Tension of a rod rotating on a horizontal table about a vertical axis

[Kaushik]

If a rod is on a table (horizontally) and rotating about an axis that passes through one of its ends and vertical to the table, what would be the tension on the opposite end of the rod (the end opposite to the axis) . In this post (Check this post out from Socratic QnA), the limits take while integrating $dT$ is T to 0. But if that is the limit then $T(L) = 0$. If $T(L) = 0$ then how does the particle in that end rotate? Because Tension on that end, $T(L)$ is $0$ so there is no force that provides centripetal acceleration.

Where am I going wrong?

 

[PeroK]

Summary: Tension of a rod rotating on a horizontal table about a vertical axis passing through one of its ends.

If a rod is on a table (horizontally) and rotating about an axis that passes through one of its ends and vertical to the table, what would be the tension on the opposite end of the rod (the end opposite to the axis) . In this post (Check this post out from Socratic QnA), the limits take while integrating $dT$ is T to 0. But if that is the limit then $T(L) = 0$. If $T(L) = 0$ then how does the particle in that end rotate? Because Tension on that end, $T(L)$ is $0$ so there is no force that provides centripetal acceleration.

Where am I going wrong?

If you model the rod as a finite set of point masses, then there is a force on the last point, depending on its mass.

If, however, you model the rod as a continuous linear mass distribution, then there is no last particle and the force reduces continuously to zero at the end of the rod.

 

[Kaushik]

If you model the rod as a finite set of point masses, then there is a force on the last point, depending on its mass.

If, however, you model the rod as a continuous linear mass distribution, then there is no last particle and the force reduces continuously to zero at the end of the rod.

Why is there no "last" particle in linear distribution of mass?

 

[PeroK]

Why is there no "last" particle in linear distribution of mass?

If the rod has a total mass of $m$ and a length of $L$ what would be the mass of the last particle? Assuming a linear mass distribution.

 

[jbriggs444]

Why is there no "last" particle in linear distribution of mass?

A particle suggests something of finite size. A linear distribution of mass suggests that there is no such thing as a smallest particle.

I suggest a course in real analysis.

 

[Kaushik]

If the rod has a total mass of $m$ and a length of $L$ what would be the mass of the last particle? Assuming a linear mass distribution.

For unit length the mass is $\frac{M}{L}$. So will it be the mass of last particle?

 

[PeroK]

For unit length the mass is $\frac{M}{L}$. So will it be the mass of last particle?

So, if the rod is $1m$ long, then the last particle has all the mass $M$?

What about all the other particles? Do they have the same mass?

Note that $\frac M L$ is a linear mass density, not a mass.

 

[Kaushik]

Note that $\frac {M}{L}$is a linear mass density, not a mass.

This is where I went wrong.

I actually don't know what the mass of the last particle could be. But according to my intuition, "Matter" is something that has mass. So the "last" particle of the rod must have mass. But from what you are saying, there is no such thing called "last" particle in linear mass distribution. I'm confused right now.

 

[A.T.]

Because Tension on that end, $T(L)$ is $0$ so there is no force that provides centripetal acceleration.

Centripetal acceleration of what? There is nothing beyond L, so there is no need for tension at L.

 

[PeroK]

This is where I went wrong.

I actually don't know what the mass of the last particle could be. But according to my intuition, "Matter" is something that has mass. So the "last" particle of the rod must have mass. But from what you are saying, there is no such thing called "last" particle in linear mass distribution. I'm confused right now.

See post #5.

Linear mass distribution is something of a mathematical trick. A rod, for example, is made of a finite number of molecules and is mostly empty space. Even the molecules are mostly empty space.

But, the simplest way to study the macroscopic motion of a rod is to consider it a continuous distribution of mass with no individual point particles.

Continuous functions can usually be differentiated and integrated, which gives you the power of calculus.

You need to grasp this idea, as it's very important.

 

[jbriggs444]

But from what you are saying, there is no such thing called "last" particle in linear mass distribution. I'm confused right now.

Correct. This is one of the behaviors of the continuum (or of the real number line) that one needs to wrap an intuition around.

A course in real analysis is an eye opener for this kind of thing.

 

[Kaushik]

Centripetal acceleration of what? There is nothing beyond L, so there is no need for tension at L.

Oh wow!

Now what about the "last" particle? Is there no "last" particle because we can keep on dividing the particle and hence, there is no defined "last particle"?

 

[A.T.]

Now what about the "last" particle?

The tension to accelerate the last finite part is not at L but at less than L.

Is there no "last" particle because we can keep on dividing the particle and hence, there is no defined "last particle"?

If the part size goes to zero, then so does their mass and thus the required centripetal force.