# Instantaniouse center of 0 velocity

1. Apr 27, 2009

### joemama69

1. The problem statement, all variables and given/known data
Solve using the mothod of instantaneous center of zero velocity
Point A = (.5,1.5), B = (1.5,1) and IC = (.5,2), w = 4 rad/s

Determine the velocity of B with respect to A

2. Relevant equations

v=wr

3. The attempt at a solution

V(a) = wr(a) where w = 4 rad/s and r(a) = .5
=2
V(b) = wr(b) hwere w = 4, and r(b) = (x^2+y^2)^1/2
V(b) = 4.4721...

How do i find the velocity of B with respect to A

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2. Apr 27, 2009

### LowlyPion

The Velocities are vectors. A is directed positive X and B is directed 45° above positive X.

So ... Vb relative to Va is Vb - Va isn't it?

3. Apr 29, 2009

### joemama69

I'm confused. My book does not mention anything about vectors in this section.

How are u able to say that A is headed possitive x and B is headed 45 above x.

But i'll take ur suggestion and see hwat happens

vector A = 2i, vector B = 3.162i + 3.162j

vector B - vector A = 2i - (3.162i + 3.162j) = -1.162i + 3.162j

Doesnt seem right.

4. Apr 29, 2009

Look at point A: if the object id 'rotating' counterclockwise about the IC, then what direction is A moving in at this instant ? Surely both points are moving to the right..right?

Now, relative motion equations say that:

{the absolute velocity of point B}=

{the absolute velocity of point A}+{the velocity of point B with respect to point A}

or in vector notation:

$$\vec V_B=\vec V_A+\vec V_{B/A}$$

Solve this for

$$\vec V_{B/A}$$

Then using the IC, find $\vec V_B$ and $\vec V_A$

Last edited: Apr 29, 2009
5. Apr 29, 2009

### LowlyPion

Looking at the picture you provided, I'd say that the arrow to the Northwest looks to be counterclockwise to the center of zero motion. Drawing the displacement of A from O, then isn't the direction of motion necessarily ⊥ to the displacement vector?

That would tell me to direct that velocity along positive X if it is to be ⊥ to OA.
If this is incorrect, then my understanding of the center of zero velocity must be in error.

6. Apr 29, 2009

### joemama69

Ok Ok I understand the vector part now and the direction they are headed.

Absol V of B = 4.4721
Absol V of A = 2

So would the Absol V of B with resp to A = 4.4721 - 2 = 2.4721

7. Apr 29, 2009

Unless it is a special case, I do not think that this is correct since $\vec V_B$ and $\vec V_A$ are vectors and thus must be added (or subtracted) as such.

I am not sure is you will get the same answer.

8. Apr 29, 2009

### joemama69

ok now im really confused. If i subtract vectors i get

vector A = 2i, vector B = 3.162i + 3.162j

vector B - vector A = 2i - (3.162i + 3.162j) = -1.162i + 3.162j

If i use the absolute velocity

Absol V of B = 4.4721
Absol V of A = 2

So would the Absol V of B with resp to A = 4.4721 - 2 = 2.4721

Is either of them right or wrong.

9. Apr 29, 2009

I could be making an error, here...but I am getting different numbers for $\vec V_B$

Using the exact same equations at the IC to find $\vec V_B$ we have

$\vec V_B=\vec V_{IC}+\vec V_{B/IC}$

$\Rightarrow \vec V_B=0+\vec V_{B/IC}$

By definition:

$\vec V_{B/IC}=\vec \omega \times \vec r_{B/IC}$

Where $\vec \omega=4 \hat{k}$ and $\vec \r_{B/IC}=1\hat{i}-1/hat{j}$

Thus,

$\vec V_B=4\hat{i}+4\hat{j}$

10. Apr 29, 2009

### joemama69

ahhhhhhhhhhh i gotcha calculator error

ok so then the Velocity of B with respect to A is..

2i+4j m/s in vector notation

3.6568 m/s in normal notation

11. Apr 29, 2009