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.

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    $\begingroup$ In your formulas, I don't see any friction. If there is no friction, it does not matter if the sphere is rotating. $\endgroup$
    – R.W. Bird
    Commented Aug 23, 2021 at 14:08
  • $\begingroup$ @R.W.Bird can you please elaborate why is it so ? Some mathematics would also be highly appreciated. Or maybe some physical intuition. $\endgroup$ Commented Aug 23, 2021 at 14:48
  • 1
    $\begingroup$ With the non-rotating sphere, you appear to assume that the sliding mass starts at the top of he sphere with a minimal downward speed . Without friction, the rotating sphere will not exert a tangential force on the mass, and its motion will be the same as in the previous case. $\endgroup$
    – R.W. Bird
    Commented Aug 24, 2021 at 14:17

2 Answers 2


The point mass only moves due to forces acting on it.

Ignoring air resistance and idealising as usual in homework exercises, the forces acting on it are:

  • gravity: acting vertically downwards
  • physical presence of sphere (it cant sink onto the sphere!), acting normal to the spheres surface at the poi t of contact, and nullifying the normal component of gravity, leaving only the parallel-to-surface component of gravity (tangent) as a net force while in contact.
  • friction with sphere, if any, acting parallel to surface of sphere (tangent) and possibly directed in the opposite direction from the particles movement, while in contact.

Your question omits friction.

If there is no friction (zero coefficient), then the sphere can rotate as it likes, none of that will affect the particle, because rotation of the sphere alone doesn't. You have to ask yourself what actual forces are acting, to cause or change motion. Rotation itself is not a force. Only forces (including friction due to spheres relative movement) do.

If there is a non-zero coefficient, then there may be friction. If there is friction, it will affect the points movement as usual. I say there may be friction, and only a "possible" direction, because it totally depends how the ball is rotating. For example if its rotation exactly matched the particles motion, so the surface was in effect not moving at the point of contact, thered be no friction. If it rotated laterally, friction would be horizontal not vertical. And so on.

  • $\begingroup$ Shouldn't there also be the centrifugal force, acting perpendicular to the axis of rotation, that we need to factor? Since, we are solving this sum, from the sphere's frame which is non-inertial, there should technically be a component of centrifugal force acting in the same direction as normal force. $\endgroup$ Commented Aug 23, 2021 at 21:46
  • $\begingroup$ I've solved a very similar problem, where we had a block sliding on a rotating inclined plane and we had to find the distance along the plane, when the block loses contact. Here we had the cosine component of gravity equal to the sine component of the centrifugal force. If we ignore the centrifugal force, the body would not fly off. So, in this sum too, we should consider the centrifugal force. $\endgroup$ Commented Aug 23, 2021 at 21:48
  • $\begingroup$ No. Centrifugal force is an illusory thing, a pseudoforce. It looks from a rotating frame as if a force is pulling something outwards or pishing it inwards, because the frame itself is not inertial (its rotating). When you look closer, you'll see that the only actual ("true") forces acting on the object are the 3 that I mentioned. Resolving those 3 will solve the problem. You don't need to try and resolve the problem in a non-inertial frame. Its asking for you to get confused which forces should be counted or not. Stick to solving it as a rotating sphere in a nonrotating frame. Much easier $\endgroup$
    – Stilez
    Commented Aug 23, 2021 at 21:51
  • $\begingroup$ What I'm inclined to believe is that the problem becomes complicated, if one factors in the centrifugal force. So the book assumed that the mass loses contact with the rotating sphere at the same angle as a non-rotating sphere, to simply the assumptions. In reality, because of this centrifugal force, the body would lose contact at a lesser angle i.e. earlier. $\endgroup$ Commented Aug 23, 2021 at 21:51
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    $\begingroup$ I see, I finally get my mistake. I was trying to solve the problem, from the perspective of the rotating frame. In that case, the original problem would have 6 forces ( in the non-inertial frame ) - 2 centripetal forces for vertical and horizontal motion, as seen from the rotating frame, 1 centrifugal force, 1 Coriolis force, gravity and normal reaction. In non-inertial frame, we can cancel out 3 of these. $\endgroup$ Commented Aug 25, 2021 at 7:39

A pseudo-force (m$v^2$/R) can be used (along with other radial forces) to determine the radial acceleration of an object in a coordinate system which is rotating with the object. In this case, the object is moving along a vertical arc about the center of the sphere. The coordinate system rotating with it is rotating about a horizontal axis (not with the sphere). In that system R is constant (as long as the object stays on the surface), and one can say that N + m$v^2$/R – mg sin(θ) = 0.


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