Torque required to accelerate a steel shaft

[rwfriestad]

Background -
I am a retired aeronautical and mechanical engineer (70 years old) and closed my consulting/manufacturing firm about 2 years ago. My technical books are still in storage and I can't get to them right now.

A friend has asked me to assist with some machine design calculations and my classes in dynamics were a loooong time ago. While "googling" around I came across this website and hope you can help me.

Problem

I need to calculate the torque required to accelerate and decelerate a 1.5 inch diameter steel shaft through a 540 degree move in 0.35 seconds from start to stop. I am presuming that the calculation would be linear with respect to the length of the shaft. Acceleration may be considered constant for purposes of calculation. Whipping is not considered a problem.

Either the equations or pointing me in the right direction to locate them on the internet would be appreciated. My local library is of very little assistance with this sort of thing and my personal memory bank no longer retains this information in storage.

 

[torquil]

I need to calculate the torque required to accelerate and decelerate a 1.5 inch diameter steel shaft through a 540 degree move in 0.35 seconds from start to stop.

Ok, the steel shaft is not acted upon by any other forces than the torque supplied to accelerate/deccelerate it?

I am presuming that the calculation would be linear with respect to the length of the shaft.

Agreed.

Acceleration may be considered constant for purposes of calculation. Whipping is not considered a problem.

I don't know what "whipping" is, but I'm assuming it means that the motion of the steel will consist solely of rotation about its axis of rotational symmetry.

Either the equations or pointing me in the right direction to locate them on the internet would be appreciated. My local library is of very little assistance with this sort of thing and my personal memory bank no longer retains this information in storage.

You need the moment of inertia of the "solid cylindrical" steel rod. If I'm not mistaken, it is

I = 0.5*M*r^2

where K is the mass of the entire rod, and r is its radius. It might be wise to double check it by looking it up on wikipedia.

You want the constant torque needed for it to "travel" 270 degrees during a time 0.175s. After that it is just a matter of reversing the torque. 270 degrees is 4.7124 radians.

So it corresponds to a constant angular acceleration of

2*4.7124 rad / (0.175s)^2 = 307.75 rad/s^2

Using the formula relating angular acceleration, moment of inertia and torque, you can calculate the needed torque tau:

tau = (moment of inertia) * (angular acceleration) = I * (307.75 rad/s^2)

the result will be given in Newton*meter if the mass is given in kg, and the radius in meters. Of course, you can redo this using inches/feet/etc and get a result in pounds*feet.

NB: I just wrote this down quite quickly, so there may be a few errors/misunderstandings.

 

[rwfriestad]

Thanks, that got me going in the right direction and jogged some little used memory cells