Calculating Torque for Stopping Rotating Body - Confirm & Comment

[mechdesignron]

I would like someone to confirm, comment on my understanding of torque calculations of a rotating body.
Formula: T=I\alpha
I=LB-FT² (1366.5) Known
\alpha=Rad/s²(43.33)
Angular acceleration from 600-0 RPM in 1.45 sec

Now I believe I need to divide by 32.2 ft/s² (accelleration of gravity)
This should give me FT-Lbs torque required to (stop) 1366.5 inertia in 1.45 sec.

Please confirm that this is correct with units.

Thank You

Ron

 

[FlexGunship]

I would like someone to confirm, comment on my understanding of torque calculations of a rotating body.
Formula: T=I\alpha
I=LB-FT² (1366.5) Known
\alpha=Rad/s²(43.33)
Angular acceleration from 600-0 RPM in 1.45 sec

Now I believe I need to divide by 32.2 ft/s² (accelleration of gravity)
This should give me FT-Lbs torque required to (stop) 1366.5 inertia in 1.45 sec.

Please confirm that this is correct with units.

Thank You

Ron

Woah there Speedy Gonzales. If you need to include gravity in there then you haven't fully defined the problem for us PFers. Recommend you describe the situation you are trying to solve for.

EDIT: Allow me to elaborate. In a uniform gravitational field, gravity won't effect the stopping torque. That is to say, that all things being equal, it takes the same torque to rotate the same inertia in any gravitational reference frame. If you are simply trying to convert to other units, then please specify.

 

[mechdesignron]

I am testing a brake

Ok, I know I have 1366.5 LB-Ft² of material spinning at 600 RPM, I stop that in 1.45 seconds. What is the torque generated to do that.

I would think that gravity would not be a factor but when I ommit the gravatational constant the results I get are not really possible.

IE: 1366.5 LB-FT²*43.33 Rad/s²=59210.5 FT-Lbs
Add the grav constant: 1366.5 LB-FT²*43.33 Rad/s²/32.2 Ft/s²=1838.8 Ft-Lbs

The later awnser makes sence, I would like someone to confirm that it is correct. Also adding the Grav constant Ft/s² helps the dimensional analysis.

The first awnser cannot be right. If I use it to calculate my COF I get a value of over 1(Impossible)

Let me know what you think.

Thanks
Ron

 

[FlexGunship]

Inertia is a property of matter, but momentum is a property of motion. Calculate your angular momentum.

This sounds suspiciously like a homework problem, so I won't go any further.

 

[mechdesignron]

This is not a homework problem,

Is there someone who understands this enough to at least tell me if what I am doing is wrong. Inertia is a bodys resistance to motion, I have a steel disc, with its weight and diameter I can calculate its inertia, it is spinning at 600 Revolutions per minute. I want to find the torque required to stop this in 1.45 seconds. Torque, rotational force applied to a body. We could also look at this as how much torque is required to take 1366.5 LB-FT^2 from 0-600 RPM in 1.45 seconds.

If you do not understand the problem, please don't comment. I occasionally submit questions to forums but often look at them for information and it is very frustratrating reading through commentary.

 

[FlexGunship]

The first awnser cannot be right. If I use it to calculate my COF I get a value of over 1(Impossible)

This is not a homework problem,

Is there someone who understands this enough to at least tell me if what I am doing is wrong.[?]

If you do not understand the problem, please don't comment. I occasionally submit questions to forums but often look at them for information and it is very frustratrating reading through commentary.

Well, Ron... I would start with your textbook.

Firstly, there's nothing that says coefficient of friction can't be over 1 (which is what led me to believe you were still a student). For example, aluminum against aluminum is over 1. Some silica rubbers against concrete are over 1. Examples are not exotic and are easy to find.

Secondly, I'll give you some help to get you on the right track, but this STILL sounds academic...

Solve for angular momentum: L = Iw (w = omega... sorry about not using Latex)
Solve for torque: t = dL/dt

Start there. If nothing else, your dimensional analysis should've told you that your answer doesn't work.

 

[tvavanasd]

Now I believe I need to divide by 32.2 ft/s² (accelleration of gravity)
This should give me FT-Lbs torque required to (stop) 1366.5 inertia in 1.45 sec.

You are correct.

The issue arises from the fact that the "lb" is a unit of force, not mass. By dividing your inertia by accel due to gravity (32.2 ft/s^2 in this case), you will convert it to a mass based inertia (US mass unit is the "slug" (m=W/g)).

By dividing by gravity, you convert inertia units from lb*ft^2 (weight based, which is confusing) to lb*ft*s^2 (mass based, which is still perhaps confusing).

Using this in your calculation will result in proper torque units of ft*lb.

If you have been exposed to SI units, you may find it simpler to convert to kg and N*m from the start, or just use them for checking.