# Search results

bump.
2. ### Falling rod against a wall

You missed a negative sign in the expression for ##\omega## which arises from the derivative of cosine.
3. ### Falling rod against a wall

Alright, but then why when using parametrization we get a negative answer?
4. ### Falling rod against a wall

$$a \hat j= \alpha \hat k \times (-0.33 \hat i + 0.33 \hat j) -6 \hat k \times (2 \hat i + 2 \hat j)$$ $$-6 \hat k \times (2 \hat i + 2 \hat j)= \begin{vmatrix} i & j & k \\ 0 & 0 & -6 \\ 2 & 2 & 0 \end{vmatrix} =6 \begin{vmatrix} i & j \\ 2 & 2 \end{vmatrix} = 12 \hat i -12 \hat j$$...
5. ### Falling rod against a wall

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.
6. ### Falling rod against a wall

I believe you missed the linear acceleration term that arises due to the quotient rule. Plus the derivative of 1/sin(x) is not 1/sin^2(x).

8. ### Falling rod against a wall

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.

21. ### Angular momentum of a mass-rope-mass system

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...
22. ### Angular momentum of a mass-rope-mass system

##m_2## or particle 2 mass duplicate, A.K.A ##m_2’=2m_2## so the system is not stationary anymore.
23. ### Angular momentum of a mass-rope-mass system

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.
24. ### Angular momentum of a mass-rope-mass system

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...
25. ### Angular momentum of a mass-rope-mass system

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...
26. ### Kinematics question on 2 planes moving relatively with each other

oh you are right, that was a typo mistake actually, thanks a lot for the help!
27. ### Kinematics question on 2 planes moving relatively with each other

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...
28. ### Kinematics question on 2 planes moving relatively with each other

But if I took the plane as my origin then my frame of reference would be moving with acceleration, and I do not know how to handle that yet.
29. ### Kinematics question on 2 planes moving relatively with each other

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...
30. ### Relative motion between cars with different types of movements

You are right, it should be -1.2. Thanks a lot for the help!