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=mar[SUP]2[/SUP], 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

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(R[SUP][SIZE=2]2[/SIZE][/SUP]+ r[SUP][SIZE=2]2[/SIZE][/SUP]) which would revise your formula to T = 1/2M(R[SUP][SIZE=2]2[/SIZE][/SUP]+ r[SUP][SIZE=2]2[/SIZE][/SUP])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.

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

Awesome JAlberts!

More equation are located here: [URL="http://www.engineersedge.com/mechanics_machines/mass_moment_of_inertia_equations_13091.htm"]Polar Moment of Inertia[/URL]

[QUOTE=JAlberts;7927]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(R[SUP][SIZE=2]2[/SIZE][/SUP]+ r[SUP][SIZE=2]2[/SIZE][/SUP]) which would revise your formula to T = 1/2M(R[SUP][SIZE=2]2[/SIZE][/SUP]+ r[SUP][SIZE=2]2[/SIZE][/SUP])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.[/QUOTE]

Thanks again