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

In summary: 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.
  • #1
ever1221
2
0
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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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?
 
  • #7
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
 
  • #8
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
 

1. What is torque and why is it important in calculations?

Torque is a measure of the force that causes an object to rotate around an axis. In other words, it is the measure of an object's ability to rotate. Torque is important in calculations because it helps us understand and predict the motion of objects.

2. How do you calculate torque?

Torque is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied. The formula for torque is T = F x d, where T is the torque, F is the force, and d is the distance.

3. Is there a difference between clockwise and counterclockwise torque?

Yes, there is a difference between clockwise and counterclockwise torque. Clockwise torque is when the rotation is in the same direction as the hands of a clock, while counterclockwise torque is when the rotation is in the opposite direction.

4. What are some real-life examples of torque?

Some real-life examples of torque include turning a doorknob, opening a jar lid, and using a wrench to tighten a bolt. In all of these examples, a force is applied at a distance from the axis of rotation, resulting in torque.

5. How does torque relate to angular acceleration?

Torque is directly related to angular acceleration, as stated in Newton's Second Law, which states that the net torque acting on an object is equal to the moment of inertia times the angular acceleration. In simpler terms, the greater the torque applied, the greater the angular acceleration will be.

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