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

• Kaushik
In summary: So, the limit as the part size goes to zero is still zero.In summary, the conversation discusses the tension of a rod rotating on a horizontal table about a vertical axis that passes through one of its ends. It is explained that in a linear mass distribution, there is no "last" particle and the force reduces continuously to zero at the end of the rod. This concept is important to understand in order to grasp the idea of a continuous distribution of mass and the power of calculus. It is also mentioned that a course in real analysis can help in understanding this concept.
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?

Kaushik said:
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
PeroK said:
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?

Kaushik said:
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.

Kaushik said:
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.

PeroK
PeroK said:
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?

Kaushik said:
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.

PeroK said:
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.

Kaushik said:
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.

Kaushik
Kaushik said:
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.

Kaushik
Kaushik said:
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.

sophiecentaur and Kaushik
A.T. said:
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"?

Kaushik said:
Now what about the "last" particle?
The tension to accelerate the last finite part is not at L but at less than L.

Kaushik said:
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.

Kaushik

## 1. What is the cause of tension in a rod rotating on a horizontal table about a vertical axis?

The tension in a rod rotating on a horizontal table about a vertical axis is caused by the centripetal force acting on the rod. This force is directed towards the center of the rotation and is necessary to keep the rod moving in a circular path.

## 2. How does the tension in the rod change as the speed of rotation increases?

The tension in the rod will increase as the speed of rotation increases. This is because the centripetal force needed to maintain the circular motion also increases with the speed of rotation.

## 3. Does the length of the rod affect the tension in this scenario?

Yes, the length of the rod does affect the tension. A longer rod will experience a greater tension than a shorter rod, as it has a larger distance to travel in the same amount of time and therefore requires a greater centripetal force.

## 4. How does the mass of the rod impact the tension in this situation?

The mass of the rod does not directly impact the tension. However, a heavier rod will require a greater centripetal force to maintain the circular motion, which in turn will result in a higher tension.

## 5. Can the tension in the rod be greater than the weight of the rod?

Yes, it is possible for the tension in the rod to be greater than its weight. This occurs when the speed of rotation is high enough to require a centripetal force that is greater than the force of gravity acting on the rod.

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