Does the block move along the pink dotted lines

[Kaushik]

Absolutely. Now draw the FBD.

I am getting $N\cos(\theta) = mg$, Am I correct? If yes, what should I do after this?

 

[kuruman]

That is the condition, yes. How would you find the angle at which this is the case? Hint: I asked you to do something in post #42.

 

[Kaushik]

That is the condition, yes. How would you find the angle at which this is the case? Hint: I asked you to do something in post #42.

$N\sin(\theta)$'s component towards the centre is causing the centripetal acceleration.
So from the condition we got, $N = mg/cos(\theta)$.
Substituting the value of $N$ in $N\sin^2(\theta) = 2mg\cos(\theta)$
We get, $\theta = \tan^{-1}(\sqrt2)$.
Is this correct?

 

[kuruman]

That is correct. Does the angle look familiar?

 

[Kaushik]

That is correct. Does the angle look familiar?

Yes.
So the angle at which the velocity's vertical component is maximum is same as the angle at which the $F_{net}$ is Horizontal.

Thanks a lot for your help @kuruman @jbriggs444

 

[kuruman]

Yes.
So the angle at which the velocity's vertical component is maximum is same as the angle at which the $F_{net}$ is Horizontal.

In retrospect and with the benefit of 20/20 hindsight, do you see why it has to be this way?

 

[Kaushik]

In retrospect and with the benefit of 20/20 hindsight, do you see why it has to be this way?

The velocity's vertical component will be maximum when acceleration along the vertical is minimum. So when $a$ along vertical is 0, the net acceleration is horizontal.

 

[kuruman]

So when a a along vertical is 0, the net acceleration is horizontal.

This just says that when one component of the acceleration is zero, there is only the other component left.

Here is what I had in mind. The vertical component of the velocity has a maximum when the vertical component of acceleration is zero. When the vertical component of the acceleration is zero, the vertical component of $F_{net}$ is zero. The vertical component of $F_{net}$ is zero when the vertical component of $N$ is equal to the weight. Therefore, when the vertical component of $N$ is equal to the weight, the vertical component of the velocity is maximum and vice-versa.

Thanks a lot for your help @kuruman @jbriggs444

Thank you for an interesting post. I hadn't seen this variation before.