How Do You Calculate Torque for a Rotating Shaft with Bearings?

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Discussion Overview

The discussion revolves around calculating the torque required for a rotating shaft supported by bearings, considering factors such as moment of inertia, angular acceleration, and friction. Participants explore the implications of these factors on the torque calculation and the conditions under which the shaft operates.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that torque can be calculated using the formula: Torque = Moment of Inertia x Angular Acceleration, and provides specific values for the shaft and bearings.
  • Another participant questions the application of the torque formula, suggesting that if there is friction, the torque needed to maintain constant speed must account for overcoming friction.
  • There is a discussion about the angular acceleration value, with some participants asserting that if the shaft is to rotate at a constant speed of 15 rpm, the angular acceleration should be zero.
  • One participant mentions the need to convert units consistently, either to US Customary or SI units, to avoid confusion in calculations.
  • Another participant expresses uncertainty about the relationship between friction and torque, suggesting that without considering friction, even a minor torque could keep the shaft accelerating.
  • Further clarification is sought regarding the calculation of angular acceleration and its derivation from rpm to rad/s, as well as the units for torque in the context of the discussion.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the correct interpretation of angular acceleration in relation to constant shaft speed and the role of friction in torque calculations. There is no consensus on the correct approach to calculating torque under the given conditions.

Contextual Notes

Participants express uncertainty about unit conversions and the implications of friction on torque calculations. There are unresolved questions about the correct application of formulas and the assumptions underlying the calculations.

Who May Find This Useful

This discussion may be useful for individuals interested in mechanical engineering, particularly those dealing with rotational dynamics, torque calculations, and the effects of friction in mechanical systems.

ever1221
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Hello Guys,

I have a 90mm=3.543" shaft going through 2 bearings that are the same shaft weighs 10lbs bearings weigh 25lbs each hub of bearing weigh 5lbs, bearing coefficient of friction is 0.15

bearing OD: 190mm ID: 90mm

ok so I thought Torque to rotate shaft is = Moment of Inertia x angular acceleration

I want the shaft to be able to rotate at an angular speed of 15rpm so angular acc is 1.57 rad/s^2

angular acc= 1.57 rad/s^2
I= 1/2 M(r^2) --> 1/2 x 10lb x (1.77^2) = 15.7 lbm.in^2 (does this have to be converted to pound force?)

T= 15.7 x 1.57

is this right? Also how do I use coefficient of friction??
 
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If there is friction then a torque is needed to keep things moving at a constant rate, the applied torque just equals the torque needed to overcome friction. You are using the formula incorrectly?
 
ever1221 said:
Hello Guys,

I have a 90mm=3.543" shaft going through 2 bearings that are the same shaft weighs 10lbs bearings weigh 25lbs each hub of bearing weigh 5lbs, bearing coefficient of friction is 0.15

bearing OD: 190mm ID: 90mm

ok so I thought Torque to rotate shaft is = Moment of Inertia x angular acceleration

I want the shaft to be able to rotate at an angular speed of 15rpm so angular acc is 1.57 rad/s^2

angular acc= 1.57 rad/s^2
I= 1/2 M(r^2) --> 1/2 x 10lb x (1.77^2) = 15.7 lbm.in^2 (does this have to be converted to pound force?)

T= 15.7 x 1.57

is this right? Also how do I use coefficient of friction??
Without friction even minor torque can also rotate the shaft. So i don't suppose it will need such complicated calculations. Just force required will be equal to force required to overcome friction i suppose.
 
ever1221 said:
I want the shaft to be able to rotate at an angular speed of 15rpm so angular acc is 1.57 rad/s^2

How do you figure this? If the shaft speed is a constant 15 rpm, then the angular acceleration is zero. Where did you get the 1.57 r/s^2 figure?

I would strongly recommend getting everything into the same units system, either US Customary or SI, but not mixed as you have it.
 
OldEngr63 said:
How do you figure this? If the shaft speed is a constant 15 rpm, then the angular acceleration is zero. Where did you get the 1.57 r/s^2 figure?

I would strongly recommend getting everything into the same units system, either US Customary or SI, but not mixed as you have it.
And i think for getting specific speed at specific acceleration friction should be considered first because without considering friction it will always be in acceleration even with a minor amount of force.
 
Kishan, I do not understand your statement,
Kishan Majethia said:
because without considering friction it will always be in acceleration even with a minor amount of force.

What are you saying, please?
 
OldEngr63 said:
Kishan, I do not understand your statement,What are you saying, please?
I want to say without considering friction and with application of minor amount of force body will always be in acceleration without considering magnitude of force except zero
 
OldEngr63 said:
How do you figure this? If the shaft speed is a constant 15 rpm, then the angular acceleration is zero. Where did you get the 1.57 r/s^2 figure?

I would strongly recommend getting everything into the same units system, either US Customary or SI, but not mixed as you have it.

thanks guys for the replies

ok so how I got angular acc is by converting rpm to rad/s which is 1.57 rad/s and I wanted it to reach that acc within 1second so acc=angular velocity/time basically that's the motor am going to be getting, one with 15rpm but torque is what am not sure ofam having problems with units, what's the units of torque here when I use the following? I need it to be in ft lbs
I= 1/2 M(r^2) --> 1/2 x 10lb x (1.77^2) = 15.7 lbm.in^2 (does this have to be converted to pound force?)

T= 15.7 x 1.57also guys can I use this for starting toque??

http://www.engineersedge.com/friction/friction_calc_protected/shaft_journal.htm
 

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