Instantaniouse center of 0 velocity

[joemama69]

Homework Statement

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

Homework Equations

v=wr

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

 

[LowlyPion]

Homework Statement

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

Homework Equations

v=wr

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

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?

 

[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.

 

[Saladsamurai]

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 $\vec V_A$

 

[LowlyPion]

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.

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.

 

[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

 

[Saladsamurai]

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

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.

 

[joemama69]

ok now I am 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.

 

[Saladsamurai]

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,

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

 

[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

 

[Saladsamurai]

2i+4j m/s in vector notation

Sounds good.

3.6568 m/s in normal notation

Want to run that by me again?

2²+4²=20

20^1/2=4.47