Recent content by Santilopez10

  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? $$ \frac{1}{2}m{V_0}^2=\frac{1}{2}m(({\dot r})^2+{r (\dot{\theta})^2)}+\frac{1}{2}k(r-H)^2$$
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