Rotating Rod

1. Jul 20, 2015

andyrk

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

2. Jul 20, 2015

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 .

3. Jul 20, 2015

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?

4. Jul 20, 2015

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 .

5. Jul 20, 2015

A.T.

The acceleration along a rotating rod isn't constant.

What prevents you from considering parts of equal mass?

6. Jul 20, 2015

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?

7. Jul 20, 2015

Qwertywerty

For the entire part behind it .

8. Jul 20, 2015

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?

9. Jul 20, 2015

Qwertywerty

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

10. Jul 20, 2015

Staff: Mentor

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

11. Jul 20, 2015

andyrk

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?

12. Jul 20, 2015

A.T.

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.

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

13. Jul 20, 2015

Staff: Mentor

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

14. Jul 20, 2015

andyrk

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. :)

15. Jul 20, 2015

A.T.

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

16. Jul 20, 2015

andyrk

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.

17. Jul 20, 2015

Staff: Mentor

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

18. Jul 20, 2015

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 ?

19. Jul 20, 2015

Staff: Mentor

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

Last edited: Jul 20, 2015
20. Jul 21, 2015

Qwertywerty

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 .

Last edited: Jul 21, 2015