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.