Average tension force acting on a simple pendulum

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Trying to answer it before solving it,

The maximum tension force is mg, so the average has to be less than this. Hence, the option (a) and (c) are not correct.

$T = mg \cos \theta$, where $\theta$ is the angle between the string and the vertical axis.

Now, $\theta = \omega t$

So, $<T> = \frac { { \int_{ T}^{2T } mg \cos { (\omega t)} dt} } {\int_{ T}^{2T } dt } = 0.$What is wrong here?

 

[haruspex]

The maximum tension force is mg

Is it? Why?

 

[Pushoam]

Is it? Why?

Because T = mg cosθ

 

[haruspex]

Because T = mg cosθ

What is the acceleration of the mass when the string is vertical?

 

[Pushoam]

What is the acceleration of the mass when the string is vertical?

It is 0. So, T = mg when the mass is not moving.

 

[haruspex]

It is 0.

So, at the bottom of the arc the mass is moving with constant velocity? Seems a bit contradictory.

 

[Pushoam]

So, at the bottom of the arc the mass is moving with constant velocity? Seems a bit contradictory.

Yes, at the bottom, there is no acceleration.
The bottom is the point of stable equlilbrium. So, the net force is 0 at this point.
The acceleration is $g\sin\theta$ and so it is 0 as $\theta$ is 0 at bottom.

I don't see what is contradicting it. Will you please give me a hint?

 

[haruspex]

Yes, at the bottom, there is no acceleration.
The bottom is the point of stable equlilbrium. So, the net force is 0 at this point.
The acceleration is $g\sin\theta$ and so it is 0 as $\theta$ is 0 at bottom.

I don't see what is contradicting it. Will you please give me a hint?

The bob passes along an arc, not a straight line. While it is moving it is therefore accelerating. What is the usual name for that acceleration?

 

[Pushoam]

What is the usual name for that acceleration?

It is centripetal acceleration which keeps a body moving into an arc.

The bob passes along an arc, not a straight line.

But, we are considering locally.
For $d\theta$ around $\theta = 0$, the arc could be taken as a straight line. Well, I am not writing this to cut your suggestion. I am writing it as I am getting this thought, too, and I want to get this thought checked by you.

 

[haruspex]

For dθ around θ=0, the arc could be taken as a straight line.

To a first order yes, but acceleration is second order.
You know what the formula is for centripetal acceleration, so use it. If your argument were correct it would give zero, but it doesn't.

 

[Pushoam]

The bob is not moving in the radial direction, so, I took the centripetal acceleration to be 0. I think this is the mistake.

I should follow Newton's law explicitly.

Taking origin of the polar coordinate system at P,

$\vec F = (T – mg \cos { \theta } )~\hat r – mg \sin { \theta } ~\hat {\theta}$ ………..(1) wrong
I am editing the following after getting the suggestion from haruspex in post 16.

I have taken the wrong sign of forces in radial direction.
The correct one is
$\vec F = (T – mg \cos { \theta } )~(-\hat r) – mg \sin { \theta } ~\hat {\theta}$ ………..(1) correct$\ddot {\vec r} = - r {\dot \theta }^2 ~\hat r + r \ddot \theta ~\hat {\theta}$, taking $\dot r = 0$ ……….(2)

$- r {\dot \theta }^2 ~\hat r = \frac { 1 } {m } (T – mg \cos { \theta } )~\hat r$.... wrong
$r {\dot \theta }^2 ~\hat r = \frac { 1 } {m } (T – mg \cos { \theta } )~\hat r$....correct

So, at $\theta = 0$, acceleration is not 0.

The radial acceleration is non- zero, $a_r = - r {\dot \theta }^2 = \frac { 1 } {m } (T – mg \cos { \theta })$....(3).....wrong

The radial acceleration is non- zero, $a_r = - r {\dot \theta }^2 = \frac { -1 } {m } (T – mg \cos { \theta })$....(3)...correctThe tangential acceleration is 0 at $\theta = 0$, $a_{\theta } = r \ddot \theta = -g \sin { \theta }$. .....(4)

 

[haruspex]

$\vec F = (T – mg \cos { \theta } )~\hat r – mg \sin { \theta } ~\hat {\theta}$ ………..(1)

$\hat r$ is away from the source of tension, no?

 

[Pushoam]

$\hat r$ is away from the source of tension, no?

Yes.

So, for calculating < Tension >, I have to express tension in terms of $\theta = \theta ( t)$ and then I have to take the average of tension over one time period T.

Solving equation (3) with the approximation $\sin { \theta } = \theta$ for small $\theta$ gives,

In equation (3), it is $\theta$ which oscillates with angular frequency $\omega = \sqrt{ \frac { g} {l }}$ where l is the length of the string.

I note that $\dot \theta$ may or may not be equal to $\omega$.

$\theta (t) = A \cos{ \omega t} + B \sin{ \omega t}$

Taking at t = T/4, $\theta (T/4) = 0$ and at t = T, $\theta (T ) = \theta _0$,$\theta (t) = \theta _0 \cos{ \omega t} \neq \omega t$

$\dot \theta = -\omega \theta _0 \sin{ \omega t} \neq \omega$

Tension, $T = m( g \cos{ \theta } + l (\dot \theta )^2)$ ...(5)
Is this correct so far?

Now, approximating $\cos{\theta } as 1$,

$<T> = mg + \frac 1 2 m g {\theta_0}^2 = mg + \frac {m { \omega}^2A^2}{ 2l}$

 

[haruspex]

$\hat r$ is away from the source of tension, no?

Yes

So what should the sign be on the $T\hat r$ term, and what on the $g\cos(\theta)\hat r$ term?

 

[Pushoam]

Homework Statement

View attachment 216556

Homework Equations

The Attempt at a Solution

Trying to answer it before solving it,

The maximum tension force is mg, so the average has to be less than this. Hence, the option (a) and (c) are not correct.

$T = mg \cos \theta$, where $\theta$ is the angle between the string and the vertical axis.

Now, $\theta = \omega t$

So, $<T> = \frac { { \int_{ T}^{2T } mg \cos { (\omega t)} dt} } {\int_{ T}^{2T } dt } = 0.$What is wrong here?

What is wrong there is:

1) The acceleration at $\theta = 0$ is 0. So, the maximum tension is mg.

The correct thing is acceleration at $\theta = 0$ is not 0.
At $\theta = 0, a_r = l {\dot {\theta}} ^2, a_{\theta }= 0$.2) $\theta = \omega t$

The above equation is valid for uniform circular motion. It is not valid every where.

I have a tendency to use it whenever I see $\omega ~ and ~ \theta$.

Here $\omega \neq \dot {\theta} , ~and~ \theta \neq \omega t$.Is this correct?

 

[haruspex]

I have a tendency to use it whenever I see ω and θ.

Right but here $\omega$ does not represent $\dot\theta$.

You did not answer my question in post #14.

 

[Pushoam]

You did not answer my question in post #14.

Sorry for this. I remained in solving the question.

So what should the sign be on the $T\hat r$ term, and what on the $g\cos(\theta)\hat r$ term?

The sign should be negative of what I wrote earlier in post 11. I edited the post 11. Will you please check it again?

$^\vec F = (T – mg \cos { \theta } )~\hat r – mg \sin { \theta } ~\hat {\theta}$………..(1) wrong
I am editing the following after getting the suggestion from haruspex in post 16.

I have taken the wrong sign of forces in radial direction.
The correct one is
$\vec F = (T – mg \cos { \theta } )~{(-\hat r)} – mg \sin { \theta } ~\hat {\theta}$………..(1) correct

Fortunately, in post 13, eq(5), I had taken the correct sign by mistake.

$T = m( g \cos{ \theta } + l (\dot \theta )^2)$

So, my final answer should remain the same i.e. option (c). Right?

 

[haruspex]

So, my final answer should remain the same i.e. option (c). Right?

Check what happens if you use a more accurate approximation for cos(θ).

 

[Pushoam]

Check what happens if you use a more accurate approximation for cos(θ).

Tension, $T = m( g \cos{ \theta } + l (\dot \theta )^2)$...(5)
Is this correct so far?

Now, approximating cosθ as1,

$<T> = mg + \frac 1 2 m g {\theta_0}^2 = mg + \frac {m { \omega}^2A^2}{ 2l}$
l

$\cos{\theta} = 1 - \frac {{ \theta}^2} 2$

$< {\theta }^2> = {\theta_0}^2/3$

So, $<T> = mg - \frac { mg{\theta_0}^2} 6 + \frac { mg{\theta_0}^2} 2= mg - \frac { mg{\theta_0}^2} 3 = mg + \frac { m {\omega }^2A^2}{3l}$

So,this means that $mg< \left <T \right > < mg + \frac { m {\omega }^2A^2}{2l}$.Thanks for taking me to more accurate approximation.

But, I am not seeing the significance of it. Is it only for mathematical calculation or is there anything to appreciate in it which I am missing?

 

[Pushoam]

Do you mean that if we take more order of correction, then <T> will go to $mg + \frac { m {\omega }^2A^2}{4l}$?

But, I have taken $\sin \theta = \theta$, so will it make sense to take more correct order (more than 1st order) of approximation for $\cos \theta$?
Shouldn't I take $cos\theta = 1$, when I take $\sin \theta = \theta$?

 

[haruspex]

Do you mean that if we take more order of correction, then <T> will go to $mg + \frac { m {\omega }^2A^2}{4l}$?

That is what I got.

Shouldn't I take $cos\theta = 1$, when I take $\sin \theta = \theta$?

Zeroth order: sin θ=0, cos θ=1
1st order: sin θ=θ, cos θ=1
2nd order: sin θ=θ, cos θ=1-½θ²
3rd order: sin θ=θ-θ³/6, cos θ=1-½θ²
Etc.

 

[Pushoam]

But this is what I am getting. Where am I wrong?

$\cos{\theta} = 1 - \frac {{ \theta}^2} 2$

$< {\theta }^2> = {\theta_0}^2/3$

So, $<T> = mg - \frac { mg{\theta_0}^2} 6 + \frac { mg{\theta_0}^2} 2= = mg - \frac { mg{\theta_0}^2} 3 = mg + \frac { m {\omega }^2A^2}{3l}$

 

[Pushoam]

Zeroth order: sin θ=0, cos θ=1
1st order: sin θ=θ, cos θ=1
2nd order: sin θ=θ, cos θ=1-½θ2
3rd order: sin θ=θ-θ3/6, cos θ=1-½θ2
Etc.

You probably want to write:
1st order: sin θ=θ, cos θ=1 - ½ θ²
2nd order: sin θ= θ - θ³/6 , cos θ=1- ½θ² +θ⁴/24
and so on...

 

[haruspex]

You probably want to write:
1st order: sin θ=θ, cos θ=1 - ½ θ²
2nd order: sin θ= θ - θ³/6 , cos θ=1- ½θ² +θ⁴/24
and so on...

No, the order is the power of θ.

 

[haruspex]

$< {\theta }^2> = {\theta_0}^2/3$

No, that results from integrating wrt θ. We want the average over time, so an integral wrt time.

 

[Pushoam]

No, that results from integrating wrt θ. We want the average over time, so an integral wrt time.

I thought that $<\theta ^2>$with respect to $\theta$ going from $-\theta _0 ~to~ +\theta_0$ is equal to $<\theta ^2>$ over one time period T. But it is not so. I don’t see why I am inclined think that they should be equal but they are not.

So, average of a function depends upon the variable with respect to which I want to calculate the average. It is not independent of the variable.

Here I am having some misconception. And I feel that I may do this mistake again if I don’t become clear about what I have understood wrongly.With respect to time, $<\theta ^2> = \frac { {\theta_0 }^2} 2$, which gives me the required result, $<T> = mg + \frac { m {\omega }^2A^2}{4l}$

 

[haruspex]

I am inclined think that they should be equal but they are not.

$\int \theta^2.dt = \int \frac{\theta^2}{\dot \theta}d\theta$

With respect to time, $<\theta ^2> = \frac { {\theta_0 }^2} 2$

Maybe, but how do you get that?

 

[Pushoam]

With respect to time,$<\theta^2> = \frac {\theta_0 ^2}2$

Maybe, but how do you get that?

$\theta (t) = \theta _0 \cos{ \omega t}$

$<\theta ^2>= {\theta_0^2 } <(\cos{ \omega t})^2>$ $= {\theta_0^2} \frac { \int_{0 }^{T }{ (\cos{\omega t}})^2 dt}{ \int_0 ^T dt}$$\int_{0 }^{T }{ (\cos{\omega t})}^2 dt =$ $\int_{0 }^{T } \frac { \cos{ (2\omega t)} +1}2 dt =$ $\frac { T} 2 +\frac { \sin{ (2\omega t )}}4|_0 ^T = \frac { T} 2 +0 = \frac { T} 2$

as ${(\cos{ \omega t})}^2 = \frac { \cos{( 2\omega t)} +1}2$

$<(\cos{ \omega t})^2> = \frac 1 2$ over one T.

$<\theta ^2>= {\theta_0 }^2 <(\cos{ \omega t})^2> = \frac { {\theta_0 }^2} 2$

I hope now this question is solved.

 

[Pushoam]

$\int \theta^2.dt = \int \frac{\theta^2}{\dot \theta}d\theta$

Thanks for it although I have not digested it yet. I will reply to you about it later.
I know that $\dot \theta = \frac {d\theta}{dt} , ~~ dt = \frac{d\theta}{\dot \theta }$.
I feel that with the help of the above eqn, I will be able to see where I am mistaking.
It makes me to see that my tendency to take average of a function independent of the variable (with respect to which average is calculated) is wrong.
But, in what way is it wrong?
I will post the answer to the above question after getting it with the help of $\int \theta^2.dt = \int \frac{\theta^2}{\dot \theta}d\theta$ to get it checked by you.

 

[haruspex]

$\theta (t) = \theta _0 \cos{ \omega t}$

$<\theta ^2>= {\theta_0^2 } <(\cos{ \omega t})^2>$ $= {\theta_0^2} \frac { \int_{0 }^{T }{ (\cos{\omega t}})^2 dt}{ \int_0 ^T dt}$$\int_{0 }^{T }{ (\cos{\omega t})}^2 dt =$ $\int_{0 }^{T } \frac { \cos{ (2\omega t)} +1}2 dt =$ $\frac { T} 2 +\frac { \sin{ (2\omega t )}}4|_0 ^T = \frac { T} 2 +0 = \frac { T} 2$

as ${(\cos{ \omega t})}^2 = \frac { \cos{( 2\omega t)} +1}2$

$<(\cos{ \omega t})^2> = \frac 1 2$ over one T.

$<\theta ^2>= {\theta_0 }^2 <(\cos{ \omega t})^2> = \frac { {\theta_0 }^2} 2$

I hope now this question is solved.

Looks good.