I've been studying rotating frames of reference, and I faced a problem regarding a particular type of sum.
Imagine there is a point mass, sliding off a sphere, and eventually looses contact. I'm supposed to find the angle with the horizontal at which this happens. This problem is relatively simple and can be solved using a combination of free body diagrams of the forces and conservation of energy.
The angle comes out to be $\sin^{-1}\frac{2}{3}$. At this angle with the horizontal, the body loses contact.
Now imagine, the sphere is rotating about a central axis, with constant angular velocity $\omega$. I'm supposed to find the centrifugal force acting on the point mass. In order to do that, I need to also find the angle at which the body loses contact. Moreover, I need to find the force with respect to a frame attached to a sphere.
I try to draw a free-body diagram of the point mass as before, and I get the following equation :
$\frac{mv^2}{R} = mg\sin\theta - N - m\omega^{2}R\cos^2\theta$
Here I've assumed $\theta$ to be angle with the horizontal (latitude), and the last term is the component of the centrifugal force acting in the direction of the normal reaction.
I need to find out the centrifugal force i.e. $m\omega^2a = m\omega^2R\cos\theta$.
Hence I need to find $cos\theta$ here.
I try to use the law of conservation of energy as before,
$mgR-mgR\sin\theta = \frac{1}{2}mv^2$
Plugging this back into our force equation, I have :
$2mg(1-\sin\theta) = mg\sin\theta - N - m\omega^{2}R\cos^2\theta$.
Normally, as in the non-rotating case, I'd put $N=0$ at the point where the mass loses contact, and then find out the angle, as the normal force would disappear there. However, I'm stuck here, because of the extra term.
I've seen in many solutions, it is written "from a non-rotating frame, we can say $\omega$ is $0$", and then ignore the final term, and then calculate the angle to be exactly the same as the non-rotating sphere case i.e. $\sin^{-1}\frac{2}{3}$.
Then they plug this back to the expression for centrifugal force i.e. $m\omega^2R\cos\theta$.
However, I'm still unable to figure out, why are they suddenly considering $\omega=0$, even in the case of the rotating sphere. It can not be because of the frame being attached to the rotating sphere, as that would make $\omega=0$, but even the centrifugal force would disappear in that case. I'm not being able to understand this, or I'm missing something extremely obvious.
Is this because, from the outside, i.e. non-rotating inertial frame perspective, the centrifugal force does not exist, as it only exists in a non-inertial frame. So, from the outside, the angle of losing contact doesn't matter if the sphere is rotating or not. Is that why we put $\omega = 0$ ? I don't think that analogy is correct. What is the correct reasoning of plugging $\omega=0$ ?
Why is the angle of losing contact, same for the rotating sphere and the non-rotating sphere.
Here is a link to the question, however, on this website, it is done in a rather convoluted way.
Any help in understanding this would be highly appreciated.