlberts!

    More equation are located here: Polar Moment of Inertia

    Quote Originally Posted by JAlberts View Post
    Assuming you used consistent units, quick combing of the basic equations for Jm (Polar Moment of inertia) for a tube and T (torque, Nm) indicates that your equation is generally correct except for one issue. For a tube about its central axis: Jm = 1/2M(R2+ r2) which would revise your formula to T = 1/2M(R2+ r2)a

    For your reassurance on your equation calculation, by calculating the two rotational velocities used to determine the rotational acceleration value in radians/sec gives me (377-52.4)/2.4 = 135.25 radians/sec^2 and using that in your equation my solution is also: T = 1.8 x 135.25 x (.045/2)^2 = .123 Nm, so all other quantities kept equal, using the above revised equation should obtain a correct value for your solution.
    Last edited by Kelly Bramble; 11-26-2013 at 09:32 AM.

  5. #5
    Associate Engineer
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    Thanks again

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