Relative Velocity coordinate system

[tronter]

(a) A point is observed to have velocity $$v_A$$ relative to coordinate system $$A$$. What is its velocity to coordinate system $$B$$ which is displaced from system $$A$$ by distance $$R$$? ($$R$$ can change in time)

I think its $$v_B = v_A - \frac{dR}{dt}$$. But I am not completely sure why this is the case.

(b) Particles $$a$$ and $$b$$ move in opposite directions around a circle with angular speed $$\omega$$, as shown. At $$t = 0$$ they are both at the point $$r = l \bold{j}$$, where $$l$$ is the radius of the circle. Find the velocity of $$a$$ relative to $$b$$.

So $$v_B = v_A - \frac{dR}{dt}$$

$$= (\sin t \bold{i }+ \cos t \bold{j)} \omega - (\cos t \bold{i} - \sin t \bold{j})$$.

Is this correct?

 

[Shooting Star]

Ra = R +Rb, by vector addition, where Ra is the posn vector of the pt wrt frame A, Rb is the posn vector of the pt wrt frame B and R is the posn vector of the origin of frame B wrt A. So, differentiating,

Va = dR/dt + Vb, which is what you've got.

An easier way to remember is that V_a/b = Va –Vb, where the latter velos are wrt the same frame, and V_a/b represents velo of point a wrt point b. This is a vector eqn.

Ra = l*[sin(wt) i + cos(wt) j], if a is moving clockwise.
Rb = l*[-sin(wt) i + cos(wt) j], if b is moving counter-clockwise.

You can now take the time derivatives, apply the formula, and see if the result tallies with your answer.

 

[tronter]

So $$v_B = -l \omega \cos \omega t \bold{i} - \omega \sin \omega t \bold{j}$$.

Is this correct?

 

[Shooting Star]

'l' is missing in the 2nd term, otherwise it's correct.