Question:

Why do the outer nuts slip first?

Answer:

Circular motion does not happen by itself. A centripetal force is required to keep the nuts in circular motion.

In this demonstration, it is friction that provides the required centripetal force. Frictional forces arise naturally because of the attempt by both surfaces (nut and spinning disc) to stick together. If the grip is strong enough, the nuts stick to the spinning disc and undergo circular motino. If the grip is not strong enough, the nuts skid off the disk in the tangential direction.

Obviously, the outer nuts are undergoing circular motion at larger radius r compared to the inner nuts. So the question we are asking is whether the required centripetal force F is larger for larger r.

The formula that comes to most students’ mind is F=mv²/r. However, on the spinning disc, the tangential speed v is also higher for outer nuts compared to inner nuts. Using F=mv²/r, both v and r are larger for outer nuts. So it is not clear whether F is larger if r is larger.

It is better to use F=mrω²  instead,. On the spinning disc, all the nuts complete each revolution in exactly the same amount of time. In other words, at any time instant, all the nuts have the same angular velocity ω. From F=mrω² , we can conclude that larger r means larger F.

As the disc spins faster and faster and ω increases, the required centripetal F to keep all the nuts in circular motion also increases. However, the outer nuts always require a larger centripetal force than the inner nuts at the same ω. This means that the outer nuts reaches the maximum frictional force first. Therefore they are the ones to slip first.

Bonus Video

Realize that if you are at the centre of a spinning disc, you do not require any centripetal force. That’s why the green bean in the centre is never going to skid off. In fact, it does not even have any tangential speed v.

• Uncategorized