If there are heavy components mounted on it, what will happen? I think it accelerates very slowly, but reaches at the highest RPM at the end because torque is equal to angular acceleration * moment inertia.
Am I correct?
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1$\begingroup$ No different than pushing something very heavy on a ice or wheels with very little force. $\endgroup$– DKNguyenMar 20 at 17:41
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$\begingroup$ @DKNguyen then case that we have a rotating shaft with high torque but little RPM is similar to it that I'm pushing something heavy on highly-frictional ground with much force? Can we compare it like this based on your analogy? It will reach its maximum RPM very quickly, but maximum RPM is low. Can I say like this? $\endgroup$– Jawel7Mar 23 at 10:59
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$\begingroup$ Yes. But the point is that in all systems, once you provide enough force or torque to cancel out friction, it behaves like a frictionless system. So any excess torque or force results in acceleration, even if it is very little. $\endgroup$– DKNguyenMar 23 at 13:22
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$\begingroup$ "because torque is equal to angular acceleration * moment inertia." This is wrong though. The RPM stops increasing when some speed dependent loss equals the torque, or when you simply can't apply the torque fast enough (i.e. you can't turn the crank fast enough even though you still have torque to spare). $\endgroup$– DKNguyenMar 23 at 15:40
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2 Answers
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As long as the torque is greater than any of the resistances the shaft is experiencing, then yes.
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All else being equal, yes.
However
Real imbalances will tend to be bigger, or need more labor to bring them down to some fixed value. Power loss due to vibration will tend to be proportional to the absolute value of the imbalance, not to the proportion of the thing's mass.
Bearings that can support something heavy will tend toward greater drag than bearings that can only support something light.
So in real life, having a big heavy "thing" will have greater losses, and will be harder to spin. Or the amount of effort -- and thus expense -- that needs to be put into getting those losses down will be greater.
