Linear velocity of rigid body with vectors

[eximius]

Homework Statement

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 .)

Homework Equations

v= ω x r

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.

 

[SammyS]

Homework Statement

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 .)

Homework Equations

v= ω x r

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.

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

 

[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.

 

[SammyS]

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.

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 .

 

[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...

 

[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?

 

[SammyS]

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...

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 .​

 

[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?

 

[SammyS]

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?

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?

 

[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?

 

[SammyS]

...

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

Answer this quoted question.