# 3d vectors applications

1. May 31, 2015

### nodtomc

The tangential velocity of a body rotating about the centre point "O" is given by v=w.r, where w is angular velocity and r = radius. give that the vectors
w = 3i - 2j + 7k
r = 2i + 5j - 3k

a) Calculate the magnitude of velocity
b) the vector product in the form i+j+k

So far I have the vector product as -29i +23k + 19k. I am a little confused when it comes to caclulating the magnitude of velocity. Do I do the magnitude of the vector product using Pythagoras? Or do I do the dot product of the w and r vectors?

I have the answers, 41.61m/s, or 25m/s from using either method.

I think i'm being a little thrown off by a few things;
the word magnitude in the question, although I'm under the assumption that magnitude of velocity is speed?)
The fact that velocity is a vector quantity.

I also have a question on work done, which again, as it's scalar quantity, I'm being a little thrown off by the fact it's a scalar quantity, so do i use dot/scalar product, or magnitude of the vector product?
EDIT: worth mentioning it's coursework so ideally if you can just hint at the right direction to follow, that would be appreciated

Thanks

2. May 31, 2015

### Staff: Mentor

Sure.
The magnitude of the velocity is called speed, right.

If the i-component of the motion is 29 (m/s?) already, there is no way the speed can be 25 (m/s?).

Work done where?

Full solutions are against the forum rules anyway.

3. May 31, 2015

### nodtomc

Sorry I should have been more clear, the work done is a seperate question among the same lines, calculating work done from force and distance vectors

4. May 31, 2015

### Kinta

Hello nodtomc,

Your vector product results looks good to me. The way you've written the velocity equation makes it seem as though you intended to write that $\vec v = \vec \omega \cdot \vec r$, but velocity, when $\vec \omega$ is angular velocity about the rotation center and $\vec r$ is the radius from the rotation center, is given by $\vec v = \vec \omega \times \vec r$. The difference between these two types of multiplication is key.

It may help to draw the problem. From experience with situations like that of this problem, I know that I often find it most convenient to have the angular velocity vector perpendicular to the plane of the page. The body's tangential velocity vector will then be, at all instances, in the plane of the page or a plane parallel to the page depending on where the radius points. If you're familiar with the right-hand rule, you can quickly confirm this yourself without any written mathematics! If you're not familiar with the right-hand rule, it basically says that to get the direction of a vector that is the result of a cross-product, align your hand and fingers with the first vector ($\vec \omega$ perpendicular to the page in this case) then, with your palm stationary, align your fingers with the second vector ($\vec r$ anywhere in this case), and then your thumb, when extended to form a $90^o$ with both your palm and fingers, will give the direction of the produced vector. This is based on the fact that taking the cross product of two vectors yields a vector perpendicular to both input vectors. When testing this yourself, remember that the radius remains constant, so your radius should trace out either a circle or a cone when it moves as the velocity vector dictates.

One interesting yet intuitive connection to make is the following: The fact that $\vec v$ is necessarily perpendicular to $\vec \omega$, which remains constant throughout the described rotation unlike $\vec r$, means that you should be able to add any scalar multiple of $\vec \omega$ to $\vec r$ without changing $\vec v$ (try this!). That is, in the picture I tried to describe above, the vector $\vec r$ from the center point O could point to any location on the sides of a cylinder of radius $r_{cylinder}$ and of infinite height along the axis perpendicular to the plane of the page.

Does this make sense? Does this help with your confusion?

Edit: I forgot to mention that, for the curious, $r_{cylinder}$ should be such that $\vec \omega \cdot \vec r_{cylinder} = 0$.

Last edited: May 31, 2015
5. May 31, 2015

### nodtomc

Yes I can see now why the way I typed the equation is wrong. It's given as v = w x r in the question. The separate work done question is written as "Work done = force x distance (f.d)". Is this the key distinction?

Therefore am I right in thinking that I should use dot product to calculate a scalar quantity such as work done, and use a cross product to work out a vector quantity such as velocity?

So for the velocity question, as I already have the cross product I would work out the magnitude of that, and for work done, I would find the dot product of the force and distance vectors?

The base of my confusion is I'm doing an electrical course with half a unit that very briefly covers mechanics, where we've not covered vectors to this degree, and this question is from my Maths unit, where they've only taught us generic vectors without any practical applications.

I think I understand what you're saying about the right hand rule. We've not covered it, but I think i understand how it translates to direction, I'm just not sure how that helps me find the answer?

6. May 31, 2015

### Kinta

Under the application of a constant force $\vec F$, work done on an object by that force after the object is displaced a distance $d$ is given by $W = \vec F \cdot \vec d$ where the direction associated with the displacement is, as you would expect, like an arrow drawn from the initial position to the final position. It may be important to note that centripetal forces are not constant.

Yes. Dot products produce scalar quantities and cross or vector products produce vectors. Keep in mind that the order in which terms appear in a cross product matters (if you're already aware of this, great! If not, try switching two vectors and see what happens). The same is not true of dot products.

For the velocity question, I can't off the top of my head think of another way to find the magnitude of the velocity other than the way you described. However, if the ordering of the question is as you first presented it (asking for the magnitude, then the whole velocity vector), there may be some simple method I'm forgetting that will give you the magnitude of the velocity without having to find the full vector first. I can say with high certainty that the dot product of $\vec \omega$ and $\vec r$ should absolutely not give you the magnitude of the velocity though because this is at a maximum when the two vectors align, in which case the body itself would be rotating with its center of mass stationary, zero velocity.
For the work done by the force, if the force is constant, yes, simply take the dot product of the two vectors as you described.

As for the right-hand rule, this is only meant to be a simple technique to get a feeling for what kind of answer to expect. If you obtain a solution for a problem to which the right-hand rule can be applied, and the direction of your vector disagrees with the right-hand rule, there's reason to believe something's fishy with the answer you obtained.

Last edited: May 31, 2015
7. May 31, 2015

### PeroK

I think you may be making a fundamental mistake from what you've learned about vectors:

1) The vector (cross) product of two vectors is a vector.
2) The scalar (dot) product of two vectors is a number.

But, that does not mean that if you are trying to find a scalar quantity, then you must use the scalar product. You may need to find the magnitude of a vector. And, that might mean taking the vector product to get a vector and then taking the magnitude of the resulting vector.

So, when you are asked to find the magnitude of the velocity, you must first calculate the velocity (vector), then take the magnitude of that. In this case, what you need is:

$v = |\vec {v}| = |\vec{\omega} \times \vec{r}|$

It seems to me that you were thinking that:

$\vec {v} = \vec{\omega} \times \vec{r}$ (Right)

and

$v = \vec{\omega} \cdot \vec{r}$ (Wrong)

But that is not correct at all. Although, you may already have figured this out for yourself.