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vantageman · Feb 9, 2011, 06:04 AM

How do I calculate the minimum torque to rotate a .925lb, 0.625' diameter shaft with a .925lb 2.19' hub 10K rpm within 3 seconds? I am trying to spec out an electric motor that I can couple up to the shaft supported by two ball bearings to spin test the hub on.

T=IxA
"I" inertia for a solid shaft, I=1/2MR^2 = 1/2*0.925lb * (1.095in)^2 = 0.55 lb-in^2
"R" - radius = (2.19in/2)= 1.095in
"M" - mass = 0.925 lb
"A" assume to use Angular Acceleration, A= 10,000 rpm * 2pi * 1/60 sec = 1047 rad/sec

T = 0.55 lb-in^2 * 1047 rad/sec * 3 seconds = 1727.55 lb-in^2 This is where it does not makes sense, because I know T units british units if lb-in. How do I get rid of one of the "in" units? Where did I go wrong?
Thanks.

ebaines · Feb 9, 2011, 07:08 AM

Two things:

1. The angular acceleration is 1047 rad/sec divided by 3 seconds, or 349 rad/s^2

2. You need to convert pound mass to pound force. You do this by using the identity

$ 1 = 32.2 \frac {lb_m}{lb_f} \frac {ft}{s^2} $

The torque calculation is then:
$ T= I \alpha = 0.55 lb_m \ in^2 \ \times \ 349 \frac {rad}{s^2} \ =\ 192 \frac {lb_m \ in^2}{s^2} =\ 192 \frac {lb_m in^2}{s^2} \times (\frac 1{32.2}\ \frac {lb_f}{lb_m} \frac {s^2}{ft}) \ \times \ (\frac {1 ft}{12 in})_\ =\ 0.5 lb_f- in $

Note that this torque only covers the hub; you need to do a similar calculation for the shaft.

vantageman · Feb 9, 2011, 07:58 AM

Thanks ebaines.
Then to calculate hp, if I remember correctly is HP = (T*frequency)/5252
HPhub = (.5 lb-in * 10000 rev/min * 1/60 min/sec * 2pi)/5252
= 0.207 lb-in/sec

ebaines · Feb 9, 2011, 08:31 AM

I believe it's HP = T*RPM/5252, where T is in lb-ft (not lb-in):

HP = 0.5 lb-in * 1 ft/12 in * 10^4 RPM/5252 = 0.08 HP