Another attempt :
View attachment 209130
Center ūf mass motion gives,
##
mg(-\hat y ) + N \hat y = m\ddot y \hat y ~~~~~~~~~~~~~~~~~ \ddot y <0. ~~~~~~~~~~~~~~(1)##
## y = l \sin \theta
\\ \ddot y = l\{\ddot \theta \cos \theta - \sin \theta {\dot \theta }^2\}
##
I can't decide here whether ## \ddot \theta ## is positive or negative.
Since the torque about both center of mass C and pivot P is anti - clockwise,
considering ## \vec \tau = I \vec \alpha : \vec \alpha = \ddot \theta \hat z, ## I decide ##\ddot >0. ~~~~~~~~~~~~~~~~(2)##
But this method is valid only if ## \vec \tau = I \vec \alpha## is valid only for fixed axis rotation.
Using (1) and (2) ,
##N- mg = ml[ \cos \theta ~\ddot \theta - \sin \theta ~ {\dot \theta }^2 ] ~~~~~~~~~~~~~~~~~~~~~~(1.1)##
Similarly,
## \ddot x =0 ~ and~ x = l \cos \theta ## gives,
##\ddot \theta = \cot \theta {\dot \theta}^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)##
(1.1) and (3) gives,
##N - mg = \frac {ml {\dot \theta }^2 \cos {2 \theta} } {\sin \theta } ~~~~~~~~~~~~~~~(4)##
Now, considering rotational motion about center of mass,
Torque about center of mass,
##Nl\cos \theta = I_{cen} \alpha ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(5)##
Now, can I take ## \ddot \theta = \alpha##?
Is this correct so far?
