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If I have a torque value of, let's say, 100 Nm, am I right in saying that it takes 100 Joules of energy to

__generate__that torque since 1 Nm = 1 Joule.

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If I have a torque value of, let's say, 100 Nm, am I right in saying that it takes 100 Joules of energy to

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Doc Al

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No, that's not right. Even though torque is measured in Nm, and a Nm is dimensionally equivalent to a Joule, that does not mean that torque and energy are the same thing or that it is correct to measure torque in Joules. (Joules are reserved for energy--never used for torque.)If I have a torque value of, let's say, 100 Nm, am I right in saying that it takes 100 Joules of energy togeneratethat torque since 1 Nm = 1 Joule.

And generating a torque doesn't necessarily require any energy at all, just as generating a force doesn't.

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Dale

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Got it. Thanks

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jtbell

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Doc Al

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Sure.

Note that work = torque*angle is the rotational equivalent to work = force*distance.

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Pay attention to the post regarding "scalar" versus "vector". For example, what happened if you've moved for the whole 2pi radians when the force is conservative? How much work have you done?

Zz.

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Doc Al

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Doc Al

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No. If you apply a torque of 100 Nm continuously through an angle of 2π radians, then the work you do is 100*2π = 200π Joules.I wasn't asking about conservative forcebut I believe the answer to the question posed by ZapperZ is 0 Joules since the distance between the starting point and the ending point is 0, then 0 * 100Nm = 0 Joules, or...?

Edit: Since

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Dale

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The work done by a conservative force around a closed path is 0 so I think that a conservative force cannot apply a continuous torque through an angle of 2pi radians. If we look at the torque on a wheel due to the conservative force of gravity we see that the net torque is 0 (otherwise it would spontaneously spin faster and faster).

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Hi There

I don't know for sure, but, if we know the circumference of the circle and apply a 10 N.m torque * 2randians (and 2rads = 2 meters of circumference travel)

Then 10n.m * 2m = 20N.m of work? And I presume that also = 20 joules?

Willy

I don't know for sure, but, if we know the circumference of the circle and apply a 10 N.m torque * 2randians (and 2rads = 2 meters of circumference travel)

Then 10n.m * 2m = 20N.m of work? And I presume that also = 20 joules?

Willy

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Doc Al

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No. The work done is torque*angle, not torque*distance. (Check the units!)I don't know for sure, but, if we know the circumference of the circle and apply a 10 N.m torque * 2randians (and 2rads = 2 meters of circumference travel)

Then 10n.m * 2m = 20N.m of work?

(Realize that this thread is several years old.)

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