# Linear velocity of rigid body with vectors

1. Jan 21, 2012

### eximius

1. The problem statement, all variables and given/known data

A rigid body is rotating with angular velocity 2 rad/s about an axis in the
direction of the vector (i + j + k) and passing through the point Q = (0, 1, -1) on
the body. Find the linear velocity of the point P = (1, 0, 1) on the body.
(You may use the result v = ω x r .)

2. Relevant equations

v= ω x r

3. The attempt at a solution

Magnitude of vector joining P and Q = |P-Q| = r

.:. r = √6

v= ω x r
v= 2 x √6
v = 4.8990m/s

I'm assuming I'm completely wrong because the amount of work isn't enough for the amount of marks. I've seen similar questions online and I've noticed that x=1+t, y=1+2t and z=3t have been used in a number of the answers. Are these general rules or have they been derived from the question somehow? Any and all help would be greatly appreciated. Thanks.

2. Jan 21, 2012

### SammyS

Staff Emeritus
You haven't taken into account the vector nature of r and ω.

3. Jan 21, 2012

### eximius

So P-Q = (1,-1,2) and w = (2,2,2). Therefore v = (6,-2,-4)m/s ?

I understand your reply but it doesn't really tell me what to do.

4. Jan 21, 2012

### SammyS

Staff Emeritus
Well, you did take the cross product of r and what you thought was ω. So, maybe I helped point you in the right direction -- which is all I wanted to do.

By the way: ω ≠ (2,2,2). The magnitude of (2,2,2) is 2√(3) . The magnitude of ω is 2 .

5. Jan 22, 2012

### eximius

Erm... So what is the answer then? Am I even right in thinking that P-Q is equal to r? How do I get v? Sorry but you're being very cryptic and I just don't get it...

6. Jan 22, 2012

### eximius

Is it something to do with using P-Q as the normal to the plane and discovering the plane's equation with N.(x-Px) = 0? Or something like that?

7. Jan 22, 2012

### SammyS

Staff Emeritus
Yes P-Q = r .

Cryptic? I did mention that you had a problem with what you had for ω.

What is wrong with ω specifically? It does have the correct direction, but the wrong magnitude.
You multiplied i+j+k by 2 . However, the magnitude of the vector i+j+k is not 1 .​

8. Jan 22, 2012

### eximius

Ahhh so the magnitude is √3. Therefore ω= 2*√3. So v = 2√3 * √6 = 6√2 = 8.49m/s ?

I'm sorry if I seemed annoyed or something. I was just confused and frustrated.

Edit: I have r in vector form, I need ω in vector form. The magnitude of ω is 2. So I need to get the vector form from this. But how?

Last edited: Jan 22, 2012
9. Jan 22, 2012

### SammyS

Staff Emeritus
First of all: 2*√3 = √(12) ≠ √6 .

The vector, (2, 2, 2) has a magnitude of 2*√3 , so multiply that by 1/√3 . What will that give you?

10. Jan 22, 2012

### eximius

The vector (2,2,2) doesn't have magnitude 2*√3, it has a magnitude of √(2^2 + 2^2 + 2^2) = √12, doesn't it? Is the intention to get the unit vector? I really don't understand at all. If you could please just tell me, I'm simply not getting it.

v and r are vectors, ω is a scalar
If P-Q = r, then r=(1,-1,2)
ω = 2rad/s in the direction of (i,j,k)

v=(6,-2,-4) from determinant method.
|v|=√(6^2 + (-2)^2 + (-4)^2) = 2*√14=7.48m/s

Am I even close?

11. Jan 22, 2012

### SammyS

Staff Emeritus