Search results

  1. Santilopez10

    Falling rod against a wall

    You missed a negative sign in the expression for ##\omega## which arises from the derivative of cosine.
  2. Santilopez10

    Falling rod against a wall

    Alright, but then why when using parametrization we get a negative answer?
  3. Santilopez10

    Falling rod against a wall

    $$a \hat j= \alpha \hat k \times (-0.33 \hat i + 0.33 \hat j) -6 \hat k \times (2 \hat i + 2 \hat j)$$ $$-6 \hat k \times (2 \hat i + 2 \hat j)= \begin{vmatrix} i & j & k \\ 0 & 0 & -6 \\ 2 & 2 & 0 \end{vmatrix} =6 \begin{vmatrix} i & j \\ 2 & 2 \end{vmatrix} = 12 \hat i -12 \hat j$$...
  4. Santilopez10

    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.
  5. Santilopez10

    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).
  6. Santilopez10

    Falling rod against a wall

    Care to show your approach?
  7. Santilopez10

    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.
  8. Santilopez10

    Falling rod against a wall

    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...
  9. Santilopez10

    Mass attached to a rotating spring

    OOOOH I did not think about radial velocity being 0 at max, you are right! thanks a lot.
  10. Santilopez10

    Mass attached to a rotating spring

    Okay, I understand what you are saying. So what quantity should I use to find my answer?
  11. Santilopez10

    Mass attached to a rotating spring

    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...
  12. Santilopez10

    Mass attached to a rotating spring

    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?
  13. Santilopez10

    Mass attached to a rotating spring

    oh my bad. $$ \frac{1}{2}m{V_0}^2=\frac{1}{2}m({\dot r}^2+{(r \dot{\theta})^2)}+\frac{1}{2}k(r-H)^2$$
  14. Santilopez10

    Mass attached to a rotating spring

    you mean writing it like this? $$ \frac{1}{2}m{V_0}^2=\frac{1}{2}m(({\dot r})^2+{r (\dot{\theta})^2)}+\frac{1}{2}k(r-H)^2$$
  15. Santilopez10

    Mass attached to a rotating spring

    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...
  16. Santilopez10

    Mass attached to a rotating spring

    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.
  17. Santilopez10

    Mass attached to a rotating spring

    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...
  18. Santilopez10

    Mass attached to a rotating spring

    Any tip on how to solve it? the system of diff equations seems too hard for my level to be honest.
  19. Santilopez10

    Mass attached to a rotating spring

    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...
  20. Santilopez10

    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...
  21. Santilopez10

    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.
  22. Santilopez10

    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.
  23. Santilopez10

    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...
  24. Santilopez10

    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...
  25. Santilopez10

    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!
  26. Santilopez10

    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...
  27. Santilopez10

    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.
  28. Santilopez10

    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...
  29. Santilopez10

    Relative motion between cars with different types of movements

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