# Thread: trying to make a connection between torque and linear motion

1. i'm trying to make a comparison between torque induced kinetic energy, and linear kinetic energy, but.. well, i guess i'm not good enough at this stuff yet.

this is what i'm currently doing for linear motion:

1.
Force = Mass * Acceleration
2.
Acceleration = Force/Mass
3.
final velocity = sqrt(initial velocity^2 + 2*Acceleration*distance)
4.
KE = 1/2 * Mass * Final_Velocity^2
5.
pressure = Force / Area

1. first find the strength which the weapon is hit, thrown with. (done once)
2. then find out how fast this strength accelerates the object.
3. find the final speed the object has when its done accelerating.
4. find the kinetic energy that the object has when its done accelerating
5. find the pressure the object does on a surface

currently for a spear, i've got an estimated throw force of 193.12 newtons, based on the acceleration length being about 1.25m, and speed of the spear being 48kph,
and weight 2.72kg.

on mythbusters, one of the guys was able to slash a 1.10kg katana at 47mph,
and a slash length of about 1,29 meters,
which i calculated to be about the same as the spear, 190 newtons.
but the katana was slashed with 2 hands, the spear was thrown with 1 hand..
and it was done by 2 different people, one a tall guy, the other a short japanese guy.
i'm trying to interpolate this, so that i can do calculations for what would happend if i put in a 2kg sword instead of a 1kg, or a 0.6kg spear, etc.

2.

3. It's not really that easy. First of all, the force you are exerting on spear or sword is not constant, the peak force is probably higher than what you calculated. I don't know where you get the 1.25 m for the acceleration length of the spear. It starts accelerating the moment the thrower starts running, with a shorter burst at the end (I think). For a katana, what do you mean by the slash length of 1.29 m? Is it the tip of the sword? I think in that case you're really more rotating it, and there the location of the centre of gravity of the sword is important (a well balanced sword swings more easily).

You mention a difference between one and two hands. That is also not that simple. The force isn't produced by the muscles of arms or hands (or at least, a large part isn't). The spear thrower starts running to build kinetic energy in his body, and then tries to transfer that energy to the spear. At the same time he swings his body to get all possible muscles to cooperate.
The swordsman does something similar by turning his torso when slashing (possibly even pulling the sword in slightly near the end to increase the rotational velocity and thus the velocity of the tip).

At first, you could guess that increasing the weight of the sword or the spear would slow it down linearly. But, since muscles aren't exactly linear, it's not going to be that easy. You also have to take into account that you also have to accelerate your arms and hands, which easily weigh more than either spear or sword.

4. that doesn't really answer my question.

but then again maybe my question was too complex. i'm trying to get the basics down, and you throw all sorts of advanced stuff on my lap.

ok, here's a simpler question:

i believe an object traveling in a circular motion at 40kph has the same kinetic energy as an object traveling 40kph in a straight line.
is this true or false?

i also believe that applying a force of 190 newtons as torque is different to applying it in a linear motion. is this true or false?

and please, give me an answer, instead of telling me "its not that easy", then going on a tirade about how difficult it is, before in the end never giving me an answer in the first place. otherwise you just sweep whatever fundament i've been trying to build away from under my feet, and leave me to build on bare ground again.

5. Originally Posted by dejawolf
that doesn't really answer my question.

but then again maybe my question was too complex. i'm trying to get the basics down, and you throw all sorts of advanced stuff on my lap.

ok, here's a simpler question:

i believe an object traveling in a circular motion at 40kph has the same kinetic energy as an object traveling 40kph in a straight line.
is this true or false?

i also believe that applying a force of 190 newtons as torque is different to applying it in a linear motion. is this true or false?

and please, give me an answer, instead of telling me "its not that easy", then going on a tirade about how difficult it is, before in the end never giving me an answer in the first place. otherwise you just sweep whatever fundament i've been trying to build away from under my feet, and leave me to build on bare ground again.
Kinetic energy is dependent on the scalar speed rather than on the vector velocity. So an object traveling at 40 kph has the same kinetic energy whether it is traveling along a line, around a circle, in a corkscrew pattern or any kinematic description.

Force and torque are different things. Force is a vector quantity that is independent of the point of application. F=ma works no matter where the force is applied.

Torque is a bit different. Torque alone makes no sense. To define torque you not only have to define the vector, but also a point about which the torque is to be taken, or equivalently if it is generated by a force the vector from the point at which you want torque to the point of application of the force (think lever arm). More precisely if you have a point P, a vector R and a force F applied at the point represented by a displacemenet from P by R then the torque T at P is given by T= R X F, where X is the vector cross product.

You can't apply a force of 190 Newtons as torque. The units of torque are Newton-meters, which is the same units that describe energy, but torque is NOT energy. Torque is a vector quantity, while energy is a scalar quantity. Torque is the cross product of force and distance while energy is the scalar (dot) product of force and energy.

While you can derive the equations for angular motion and conservation of angular momentum from the linear equations for a system of particles (it is pretty easy and is one of the first things you would see done in a book on classical mechanics, such as the text by Goldstein), I don't think you can reverse the process.

Here is a Wiki article on torque. http://en.wikipedia.org/wiki/Torque

6. Originally Posted by DrRocket
Originally Posted by dejawolf
that doesn't really answer my question.

but then again maybe my question was too complex. i'm trying to get the basics down, and you throw all sorts of advanced stuff on my lap.

ok, here's a simpler question:

i believe an object traveling in a circular motion at 40kph has the same kinetic energy as an object traveling 40kph in a straight line.
is this true or false?

i also believe that applying a force of 190 newtons as torque is different to applying it in a linear motion. is this true or false?

and please, give me an answer, instead of telling me "its not that easy", then going on a tirade about how difficult it is, before in the end never giving me an answer in the first place. otherwise you just sweep whatever fundament i've been trying to build away from under my feet, and leave me to build on bare ground again.
Kinetic energy is dependent on the scalar speed rather than on the vector velocity. So an object traveling at 40 kph has the same kinetic energy whether it is traveling along a line, around a circle, in a corkscrew pattern or any kinematic description.

Force and torque are different things. Force is a vector quantity that is independent of the point of application. F=ma works no matter where the force is applied.

Torque is a bit different. Torque alone makes no sense. To define torque you not only have to define the vector, but also a point about which the torque is to be taken, or equivalently if it is generated by a force the vector from the point at which you want torque to the point of application of the force (think lever arm). More precisely if you have a point P, a vector R and a force F applied at the point represented by a displacemenet from P by R then the torque T at P is given by T= R X F, where X is the vector cross product.

You can't apply a force of 190 Newtons as torque. The units of torque are Newton-meters, which is the same units that describe energy, but torque is NOT energy. Torque is a vector quantity, while energy is a scalar quantity. Torque is the cross product of force and distance while energy is the scalar (dot) product of force and energy.

While you can derive the equations for angular motion and conservation of angular momentum from the linear equations for a system of particles (it is pretty easy and is one of the first things you would see done in a book on classical mechanics, such as the text by Goldstein), I don't think you can reverse the process.

Here is a Wiki article on torque. http://en.wikipedia.org/wiki/Torque
ok thanks. torque is not energy, thats very helpful.
whats the unit of energy, joules?

so to apply a force of 190 newtons as torque, i need to multiply it by the radius.
but how do i translate that then into kinetic energy?

7. Originally Posted by dejawolf

ok thanks. torque is not energy, thats very helpful.
whats the unit of energy, joules?

so to apply a force of 190 newtons as torque, i need to multiply it by the radius.
but how do i translate that then into kinetic energy?
The units of energ are force x distance. A Joule is one such unit. A Joule is a Newton-meter.

You can't translate torque to kinetic energy. That was the point of the earlier post. Torque is to angular motion as force is to linear motion. You can't translat force to energy either.

You can translate angular velocity and momentum of inertia to kinetic energy, just like you can translate linear motion and mass to kinetic energy and it is done the same way.

http://en.wikipedia.org/wiki/Rotational_energy

8. Originally Posted by dejawolf
that doesn't really answer my question.
I'm sorry, but I didn't have enough info to answer it. I tried to list some of the required information to answer your question. I also tried to warn you from drawing hasty conclusions based on an incomplete model.
Originally Posted by dejawolf
but then again maybe my question was too complex. i'm trying to get the basics down, and you throw all sorts of advanced stuff on my lap.

ok, here's a simpler question:

i believe an object traveling in a circular motion at 40kph has the same kinetic energy as an object traveling 40kph in a straight line.
is this true or false?
in general: false
again, and you're going to hate me for this: it's not that easy

Let's start with the basics: the movement of an object has two components: a translation (movement of the centre of gravity of the object) and a rotation (movement around the centre of gravity of the object).
These two components are independent, and both contain kinetic energy. Like DrRocket already explained, the translational kinetic energy is only dependent on the size of the translation, and independent of what trajectory the (centre of gravity of) the object is following.
The rotational kinetic energy is only dependent on the rotational speed (sometimes called pulsation) of the object around it's own centre of gravity.
The total kinetic energy of the object is the scalar sum of both energies.

It is not entirely clear to me what you mean by "circular motion". If it is a pure translation, the kinetic energy is in both cases the same, if the object is rotating, the rotating object has more kinetic energy (note that the rotation is seen from a non-rotating frame of reference, so if your swinging around a ball on a string, the ball will be rotating; a slashing katana is also rotating, and the centre of gravity is usually pretty close to your hand, so the translational velocity is a lot smaller.)
Originally Posted by dejawolf
Otherwise you just sweep whatever fundament i've been trying to build away from under my feet, and leave me to build on bare ground again.
I'm again sorry for that, but if the fundament under your feet will lead you to false conclusions, it's better to build a new one.

DrRocket covered the rest. If you have troubles with the concepts of force, translation, torque, rotation or translational/rotational kinetic energy, don't hesitate to ask.

Once you understand that correctly, you'll need the concept of impulse to answer your initial question.

9. Originally Posted by Bender

It is not entirely clear to me what you mean by "circular motion".
by circular motion, i mean, moving along the circumference.
i'll look at the other stuff.

10. Originally Posted by dejawolf
Originally Posted by Bender

It is not entirely clear to me what you mean by "circular motion".
by circular motion, i mean, moving along the circumference.
i'll look at the other stuff.
If all the mass is in fact concentrated on the circumference, then the object traveling at 40 kph at the circumference of the circle does have the same kinetic energy as if it were traveling in a straight line at the same speed. But, you started out discussing the swing of a sword, and the sword's mass is not concentrated on the circumference. That's why you have to define a moment of inertia for the axis of rotation and use rotational speed instead of linear speed.

11. ok, so i guess i was not fundamentally wrong in one of my assumptions
what about a pickaxe instead of a sword, with the same tip shape as the javelin?
with the same kinetic energy, i'd assume they'd punch through the same amount of armour.
also, should i concentrate on the axehead, or part of blade that will do the cutting when calculating kinetic energy, but the whole weapon when calculating the energy required to make it swing?
i know that the further to the center of the object you are, the slower the wwing will be, and thus the kinetic energy will be smaller too.

12. Originally Posted by Bender
If you have troubles with the concepts of force, translation, torque, rotation or translational/rotational kinetic energy, don't hesitate to ask.

Once you understand that correctly, you'll need the concept of impulse to answer your initial question.
is this correct then:

force = acceleration induced on a mass M*A
translation = movement of this mass along a linear path
torque = rotational force F*R
rotation = "turning" motion on an axis.
translational kinetic energy = the energy of an object in motion

rotational kinetic energy = the sum of the kinetic energies of its moving parts... hrm.

i guess i'll throw myself at the rotational kinetic energy part.

i guess i've been going a bit backwards on the angular part. i've simple taken the sector of a circumference, measured its length, and measured the time it took for it to move that length, to find out its kinetic energy. but i guess thats wrong.

seems like the people who post on wiki, are having a competition on how they can explain things in the most contrived way possible.

anyways, thanks for the help!

13. apparently, here's how i find angular velocity:

what does the |r| mean around radius?

14. actually looking at a khan academy video, why is the |r|^2 even there?

since:

velocity = (2*pi*r)*(m/s) = angular velocity * radius

whats the deal with that?

15. Originally Posted by dejawolf
apparently, here's how i find angular velocity:

what does the |r| mean around radius?
It means absolute value, but that formula is what you would use to treat the angular velocity as a vector. No sense making it more complicated than you have to, so you can just work with scalar quantities, in which case the angular velocity is just the velocity of a point divided by the radius of that point from the axis of rotation.

If you know how fast something is spinning (frequency f in revolutions per minute), then the angular velocity is 2*pi*f.

In your problem about the swinging weapon, which parameters do you think you know and which do you want to solve for?

16. here's the figures i have, and i'm trying to solve for swing speed, and kinetic energy.
with 192 newtons as the swing strength.

although i'd also like to go in reverse, with known swing speed and weight of the weapon.

should weight of the arm be included in the calculation?

here that is:

Upper arm (right): mean 1.843kg s.d. 0.218kg.
Upper arm (left): mean 1.888kg s.d. 0.299 kg.
Forearm (right): mean 1.113kg, s.d. 0.271kg.
Forearm (left): mean 1.088kg s.d. 0.185kg
Hand (right): mean 0.400kg s.d. 0.091kg
Hand (left): mean 0.374kg s.d. 0.062kg
That gives a total mean arm mass of 3.35kg (for right and left).

17. how do i calculate the point of balance?
if i'm not mistaken, thats the point where the weapon tilts neither way, but stands still.

18. Originally Posted by dejawolf
is this correct then:

force = acceleration induced on a mass M*A
translation = movement of this mass along a linear path
the path doesn't have to be linear. A translation can be along any path.
Originally Posted by dejawolf
i guess i'll throw myself at the rotational kinetic energy part.
Actually, in the case of the axe, with most of the weight at the end, you'll make a better approximation with the translational kinetic energy.

Originally Posted by dejawolf
should weight of the arm be included in the calculation?
In the case of the axe, you can probably make a decent approximation with only the weight of the axe head.

In my experience with manually splitting wood with a splitting axe, the kinetic energy of the axe head is all that matters (besides actually hitting the log of course )

The complicated part is the trajectory of the axe (both of the handle and the head). The strength of the arms is mainly important to prevent the axe from slipping from your hands. I think most is generated in body and shoulders, and even some in the legs.

About calculating the centre of gravity: take an weighted average of all parts. e.g. if the centre of gravity of the handle (0.2 kg) is halfway at 0.5 m, and the CoG of the head (0.8 kg) is at 1 m, the CoG of head and handle combined is (0.5*0.2+1*0.8)/1=0.9 m.

19. Originally Posted by Bender
[(0.5*0.2+1*0.8 )/1=0.9 m.
hey, thats a very cool formula!

the real problem for me though is going about the acceleration portion of things.
should i use translational or rotational force when calculating the acceleration, to get the final velocity, so that i can derive the kinetic energy?

20. I don't know enough about biomechanics to give a proper answer to that. While making something swing, part of the force (perpendicular to the motion) goes to the centripetal force (to keep the axe on a circular trajectory), part (parallel to the motion) goes to increasing the speed, and thus the kinetic energy.
I don't know if you also exert a torque on the axe. If so, it adds to the acceleration of the axe. , with the distance between then centre of rotation and the centre of gravity.

btw, "rotational force" is called torque or moment, and I think it's a confusing term, because a torque isn't really a force.

21. Originally Posted by Bender
I don't know enough about biomechanics to give a proper answer to that. While making something swing, part of the force (perpendicular to the motion) goes to the centripetal force (to keep the axe on a circular trajectory), part (parallel to the motion) goes to increasing the speed, and thus the kinetic energy.
I don't know if you also exert a torque on the axe. If so, it adds to the acceleration of the axe. , with the distance between then centre of rotation and the centre of gravity.

btw, "rotational force" is called torque or moment, and I think it's a confusing term, because a torque isn't really a force.
If you think about how one swings an axe, you will see that you do apply a torque.

Normally you start with one hand on the bottom of the handle and the second hand nearer the axe head. Then as you swing the top hand slides down to meet the lower hand. While they are separated the two hands apply a "couple" which provides torque as well as net force to produce both translation and rotation. That rotation about the base of the handle is very important, as it results in most of the speed of the head of the axe and therefore most of the energy. That is why you swing an axe and don't "push" it. The top hand is doing most of the work, which the bottom hand applies a reverse force, producing a couple and the torque, and keeping the axe from flying off into the surrounding countryside.

The description depends on how you choose your coordinates. The "natural" description with the axe head moving in a more or less circular arc, has the rotational axis near the bottom of the handle and quite a distance from the center of mass, which is near the head, so you can describe the motion as largely rotation about that axis (reality is, of course, a bit more complicated). The usual kinematic description used for rigid body dynamics would consider rotation about the center of gravity and translational motion of the CG, and that results in a somewhat more complicated description of the motion.

The former description is easy to visualize. The latter description is more flexible and is easier to apply to the real motion of the axe which does not have a fixed axis.

22. i think the amount of torque is dependant on whether the axe is one handed or 2 handed though. with a 2-handed weapon, you can use one hand as a lever, and the other hand as a crank or preferably crank with both hands in opposite power directions, but with a one handed weapon, you crank it with your shoulder, elbow, and hand joint only.
and yes, preferably, you do move when you strike as well. taking a step forward as you strike increases the strike velocity.
the elbows doesn't tend to move a lot when you strike with both hands though.
and for one handed axes, the speed actually depends on how far out you hold your arm. try and hold your lower arm up to your upper arm, and swing up and down.
and then try it with your arm stretched out.
when its stretched out you swing slower.

23. There are probably different techniques, depending on the size and weight of the axe. My most recent experience is with a splitting maul. It's probably as large and heavy as axes get. I got most effect by swinging it behind my back and then pulling it in a straight wide arc over my head. The initial swing with both hand separated, the final arc with both hands together near the end.
Originally Posted by dejawolf
and for one handed axes, the speed actually depends on how far out you hold your arm. try and hold your lower arm up to your upper arm, and swing up and down. and then try it with your arm stretched out. when its stretched out you swing slower.
Are you sure the translational speed is slower? The rotational speed will be slower but the weight is further away (distance ) from the centre of rotation. And:

24. [quote="Bender"]There are probably different techniques, depending on the size and weight of the axe. My most recent experience is with a splitting maul. It's probably as large and heavy as axes get. I got most effect by swinging it behind my back and then pulling it in a straight wide arc over my head. The initial swing with both hand separated, the final arc with both hands together near the end.
Originally Posted by dejawolf
and for one handed axes, the speed actually depends on how far out you hold your arm. try and hold your lower arm up to your upper arm, and swing up and down. and then try it with your arm stretched out. when its stretched out you swing slower.
Precisely. When you are splitting large logs you want to get as much head speed as possible. That is done with the technique described with the hands starting separated to get torque and then sliding together. If you do this correctly your feet leave the ground (or nearly so) just as the head of the maul strikes the log.

With a splitting maul, used to split wood rather than as a weapon, there is little translational movement of the body, since accuracy is important. You don't charge the log.

It is the torque that is important. Think about what happens if you do nothing but apply a force to the end of the handle -- hold it upside down and try that. Not much happens other than the handle pivots about the head. Only when you apply a couple through separated hands can you efficiently chop with an axe or a maul.

A one-handed hatchet works because you have enough strength in your wrist and can apply torque over the width of your hand to move the hatchet in an arc and keep the blade perpendicular to the velocity vector. You can do something similar with a saber, but it is more difficult with a Katana, and that is why a Katana is commonly used two-handed.

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