What is torque divided by angular velocity?

Jun 2017
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What is the best way to simplify torque as a proportion of angular velocity? Is that even a thing? Is there something close?
 
Apr 2017
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So... torque is measured in Newton meters ....

We have an arm fixed at one end ,but free to rotate about that anchor point.

We can apply a torque of 1Nm , which will want to make the arm rotate about the anchor point ... this could be a force of 1N , 1 m from the pivot ...or 2 N half a meter from pivot , of 4N a quarter meter from pivot ...etc .. all exert a torque of 1Nm

torque = f x (distance of force from anchor pt.)

Whether this arm has angular velocity or not is irrelevant .

It's very similar to linear force acting on a mass , it doesn't matter if the mass already has a velocity , the aplied force will want to change this already present velocity.
 
Aug 2008
113
35
What is the best way to simplify torque as a proportion of angular velocity? Is that even a thing? Is there something close?
Is there a problem statement associated with this query? May help if you posted from where you're getting this idea.
 
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Apr 2016
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I read youre question wrong first. Since it seems that you are asking of the ratio of torque and angular velocity I did not come up with a simple construct. But if you want a relation you can find it in the attachment.
 
Last edited:
Jun 2016
1,189
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England
applying a torque will generate an angular acceleration
so a change of angular velocity with respect to time.
the rate of change of angular velocity will depend of course on the size or the torque,
but also on the rotational inertia of the object being acted upon:
https://en.wikipedia.org/wiki/Moment_of_inertia

Note that there is a close match to the "standard" Force = Mass x Acceleration.
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
The correlation of torque and angular velocity is:

\(\displaystyle \vec \tau = I \frac {d\vec {\omega}} {dt} \)

where \(\displaystyle \vec {\tau} \) is torque, \(\displaystyle \vec {\omega} \) is angular velocity, and I is moment of inertia. For cases of constant torque you could integrate this to get:

\(\displaystyle \tau \Delta t = I \Delta\omega \)

If the initial angular velocity is 0, then this becomes:
\(\displaystyle \tau \Delta t = I \omega \)

Rearrange to get:

\(\displaystyle \frac {\tau}{\omega} = \frac I {\Delta t }\)

So, for this simple case torque divided by angular momentum = moment of inertia divided by time. Not a very useful equation, but I hope it helps!