# Thread: Calculating Acceleration and Torque Required in an Output Shaft

1. ## 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  Reply With Quote

2. 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.  Reply With Quote

3. Fantastic, thanks very much for taking the time to help me  Reply With Quote

4. Awesome JAlberts!

More equation are located here: Polar Moment of Inertia Originally Posted by JAlberts 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.  Reply With Quote

5. Thanks again  Reply With Quote

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•