# Weightlessness at the highest point of a vertical circular motion

Suppose there is a mass fixed at one end of a piece of massless rope. The mass is whirled in a vertical circular path, holding the rope from the other end. The velocity provided is the minimum velocity required to keep travelling along the circular path. So, at the highest point of the path the force acting to provide the centripetal acceleration is only the weight of the fixed mass. My questions are,

1. What keeps the mass from falling directly down, at its highest point? Is it inertia?

2. If the mass was travelling in a vertical circular path inside a ring, with no rope attached, the only force acting to provide centripetal force is the perpendicular reaction of the mass. What is the reaction of this force? Is it still weight? But both the weight and perpendicular reaction acts in the same direction at the highest point?

• Why do you think the mass doesn't fall down? Commented May 17 at 17:56
• A comment on your point 2 - about weight being the reaction force. The force from the ring is not in reaction to the weight. In textbooks, you'll find problems with an object at rest on the floor, where the normal force is the same magnitude but opposite direction as the weight. A familiar phrase, but this is not an instance of N3L. These two are not an action-reaction pair. They both act on the same object! The reaction to the weight (the force of gravity that the Earth exerts on the object) is the force of gravity that the object exerts on the Earth, and... 1/2 Commented May 18 at 3:35
• ...the reaction to the normal force (the force of the floor on the object) is the force the object exerts on the floor (which is not its weight, and is not gravitational in nature - the two are repelling each other when in contact). But in these textbook problems, we're only interested in the forces that act on the object (so, the weight and the normal force), and the only reason we can infer they cancel out is because (1) they do act on the same object, and (2) the object is not changing its velocity (which in this case happens to be zero) - i.e., it's the N1L that allows the inference. 2/2 Commented May 18 at 3:35
• @DJohnM I meant the force that keeps the mass from falling directly down. Commented May 18 at 12:44

It's easiest to see what is happening if one looks at the object from the perspective of an outside observer:

What keeps the mass from falling down, at its highest point?

Nothing. Nothing keeps the mass from falling down. If you really think about it, 100% of the times where the object reaches the highest point, it immediately begins accelerating downward (falling). The only thing that's obscuring the fall is that the object has a horizontal velocity at that time, so it's path isn't straight down: it's down and sideways. But it's absolutely falling.

Really the only thing that stop it from "falling" in any sense of the word is that the rope/ring "catches" it on the bottom half of the swing. But at the moment at the top of the ring, it really is just "falling."

If the mass was travelling in a vertical circular path inside a ring, with no rope attached, the only force acting to provide centripetal force is the perpendicular reaction of the mass. What is the reaction of this force? Is it still weight? But both the weight and perpendicular reaction acts in the same direction at the highest point?

When talking about reactions, remember that the equal and opposite reaction is always between the same two objects, but the roles of subject and object are switched. So if you talk about a force applied by the ring to the mass, the reaction force must be an equal and opposite force applied by the mass to the ring. No other force can be called its "reaction." It cannot be weight because weight is not a force applied by the mass to the ring.

It is very easy to confuse two concepts. Reaction forces are always equal and opposite, and the subject and object switch roles. Weight might balance the force of the ring on the mass, but that is not the same as a reaction force.

In your specific example, at the top of the ring you stated that the force on the mass by the ring is at a minimum. This minimum is at the point when the object experiences "weightlessness" or "freefall" where there are no (net) forces other than gravity acting on it. So in this very specific example of the mass at the top of the ring, the force of the ring on the mass is 0.

Now, with all of that inertial thinking out of the way, we can talk about the behavior in a rotating frame which follows the mass. In this rotating frame, the motion of the mass is 0 (sort of by definition: we defined the rotating frame such that, in that frame, it isn't moving). There are no new forces when we think about it this way. There is still gravity, a force from the rope/ring on the mass, and its equal and opposite force of the mass on the rope/ring. Changing reference frames does not change the true forces in the system.

The next few paragraphs are more math, just pinning down the equations of motion. For an intuitive understanding, they can be skipped, and one can pick up after them.

However, in a rotating frame, the equations of motion are different than they are in an inertial frame. There is a centrifugal acceleration. Instead of the equations of motion being $$\Sigma F=ma$$, they are almost $$\Sigma F=ma + (\Omega\times(\Omega\times r))+(2\Omega\times v_r)$$ where $$\Omega$$ is the angular rate of the frame (w.r.t the inertia frame), and $$v_r$$ is the velocity of the object in the rotating frame. The first of those terms is the centrifugal term, which you are thinking about in this problem, and the second is the Coriolis term which shows up in other problems.

I switched into vector notation there to line up with the variables used on Wikiepdia's page on rotating frames. They work in all cases. In this specific case, the motion is in one 2-d plane, and the angular rate vector is at right angles to that. In that special case, the cross products ($$\times$$) are simple. Let $$\Omega=\omega\hat z$$, where $$\omega$$ is a scalar (which is probably what you're used to). $$(\Omega\times(\Omega\times r))$$, the centrifugal term, computes to $$\omega^2 |r|$$ in the radial direction, $$\hat r$$ (the intuitive direction for the centrifugal term). The Coriolis term, $$(2\Omega\times v_r)$$ becomes $$2\omega v_r$$ at right angles to the velocity. And in your particular case, there's no velocity for this particular object in this particular frame ($$v_r=0$$), so we'll ignore that term. Thus we get $$\Sigma F= ma + \frac{\omega^2} {|r|}\hat r$$

If you skipped over the math, the takeaway from the above paragraphs is that in a rotating frame, the equation of motion is not $$F=ma$$. It's $$F=ma + C$$, where $$C$$ is the above expressions capturing centripetal and Coriolis accelerations. And one should note that these are accelerations not forces. Sometimes we will hand-wave them into forces, simply by multiplying in the mass and dragging them to the left side of the equation. But they are not forces. They are pseudo-forces. The difference is that every force has an equal and opposite reaction force, but that is not true for pseudo-forces. Every time someone speaks of a centifugal forces, remind yourself that it's actually a pseudo-force.

Now in the case of your example, the centripetal force from the ring is 0. Gravity is also centripetal at this one particular moment in time. These are the only true forces on the object. If we were thinking in inertial terms, we'd say $$\Sigma F=ma$$, and since the sum of the forces is $$-mg$$, acceleration must be $$a=-g$$. But since we're in a rotating frame, where' there's $$\Sigma F=ma + C$$, we have $$-mg=ma + C$$ where $$C$$ is the centrifugal acceleration associated with the rotating frame. Since in our particular example the object is not accelerating in the rotating frame, $$a=0$$, we get $$-mg=C$$. The force of graity at the top of the circle is exactly the same magnitude as the centrifugal acceleration , just in the opposite direction.

• Nice explanation of how it doesn't stop falling. The string can still be putting tension on whilst we're in the top half though - but that's because the mass is going sideways into the string and not because it's going down. Commented May 18 at 6:47
• This might add some intuition about rotating frames - Coriolis Force: Direction Perpendicular to Rotation Axis Visualization Commented May 18 at 17:52

Perhaps you mean weightlessness in the title? The mass of an object does not change as it follows a trajectory.

Suppose you throw the mass. If follows a curved trajectory. When it reaches the peak, what keeps it from falling down? Nothing. It is traveling horizontally at that point. It begins to fall.

If you throw it with a lot of horizontal speed, the trajectory has a relatively flat curvature at the peak. You could choose a horizontal velocity to match your mass on a string at its peak. The shapes of the free and circular trajectories differ over most of their path, but they are the same at the peak.

A circular mass on a rope and a circular mass on a ring have the same trajectories, and the same reaction forces over their entire trajectories, provided the speeds are fast enough to keep the rope staight.

• Ah yes! I meant weightlessness. Commented May 17 at 15:22
1. At the highest point, the mass doesn't fall because the centripetal force required for circular motion ($$mv^2/r$$) is exactly provided by the weight of the mass ($$mg$$). The tension in the rope is zero for the minimum velocity required to maintain the circular path at the highest point. Inertia (the tendency of the mass to continue in its path) and the centripetal force (provided by gravity at the highest point) keep the mass in its circular path. Inertia ensures the mass keeps moving, while the gravitational force (weight) provides the centripetal acceleration that changes the direction of this motion, keeping it on a circular path.

2. If the mass travels inside a vertical ring with no rope, the only force providing the centripetal acceleration at the highest point is the weight of the mass ($$mg$$). For the minimum velocity required to maintain the circular path at the highest point, the normal reaction force exerted by the ring on the mass is zero (just like tension in the rope). Inertia keeps the mass moving in a circular path, while the centripetal force (due to gravity at the highest point) directs this motion.

When a mass is whirled in a vertical circle to maintain its circular path, the centripetal force provided by the forces acting on the mass is a component of its weight towards the centre ($$mg\cos\theta$$) and the tension ($$T$$) in the rope, i.e., $$T+mg\cos\theta=\frac{mv^2}r$$, where $$\theta$$ is the angle the rope makes with the vertical axis, $$v$$ is the velocity, and $$r$$ is the radius of the circular path.

• For the minimum velocity at the highest point ($$\theta=0^\circ$$) required to maintain circular motion, the tension in the rope is zero at the highest point ($$T_\text{top}=0$$). This means the only force providing the centripetal force at this point is the weight of the mass. So, $$mg=\frac{mv_\text{top}^2}{r}\implies v_\text{top}=\sqrt{gr}$$.
• At the lowest point ($$\theta=180^\circ$$), $$T_\text{lowest}-mg=\frac{mv_\text{lowest}^2}{r}$$. From energy conservation, $$\frac12mv_\text{lowest}^2=\frac12mv_\text{top}^2+mg\cdot2r=\frac{5}{2}mgr\implies v_\text{lowest}=\sqrt{5gr}$$, and $$T_\text{lowest}=6mg$$.
• At the horizontal point ($$\theta=90^\circ$$), $$T_\text{hor}=\frac{mv_\text{hor}^2}{r}$$. And, $$v_\text{hor}=\sqrt{3gr},\ T_\text{hor}=3mg$$.

Both cases involve similar principles of physics. The tension in the rope ($$T$$) and the normal reaction force ($$N$$) exerted by the ring act towards the circular path's centre. For the minimum velocity required to maintain the circular path at the highest point, $$T$$ (rope) and $$N$$ (ring) are zero and only the weight ($$mg$$) of the mass $$m$$ provides the centripetal force.