Consider a solid sphere with no friction, on the Earth's surface. Consider a point mass on top of the sphere. The sphere has just a small amount of initial speed so that it starts moving down the sphere.

The gravitational force points towards the center of the Earth and has components along the tangential and radial directions of the trajectory of the point mass.

$$F_g=-mg\cos{\theta}\ \hat{r}+mg\sin{\theta}\ \hat{\theta}$$

Let's measure $\theta$ as the angle formed with the vertical diameter of the sphere, such that the point mass starts at $\theta=0$.As $\theta$ increases, the tangential component of the gravitational force increases and the radial component decreases.

There is also a normal force from the surface of the sphere onto the point mass. As far as I can tell, its magnitude is $mg\cos{\theta}$ and it is a reaction to the force that the point mass applies onto the sphere as a result of the radial component of gravity, and thus fully offsets the latter.

But since the point mass moves in a circular trajectory (at least for a while), it must have a resultant force that has a component in the radial direction towards the center of the sphere changing its velocity vectors direction. I can't understand how to reconcile the fact that the normal force and the radial gravitational force cancel each other, and the fact that there is a force generating the radial acceleration.

I guess that the motion in this case is caused by the fact that the tangential force is changing direction. Is this so? I think this is probably similar or analogous to analyzing a pendulum, but I haven't done that yet.

  • 1
    $\begingroup$ How do you know the normal force is $mg\cos \theta$? How does that square with the fact that the object will eventually depart the sphere (and thus will become zero)? $\endgroup$
    – BowlOfRed
    Sep 30, 2021 at 16:40
  • $\begingroup$ The radial component of gravity is decreasing as $\theta$ increases. While the point mass is on the circular trajectory, it has radial acceleration $-\frac{v^2}{r}$. Using Newton's 2nd Law we can find the force required to generate such an acceleration. At the start of the movement, radial force is above such a required radial force, but at some point the radial force is smaller than the required force to keep the original circular trajectory. At this $\theta$ the object departs the sphere. $\endgroup$
    – evianpring
    Sep 30, 2021 at 16:52
  • $\begingroup$ As I mentioned in the question, as far as I can tell, the normal force is obtained using Newton's 3rd Law together with the radial component of the gravitational force, which is $mg\cos{\theta}$. $\endgroup$
    – evianpring
    Sep 30, 2021 at 16:53
  • $\begingroup$ But the object is accelerating. So you can't assume the normal force cancels the gravitational force. $\endgroup$
    – BowlOfRed
    Sep 30, 2021 at 16:57
  • $\begingroup$ I decomposed the gravitational force into a tangential and a radial component. I said that the normal force cancels the radial component. The tangential component of the gravitational force is accelerating the object. $\endgroup$
    – evianpring
    Sep 30, 2021 at 16:58

2 Answers 2


As far as I can tell, [the normal force's] magnitude is $mg \cos \theta$ and it is a reaction to the force that the point mass applies onto the sphere as a result of the radial component of gravity, and thus fully offsets the latter.

A scale in an elevator accelerating downward measures the normal force from the floor. It is a reaction force to a person's gravitational pull. But it does not have magnitude $mg$. The difference between it and $mg$ provides the downward acceleration of the person.

Why wouldn't the sphere exert a force in the opposite direction and equal magnitude?

It does. It's just that neither of these forces is equal to the radial component of $mg$.

Given the pathway (the circumference of the sphere) and the speed (can be calculated by energy arguments), you know the acceleration. Given the net acceleration, you can calculate the net force. Given the constant $mg$, you can determine the normal force. When the normal force goes to zero, the object can no longer remain on the sphere and departs with a ballistic trajectory.

  • $\begingroup$ A few things in your answer left me in doubt. First of all, I surmise you mean a scale in an elevator with a person on it. When the elevator is at rest, the person pushes down on the scale with a force of magnitude $mg$, and this is equal to the magnitude of the normal force on the person. If the elevator accelerates downward, we need to assume it is "glued" to the scale", and thus applies a downward force on the scale. The normal force is now equal to the difference between $mg$ and this downward force. $\endgroup$
    – evianpring
    Sep 30, 2021 at 23:27
  • $\begingroup$ You've made the claim that neither the normal force nor the radial force applied by the point mass on the sphere is equal to the radial component of $mg$. And the original question is "What is the resultant radial force...". So can you show me the calculation that you simply described in words, to calculate the net force? $\endgroup$
    – evianpring
    Sep 30, 2021 at 23:29
  • 1
    $\begingroup$ I would assume the acceleration and calculate the net force directly from that. See the calculations in physics.stackexchange.com/questions/262282/… $\endgroup$
    – BowlOfRed
    Sep 30, 2021 at 23:39

The particle mass obeying circular motion along the sphere obeys the following 2nd law equations $$N-mg\cos{\theta}=-\frac{v^2}{R}\tag{1}$$ $$$$

From energy conservation we know $$\frac{mv^2}{2}-mgR(1-\cos{\theta})=0$$ $$v=\sqrt{2gR(1-\cos{\theta})}\tag{2}$$

Inserting this speed into $(1)$ $$N=mg\cos{\theta}-\frac{2mgR(1-\cos{\theta}}{R}$$ $$N=mg(3\cos{\theta}-2)$$

We can see that the normal force becomes zero when $\cos{\theta}=\frac{2}{3}$.

Let's consider $$m=1kg$$ $$g=9.8m \cdot s^{-2}$$

And plot the normal force and the radial gravitational force as a function of $\theta$

radial gravitational force and normal force as function of theta

In the radial direction, magnitude of acceleration is increasing in $\theta$ (because speed is increasing in $\theta$), however it has a negative sign.

Radial gravitational force is decreasing, and if we think about this as a given, then the normal force has to decrease even faster to balance out the 2nd law equation, ie, the difference between normal and radial gravitational force has to decrease proportionally to radial acceleration increase.

At some point ($\theta=\cos^{-1}{\frac{2}{3}}$) the normal force reaches zero. Since it can't become negative in this setup, the resultant force in this radial direction isn't enough to keep the particle mass on a circular trajectory with this speed and radius (equivalently, this radial acceleration).

As an extra, what would happen if the particle mass were still sliding down the sphere but it were attached to a massless string fixed at the sphere center (that can somehow stay attached to the particle as it slides down).

We'd have an extra force acting parallel and in the same direction as the radial component of the gravitational force. This force is a tension force on the string. But would it be an extra force or would it just replace what was previously the normal force?

I believe what we would have is a case where $T=0$ from $\theta=0$ to $\theta=\cos^{-1}{\frac{2}{3}}$ and then $T$ would start increasing. It is simply the previous analysis, but in which $N$ can be negative. A negative $N$ would be the $T$ required to keep the particle on the circular trajectory.

$$-T-mg\cos{\theta}=-2mg(1-\cos{\theta}),\ \theta>\cos^{-1}{\frac{2}{3}}$$ $$T=mg(2-3\cos{\theta})$$

We can see that for $\theta=\pi$



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