Electricity and magnetism
Cross Product and Torque The cross product and the direction of torque.
Cross Product and Torque
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- In all of the torque problems I've done so far in the
- physics playlist, we only just figured out the magnitude of
- torque, frankly because that's what normally matters, but
- torque is actually a vector and it's
- direction can be found.
- And that is because torque is defined as the cross product
- between the radial distance from your axis of rotation and
- the rotational force being applied.
- So these are both vectors.
- So let's take a look at how I taught you vectors the first
- time, and then I'll show you how that's really the same
- thing as what we're doing here with the cross product.
- Except now with the cross product, besides just the
- magnitude for torque, we're also getting the direction.
- But then we'll also see that direction is a little bit--
- it's just the definition of the direction of torque.
- I don't know how intuitive it really is.
- But what did I teach you before about torque?
- Well, let's say I had some arm, and let's say this could
- be the hand of a clock or it's pinned down to the wall there.
- So it would rotate around this object.
- Let's say it's some distance, r, from the pivot.
- Let's say that distance is 10.
- This is the same thing as r, and the magnitude of r is
- equal to 10.
- At some distance 10 from the pivot, I apply some force F,
- and F I will do in yellow.
- I apply some force F.
- Let me draw it straight.
- I apply some force F at some angle.
- That's my force F.
- It's also a vector.
- It has magnitude and direction.
- Let's say that this is 10 meters, and let's say that I
- apply a force of 7 newtons.
- Let me make it more interesting.
- Let's say I apply a force of square root of 3 newtons.
- And I just threw that out there because I think the
- numbers will all work out.
- And let's say that the angle between my force and the lever
- arm, or the arm that's rotating-- let's stick to
- radians this time.
- Let's say it's pi over 3, but if you need to visualize that,
- that's 60 degrees.
- pi over 3 radians is equal to theta.
- And so just based on what we already know about moments or
- torque, what is the torque around this pivot?
- Or how much torque is being applied by this force?
- And when we learn torque or we learn moments, we realize
- really the only hard part about these problems is that
- you don't just multiply the entire rotational force times
- the distance from the axis of rotation.
- You have to multiply the component of that force that
- is actually doing the rotation, or the component of
- the force that is perpendicular to this rotating
- arm, or perpendicular to this moment arm.
- So how do we figure that out?
- Well, the component of this force that is perpendicular to
- this arm-- I can visually draw it here.
- Let's see, it would look something like this.
- I could draw it there.
- I could also draw it here, right?
- This would be the component, or this would be the component
- that is perpendicular to this rotating arm, and the
- component that is parallel would be this, but we don't
- care about that.
- That's not contributing to the rotation.
- The only thing that is contributing to the rotation
- is this component of the force.
- And what is the magnitude of this vector right here?
- The component of vector F that is perpendicular to this arm.
- Well, if this angle-- let me draw a little
- triangle down here.
- If this is square root of 3, this is pi over 3 radians, or
- 60 degrees, and this is a right angle, it's pi over 3.
- I know it's hard to read.
- What is this length right here?
- Well, it's a 30-60-90 triangle, and we know that
- this length here-- I mean, there's a couple of ways you
- can think about it.
- Now that we know trigonometry, we know that this is just the
- square root of 3 times the sine of pi over 3, or the sine
- of 60 degrees, and so that equals the square root of 3.
- Sine of pi over 3, or sine of 60 degrees, is square
- root of 3 over 2.
- So the square root of 3 times the square root of 3 is just
- 3, so that equals 3/2.
- So the magnitude of this force vector that is perpendicular,
- the component that is perpendicular to the arm, is
- 3/2 newtons, and now we can figure out the magnitude of
- the torque.
- It's 3/2 newtons times 10 meters.
- So we know the magnitude of the torque, and I'm being a
- little bit more careful with my notation right now to
- remind you that torque actually is a vector, or you
- can almost view it as, they use this term pseudovector,
- because it's kind of a-- well, anyway, I won't go into that.
- So what is the magnitude of the torque vector?
- Well, it's 3/2 newtons times the distance, and remember, I
- just drew this vector here just to
- show you the component.
- I could just shift the vector here because this is actually
- where the force is being applied.
- You could draw that same vector here because you can
- shift vectors around, so this is also 3/2 newtons and maybe
- that makes it a little bit clearer.
- So it's 3/2 newtons times the distance that you are from
- your pivot arm, so times 10 meters, and so
- that is equal to what?
- 15 newton meters.
- So the magnitude of the torque is 15 newton meters.
- But all we did now-- and hopefully this looks a little
- bit familiar.
- This is what we learned when we learned moments and torque,
- but all we did now is we figured out the magnitude of
- the torque.
- But what if we wanted to know the direction?
- And that's where the cross product comes in.
- So what was the definition of the cross product?
- Cross product: r cross F, that is equal to magnitude of r
- times the magnitude of F times sine of the smallest angle
- between them times some vector that is perpendicular to both.
- And this is really where it's going to help, because all of
- these right here, these are all scalar quantities, right?
- So these don't specify the direction.
- The direction is completely specified by this unit vector,
- and a unit vector is just a vector of magnitude 1 that's
- pointing in some direction.
- Well, look, this cross product, this part of it, the
- part that just gives us magnitudes, we just calculated
- that using what we knew before of torques.
- The magnitude of our force vector times sine of theta,
- that gave us the component of the force vector that is
- perpendicular to the arm.
- And we just multiply that times the magnitude of r, and
- we got the magnitude of the torque vector, which was 15.
- We can leave out the newton meters for now.
- 15, and then its direction is this vector that we
- specified by n.
- We can call it the normal vector.
- And what do we know about this vector?
- It's perpendicular to both r-- this is r, right-- and it's
- perpendicular to F.
- And the only way that I can visualize in our
- three-dimensional universe, a vector that's perpendicular to
- both this and this is if it pops in or out
- of this page, right?
- Because both of these vectors are in the plane that are
- defined by our video.
- So if I'm a vector that is perpendicular to your screen,
- whatever you're watching this on, then it's going to be
- perpendicular to both of these vectors.
- And how do we figure out if that vector pops out or pops
- into the page?
- We use the right hand rule, right?
- In the right hand rule, we take-- r is our index finger,
- F is our middle finger, and whichever direction our thumb
- points in tells us whether or not we are-- the direction of
- the cross product.
- So let's draw it.
- Let me see if I can do a good job right here.
- So if that is my index finger, and you could imagine your
- hand sitting on top of this screen.
- So that's my index finger representing r, and this is my
- right hand.
- Remember, it only works with your right hand.
- If you do your left hand, it's going to be the opposite.
- And then my middle finger is going to go in the direction
- of F, and then the rest of my fingers are-- and I encourage
- you to draw this.
- So if I were to draw it-- let me draw my nails just so you
- know what this is.
- So this is the nail on my index finger.
- This is the nail on my middle finger.
- And so in this situation, where is my thumb going to be?
- My thumb is going to be popping out.
- I wish I could-- that's the nail of my thumb.
- Hopefully, that makes some sense, right?
- That's the palm of my hand.
- That's the other side of my-- and I could keep drawing, but
- hopefully, that makes some sense.
- This is my index finger.
- This is the middle finger.
- My thumb is pointing out of the page, so that tells us
- that the torque is actually pointing out of the page.
- So the direction of this unit vector n is going to be out of
- the page, and we could signify that by a circle with a dot.
- And I'm almost at my time limit, and so there you have
- it: the cross product as it is applied to torque.
- See you in the next video.
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