Rotating Rod Tension: Variations with Rotation Angle

[andyrk]

Why does tension across a rod rotating horizontally vary when it is rotating about one end?

 

[Qwertywerty]

Consider any point of the rod at a distance x from center of rotation - The tension exerted at that point is responsible for the circular motion of the rest of the rod ( l - x ) beyond that point .

At different points along the rod , the mass of l-x rotated varies and hence , the tension required for that mass' ( l - x ) circular motion changes too , being max at the point closest to the axis of rotation .

I hope you can understand this .

 

[andyrk]

Oh you mean, each part needs to have the same acceleration despite having different masses. So tension has to be different. Is that correct?

 

[Qwertywerty]

Let mass of rod be m . Let it's total length be l . Consider a point at a distance x from the center of the rod . If you now consider the whole mass of length l-x to be one system , then the tension acting on that would be :

Mass of system * distance of Com of system from axis of rotation * w∧2 .
=(M* (l-x) / l )*( x + (l-x)/2 )*w∧2 .

Read this patiently and you will see that tension varies with x .

 

[A.T.]

each part needs to have the same acceleration

The acceleration along a rotating rod isn't constant.

despite having different masses

What prevents you from considering parts of equal mass?

 

[andyrk]

So the tension at any point..does that act as the centripetal force for the entire rod behind that point..or does it act as the centripetal force just for that point?

 

[Qwertywerty]

For the entire part behind it .

 

[andyrk]

But if the tension is at x, then for the rod behind it of length l-x, as centripetal acceleration is ω²x, how can we assume that x is same for all points in the rod behind the point at which tension was calculated?

 

[Qwertywerty]

Please refer to the post of mine in which I calculate Tension . Centripetal acceleration is not (w∧2)*x .

 

[Chestermiller]

What you want to do is to focus on a small segment of the rod between x and x + dx. The mass of this segment is mdx/L, where L is the length of the rod and m is its mass. If you do a force balance on this segment of the rod, you obtain:

$$T(x+dx)-T(x)=-(ω^2x)\left(m\frac{dx}{L}\right)$$
or
$$\frac{dT}{dx}=-\frac{m}{L}ω^2x$$

Note that the tension T decreases with radial position x.

Chet

 

[andyrk]

What you want to do is to focus on a small segment of the rod between x and x + dx. The mass of this segment is mdx/L, where L is the length of the rod and m is its mass. If you do a force balance on this segment of the rod, you obtain:

$$T(x+dx)-T(x)=-(ω^2x)\left(m\frac{dx}{L}\right)$$
or
$$\frac{dT}{dx}=-\frac{m}{L}ω^2x$$

Note that the tension T decreases with radial position x.

Chet

Yes. So what you are saying is that every point has different centripetal acceleration. So how can tension at one point play the role of centripetal force for all points behind it? Isn't it supposed to be just for that point only?

 

[A.T.]

So what you are saying is that every point has different centripetal acceleration.

While true, it's not the key to variable tension. You should first try to understand a rod accelerated linearly by pulling one end. Here, the acceleration is constant along the rod, but the tension still varies.

So how can tension at one point play the role of centripetal force for all points behind it?

What else would provide the centripetal force to the rest of the rod beyond that point?

 

[Chestermiller]

Yes. So what you are saying is that every point has different centripetal acceleration.

Yes.

So how can tension at one point play the role of centripetal force for all points behind it? Isn't it supposed to be just for that point only?

I don't quite follow what you are saying here. I know you are aware that a geometric point has zero mass, so no net force is required to accelerate it. If you focus on a segment of the rod between x and x + dx, it does have mass, and the net force acting on it is T(x+dx) - T(x). So the tension is changing along the rod to provide a net force on every differential segment. If you add up the centripetal forces necessary to accelerate all the segments of the rod outboard of location x, you get T(x):

$$T(x)=\frac{mω^2L}{2}\left[1-\left(\frac{x}{L}\right)^2\right]$$

Hope this makes sense.

Chet

 

[andyrk]

Yes.

I don't quite follow what you are saying here. I know you are aware that a geometric point has zero mass, so no net force is required to accelerate it. If you focus on a segment of the rod between x and x + dx, it does have mass, and the net force acting on it is T(x+dx) - T(x). So the tension is changing along the rod to provide a net force on every differential segment. If you add up the centripetal forces necessary to accelerate all the segments of the rod outboard of location x, you get T(x):

$$T(x)=\frac{mω^2L}{2}\left[1-\left(\frac{x}{L}\right)^2\right]$$

Hope this makes sense.

Chet

I got what I was looking for. Basically, that tension force isn't acting on the the particle at x or x + dx etc. Instead it is acting at the COM of the remaining part of the rod so the rod gets treated as a single point and hence we have the centripetal force equation for a single particle and not the other way round like: one tension force and different particles have different x so which x do we put in the centripetal force equation, which is what was confusing me initially. Makes sense now. :)

 

[A.T.]

nstead it is acting at the COM of the remaining part of the rod

Not really, tension acts at the inner end of the remaining part of the rod.

 

[andyrk]

Not really, tension acts at the inner end of the remaining part of the rod.

Of course it does. But when we want to write the centripetal force equation, we say the mass is concentrated at the COM...and so we just have one x instead of many x's all along the rod.

 

[Chestermiller]

Of course it does. But when we want to write the centripetal force equation, we say the mass is concentrated at the COM...and so we just have one x instead of many x's all along the rod.

So Andy, when you apply this algorithm to get the tension at location x, T(x), do you get the same answer as I obtained in post #13 by integrating the differential equation that takes into account the "many x's all along the rod?"

Chet

 

[Qwertywerty]

When we say Com , we mean the Com of the part on which tension acts as a centripetal force .

Chet , yes you do .
Could you please tell me how to quote somebody ?

 

[Chestermiller]

When we say Com , we mean the Com of the part on which tension acts as a centripetal force .

Chet , yes you do .

Ah. I see it now. Thanks. Still, I would feel uncomfortable doing the problem this macroscopic way, and not by the microscopic balance.

Could you please tell me how to quote somebody ?

Look at the strip at the bottom of each post, with a group of words shown in very light lettering. One of the words is Reply. Click on Reply, and the post will be copied into your "Have something to add" area, including with proper LaTex punctuation.

Chet

 

[Qwertywerty]

Ah. I see it now. Thanks. Still, I would feel uncomfortable doing the problem this macroscopic way, and not by the microscopic balance.

Look at the strip at the bottom of each post, with a group of words shown in very light lettering. One of the words is Reply. Click on Reply, and the post will be copied into your "Have something to add" area, including with proper LaTex punctuation.

Chet

I initially solved it using ∫ . I got the answer to be the same as that due to centripetal acceleration of the Com , and then reasoned it out using the basic formula of (M*A)com = dm1 * w∧2 *x1 + ... and saw that it gives the same result ( Probably because of linear relationship between Fc and radius of circular motion ).

Ah - It works . Thank you .

 

[andyrk]

So Andy, when you apply this algorithm to get the tension at location x, T(x), do you get the same answer as I obtained in post #13 by integrating the differential equation that takes into account the "many x's all along the rod?"

Chet

I didn't try it but I think it should.
Also I was just a bit confused as to why is the tension force T(x) same at x as well as COM? If it isn't then why do we say that centripetal force acting at COM is T(x)?

 

[A.T.]

why do we say that centripetal force acting at COM is T(x)?

You say this, and I already told you it's wrong. If you hold a bag by the handles and swing it around, you apply the centripetal force at the handles, not at the COM of the bag. But the COM acceleration determines how much centripetal force you have to apply.

 

[andyrk]

But the COM acceleration determines how much centripetal force you have to apply.

Why?

 

[A.T.]

Why?

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/n2ext.html

 

[Chestermiller]

I didn't try it but I think it should.

Twertywerty already confirmed this, and I have shown it also.

Also I was just a bit confused as to why is the tension force T(x) same at x as well as COM? If it isn't then why do we say that centripetal force acting at COM is T(x)?

T(x) is not the tension at the center of mass. T(x) is the tension at any arbitrary location x along the rod. The center of mass of the portion of the rod outboard of location x is at (x + L)/2. The mass of the rod outboard of location x is (L-x)m/L. So,

$$T(x)=\left(ω^2\frac{(x+L)}{2}\right)\frac{m(L-x)}{L}=\frac{mω^2L}{2}\left[1-\left(\frac{x}{L}\right)^2\right]$$
The tension at the COM, x = L/2, is $\frac{3}{8}mω^2L$

Chet

 

[andyrk]

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/n2ext.html

Thanks!