Seems like compared to your answer, the term that is giving me problems is ##\vec \omega \times \vec v_{A/B}##. If only ##\vec v_{A/B}## would be ## -2 \hat i + 2 \hat j## instead of ## 2 \hat i + 2 \hat j## then I could get the negative sign. To be honest I do not see where I committed a mistake.
A/B means “A respect to B”, maybe I was not so clear. Otherwise I do not know whats the problem. ##\cos{\frac{\pi}{4}} 0.46=0.33## same for sin. Then A is located -0.33 meters to the left and 0.33 meters upwards.
we know that the center of instantaneous 0 velocity lies in the interception of 2 perpendicular lines to 2 points, which in this case lies above B. The velocity of any point of the rod can be described relative to the center of instantaneuous 0 velocity ##(Q)## as: $$\vec v_{P/Q}=\vec \omega...
By cyclic answer I meant this:
first we know that the initial velocity is only in the transverse direction so ##\vec v_0= V_0 \hat{\theta} \rightarrow \dot{\theta}_0= \frac{V_0}{H}## and from angular momentum we get ##\dot{\theta}=\frac{H V_0}{r^2}##.
Second, using linear momentum...
Okay but I am still missing the second answer. I tried to use conservation of linear momentum and energy but I got a ciclyc answer. Any tips on how should I approach it?