# Inertia on a rotating disc?

If I toss a ball upwards in a train moving with uniform velocity, the ball will land right back in my hand. This is because the ball has inertia and it continues to move forward at the speed of train even after leaving my hand.

Now consider I'm standing on the outer edge of a rotating disc (merry-go-round). If I toss a ball upwards, it doesn't fall back in my hand. Why? Doesn't it have a rotational inertia (is that even a term?) to continue rotating even after I let go of it? Is the ball going to land on a new location on the disc? Or is it going to fall away from the disc? At least the ball should have inertia of tangential velocity at which I tossed the ball upwards, right? So the ball should fall away from the disc? Can someone describe what happens in this situation?

• there is video at sling wikipedia article about what would happen – Matija Nalis May 17 '18 at 15:02
• This may sound confusing at first but a centripetal force (a force which is directed towards the centre) and not a tangential force is necessary to keep an object in circular motion. After you release the ball in the air, it has got an upward and tangential velocity, but no centripetal force is acting on it which can cause the ball to perform circular motion. But your body will still perform circular motion because the merry-go-round seat on which you are sitting on will push you inwards. – Apoorv Potnis May 17 '18 at 16:03

The first thing to note is that on the train you and the ball are moving with constant velocity whereas on the disc you and the ball are undergoing a (centripetal) acceleration.

The vertical motion of the ball is the same in both cases as the only vertical force acting is the same in both cases - the gravitational attraction between the ball and the Earth.

When the ball is released if no other force acts on it in the horizontal plane then due to the ball’s inertia it will continue on with the horizontal velocity it had at the instant it was released by your hand.
The direction of motion of the ball will be at a tangent to the circle along which ball was travelling before it was released so the ball will move away from the disc.

In the case of the train that constant horizontal velocity is the same as your velocity and so the ball will return back to you.
On the disc you will still be accelerating in the horizontal plane whereas the ball will move with constant horizontal velocity, so in the horizontal plane the motion of the ball and you will differ.

The idea of a rotational inertia is inappropriate for the ball when you are on the disc as there is nothing trying to make the ball rotate after the ball has been released.

On the merry-go-round, you are constantly accelerating inwards with acceleration $a=\frac{v^2}{R}=R\omega^2$. When you release the ball, it is no longer accelerating (horizontally) and will move in a straight line (horizontally). So whereas you accelerate inwards to maintain your circular motion, the ball follows a straight line and lands away from the disc.

By the way, what you call rotational inertia is called moment of inertia and for a particle of mass $m$ at radius $R$ it is $I=mR^2$. When you release the ball, it is no longer affected by any force $F$ and therefore there is no torque $\tau=RF$ from your hand and consequently it would have constant angular momentum $L=\omega I$.

• The constant angular momentum description here is correct for each particle of the ball which is an extended object (calculus gets involved because "spherical surface"). What it means is that the ball when released will be rotating along its own vertical axis (as it travels or translates tangentially to the merry-go-round). Fortunately in this case the rotation can be trivially calculated from purely geometric/kinematic arguments and you don't need to worry about the moment of inertia. – Euro Micelli May 17 '18 at 12:56
• Isn't moment of inertia mr^2? – yathish May 17 '18 at 19:58

This example might help visualize your question:

Water drops are ejected from a rotating nozzle. Once a drop is ejected from the nozzle, it follows a parabolic trajectory (let's ignore air drag) in a vertical plane. The drops fall outside of the vase shape but we don't see them well.

We don't see this trajectory when we take a picture, though. What we see is the locations of all the drops at a given instant ("streaklines"). They seem to have the "rotational inertia" you're talking about even though it's just an illusion.

From Newton's laws, we know that an object in motion will travel in a straight line unless acted upon an external force. On the merry-go-round, you are constantly changing direction; i.e., your direction is always tangential to the center of rotation. If you draw a velocity vector at different times on the merry-go-round you will see this clearly. Thus the moment you release the ball it continues in a straight line, with its initial velocity, tangential to the center of rotation at the moment of release.

There is something called "rotational inertia" more commonly referred to the "moment of inertia", however, it is not as you describe. It a measurement of the inertia of a rotating object--that is the resistance to changes in angular acceleration (increasing or decreasing rotation speed). You can think of it this way, a very light piece of material (say foam) will be easy to spin, however, a large block of granite will be harder to spin. When I say easy and harder, I really mean easy/harder to accelerate into a spinning motion. Similarly, trying to stop the granite while spinning will be harder as it has a higher moment of inertia (because of density in this case). This is the principle behind flywheels, the storage of energy in a rotating member.

Even looking at these cases, the individual particles (atoms) all experience centripetal acceleration towards the center because the atoms are connected. If they were not connected the atoms would fly away in a straight line just like the ball does. In your example.

See this for example: high speed disintegration of a compact disk.

So we can say that Newton's Law describes the fact that all particles want to travel in a straight line. Rotation is possible because we exert a force on a particle to change its direction, but the particle itself wants to continue on a straight line path.

When you will throw the ball upwards. You would have given it a velocity. The ball will also have a tangential velocity which is equal to the velocity of the merry go round. So the ball will land outside the merry go round and will behave like a ball in projectile.

## protected by Qmechanic♦May 17 '18 at 7:29

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