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: $$\vec v_{P/Q}=\vec \omega...
By cyclic answer I meant this:
first we know that the initial velocity is only in the transverse direction so ##\vec v_0= V_0 \hat{\theta} \rightarrow \dot{\theta}_0= \frac{V_0}{H}## and from angular momentum we get ##\dot{\theta}=\frac{H V_0}{r^2}##.
Second, using linear momentum...
Okay but I am still missing the second answer. I tried to use conservation of linear momentum and energy but I got a ciclyc answer. Any tips on how should I approach it?
Now I realise that in the expression of the energy, ##\frac{1}{2}mv^2=\frac{1}{2}m({\dot r}^2+{r \dot{\theta}^2)}## so then $$\frac{1}{2}m{V_0}^2=\frac{1}{2}m({\dot r}^2+{r \dot{\theta}^2)}+\frac{1}{2}k(r-H)^2$$ and substituing our already known expression of ##\dot{\theta}## we can Isolate for...
Using the energy equation I got: $$r=\sqrt{\frac{m}{k}({V_0}^2-v^2)}+H \rightarrow \dot r= -\frac{\dot v v}{r-H}$$ Should I use this? it still has modulus of acceleration and velocity in its expression.
So I wrote ##\dot{\theta}## in terms of my angular moment to get: $$\dot{\theta}=-\frac{\sqrt{3}}{2} \frac{H V_0}{r^2}$$ and plugged it in the velocity equation in polar coordinates to get $$ \vec v= \dot r \hat r-\frac{\sqrt{3}}{2} \frac{H V_0}{r}\hat{\theta}$$ but I am still missing an...
a)
Our force can be represented as: $$\vec F= -k(r-H) \hat r$$ then the equations of motion are: $$\hat r: \ddot r -r {\dot{\theta}}^2=-\frac{k}{m_1}(r-H)$$ $$\hat{\theta}: r \ddot{\theta} + 2 \dot r \dot{\theta}=0$$
Plus we know that angular momentum is constant then $$|\vec L|=m r^2...
Yes, we have studied it. The rate of change of the angular momentum is the torque applied to the system, but torque is 0 in this case as the position and force vectors are parallel with respect to our origin (the hole). Then angular momentum is constant, and with the initial conditions we can...
the system is not stationary, r is not constant. What I thought would be taking the initial angular momentum and because the force is central then angular momentum is conserved, then the initial angular momentum is constant.
I do not know what ##I## is, if it is the moment of inertia, then it has not been taught yet. We have already developed the equation of angular momentum for central forces: $$\vec L= m r^2 \dot{\theta} \hat k$$
but still I need to find and expression for r and ##\omega##= ##\dot{\theta}## as...
1) the motion equations for ##m_2## are: $$T-m_2 g=0 \rightarrow T=m_2 g$$
##m_1##: $$T=m_1\frac{v^2}{r_0} \rightarrow \vec {v_0}=\sqrt{\frac{r_0 g m_2}{m_1}}\hat{\theta}$$
2) This is where I am stuck, first I wrote ##m_2## motion equation just like before, but in polar coordinates...
part B isn't really hard considering angular acceleration is nothing but $$\ddot{\theta}(t)$$ but still the hard part is getting right the polar equations.
Let say I place my origin at A, then plane A should be seeing B deaccelerating at a rate ##\vec a= -1.22## right? then the formula for the...
So my problem isn't actually finding the components, but knowing if the initial approach I took is correct. So what I did was:
At first I found that at the same instant, ##x_{B/A}=10500 m## so then I wrote the equation of motion for plane B respect to A:
so $$\vec a_{B/O}- \vec a_{A/O}=\vec...