Question

Help

Answer 1

m

Answer 2

You must love Trojan?

Answer 3

Does this lead us anywhere?

Answer 4

@alivejeremy , O.O

Answer 5

@tkunny, not sure indeed. :/

Answer 6

I am studying general motion.

Answer 7

Tip from me

Answer 8

\[a_B =a_A + \alpha \times r_{B/A} - \omega^2r_{B/A}\]

Answer 9

Maybe you can switch between polar , and Cartesian coordinates as that is what they are mixing up here. It's asking about **instantaneous** so you can use the centre of the rod as the Origin of both axes at this very instant.....

To line it up so we can say that polar \(\theta = 0\), look at the right end of the rod, and using symmetry to reverse the velocity and acceleration vectors..... so linear motion at that end is to right and up


(\(x = r \cos \theta, y = r \sin \theta\))


Then we have:

\(y = r \sin \theta \implies \dot y = \dot r \sin \theta + r \cos \theta ~ \dot \theta\)


At this instant: \(\theta = 0 \implies \dfrac{2}{\sqrt{2}} = 5 \cos 0 ~ \dot \theta \implies \dot \theta = \dfrac{\sqrt{2}}{5 }\)

And ...

\(x = r \cos \theta \implies \dot x = \dot r \cos \theta - r \sin \theta ~ \dot \theta\)

\( \implies \dot r = \dfrac{1}{ \sqrt{2}}\)

Next up:

\(\ddot y = \ddot r \sin \theta + 2 \dot r \cos \theta ~ \dot \theta - r \sin \theta ~ \dot \theta^2 + r \cos \theta ~ \ddot \theta\)

Kicking out the sine terms again as here \(\theta = 0\):

\(\dfrac{3}{\sqrt{2}} = 2 \dfrac{1}{ \sqrt{2}} \dfrac{\sqrt{2}}{5 } + 5 \ddot \theta\)

\( \ddot \theta = \dfrac{3}{5 \sqrt{2}} - \dfrac{2}{ 25} \)


it's the same idea as:



\(\vec { \dot r } = \left( \begin{matrix} \dot x \\ \dot y \end{matrix} \right) = \dot r \hat r + r \theta \hat \theta \)

from whence:

\( \ddot{\vec r} = \left( \begin{matrix} \ddot x \\ \ddot y \end{matrix} \right) = (\ddot r - r \dot \theta^2 ) \hat r + ( r \ddot \theta + 2 \dot r \dot \theta
) \hat \theta \)

it might even work :-)