Sliding Rod

Nice one! Nice approach

When you differentiated the result of Pythagoras then you got your component of velocity of CM not velocity of CM it will also have a horizontal component.

Here is my solution :

The distance between the lowermost point of the rod ( the point in contact with the ground ) and COM of the rod is fixed and is equal to $\frac{L}{2}$

Further, let $x$ be the distance of the lower most point of contact (say point O) from its initial position and let $y$ be the distance of COM from the ground.

So, we have the following as a result of Pythagoras theorem:

$x^2+y^2 = \frac{L^2}{4}$

Differentiate both sides of the equation to get the following relation:

$x\frac{dx}{dt}=-y\frac{dy}{dt}$

Let the velocity of point O be $v_{o}$ and that of COM be $v_{c}$

So, we have :

$v_{o}=v_{c}\tan\theta$

Velocity of point O in the frame of COM can be calculated easily using vectors and is equal to $v_{c}\sec\theta$ .

We also have the following relation about COM :

Velocity of point O w.r.t COM = $\omega$ *(Distance of point O from COM)

So,

$v_{c}\sec\theta = \omega \frac{L}{2}$

So, $\omega = \frac{2v_{c}}{L}\sec\theta$

Using the energy conservation equation:

$\frac{1}{2}mgL(1-\sin\theta) = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

We get

$\omega = 2\sqrt{\frac{3g(1-\sin\theta)}{L(4-3\sin^2 \theta)}}$

Differentiating the above equation w.r.t time we get

$\alpha = \omega \frac{d\omega}{d\theta}$

Also

$N\cos\theta \frac{L}{2}= I\alpha$

Using the above two equations we get :

$N= \frac{mg[3(\sin\theta - 1)^2+1]}{(4-3\sin^2 \theta)^2}$