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I have a simple pendulum of length $l$,

(The string is massless)

What is its time period?

Simple, you say, it's $2 \pi \sqrt{(l/g)}$.

Are you sure about that?

Well, for a small angle that's true

Is it?

The limit as amplitude approaches $0$?

Sure that's better, but really is it? What if the ball of the pendulum wasn't a point mass?

Ah, got it, so let's go back to the derivation, the instantaneos axis of rotation of the Bob is fixed, and it's perpendicular to the plane of motion, through the hinge point The torque about that point is mg sin a (a being the angle of the massless string with the vertical at that time) and moment of inertial is $\frac{2}{5} m r^2 + ml^2$

Hmm?

$\frac{2}{5} m r^2 + m (l+r)^2$

That's better Now we can go on working through the details, but it's better to directly invoke the result $T = 2 \pi \sqrt{I/mgd}$ (I is moment of inetrial about hinge axis, d is distance of axis from COM) This is all fine, but there is a second derivation, often presented in a high school of the simple pendulum time period.

The one with tension?

Precisely, and my question is that when we do that, we don't get this corrected formula but rather just $2 \pi \sqrt {(l+r) /g}$

Really?

Yes, let me demonstrate, suppose at a time t the tension is the string is T and angle is a, clearly $T = mg\cos(a)$ And horizontal force on COM is directed towards the mean position and has value $T \sin(a) = mg \tan(a)$

$\tan(a)$, as $a\rightarrow 0$, is $\sin(a)$ which is $x/(l+r)$ where $x$ is $x$-displacement of COM (center of mass) from the mean. And we see this reduces to the standard differential equation of an SHM and we get the above mentioned result.

So, why do these $2$ methods give these different results?

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    $\begingroup$ You really could post your question less as a dialogue and more as statements of arguments or questions. And please use MathJax to write equations etc. $\endgroup$
    – Stuti
    Feb 9 at 13:37
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    $\begingroup$ I fixed what I could, unfortunately I am not experienced in mathJax so there is still some bad formatting $\endgroup$ Feb 9 at 13:54
  • $\begingroup$ Actually, $T=mgcos\alpha$ is wrong. It should be $Tcos\alpha=mg$, and similarly for the second equation. $\endgroup$
    – Stuti
    Feb 9 at 14:01
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    $\begingroup$ $~m\,(l+r)^2~$ is not correct . the rod length is l, and at the end of the rod you fixed a sphere (ball) of radius r. $\endgroup$
    – Eli
    Feb 9 at 14:01
  • $\begingroup$ @ThatApollo777 I just submitted some new Latex changes. $\endgroup$
    – MathZilla
    Feb 9 at 14:02

4 Answers 4

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you obtain the period $~T~$ from this equation

$$ I_A\ddot\phi+M\,g\,l\,\phi=0 $$

from here

$$\omega=\sqrt{\frac{M\,g\,l}{I_A}} \quad,T=\frac{2\pi}{\omega}$$

where $~I_A~$ is the inertia about the pendulum rotation point A

Case I

rod length l, rod mass zero, at the end of the rod is mass point (mass m).

$$I_A=m\,l^2\quad,M=m\\ \omega^2=\frac{m\,g\,l}{m\,l^2}=\frac gl$$

Case II

rod length l, rod mass zero, ball fixed, at the end of the rod (mass m_b ,radius r).

$$I_A=\frac 25\,m_b\,r^2+m_b\,l^2\quad,M=m_b\\ \omega_{II}^2=\frac{m_b\,g\,l}{I_A}=\frac{m_b\,l^2}{I_A}\omega^2$$

Case III

rod length l, rod mass $~m_r~$, ball fixed, at the end of the rod (mass $~m_b~$ ,radius r).

$$I_A=\frac 25\,m_b\,r^2+m_b\,l^2+I_r+ m_r\left(\frac l2\right)^2\\ $$ with the rod inertia at the center of mass $$I_r=\frac{1}{3}\,m_r\,l^2\\ I_A=\frac 25\,m_b\,r^2+m_b\,l^2+\frac {7}{12}\,m_r\,l^2\\ M=m_b+\frac{m_r}{2}$$ $$ \omega_{III}^2=\frac{M\,g\,l}{I_A}=\frac{M\,l^2}{I_A}\,\omega^2$$

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How can you write just the net translational equations of motion and expect to see the effect of rotation ?

I would say your analysis of simple pendulum given at the end is correct, but treats the bob as a point mass located at distance $l+r$ from pt of suspension.

Hope u get it

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  • $\begingroup$ I do not expect to see the effects of rotation, but shouldn't the translation of COM have same time period as the rotation of bob $\endgroup$ Feb 9 at 14:22
  • $\begingroup$ I don't agree with the first point. In ur first calculation u did consider rotational effects, but u didn't consider rotational effects in the lasr calculation. But I see your point, I agree both will be equal but not equal to $2\pi \sqrt{\frac{l+r}{g}}$ $\endgroup$ Feb 9 at 14:30
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I am not aware of what the stack exchange procedure is if I figure out the answer on my own(credit where its due, user Stuti Gupta helped me realise my error), so I will just post it here

  1. Sin a cos a =/= tan a (facepalm)

But that doesn't matter much as cos a goes to 1

  1. The tension isn't mg cos a!

I forgot to consider centripetal force which will be mv^2/(l+r)

As then this whole approach falls apart as the final differential equation has a v term in it and hence isn't the standard shm equation

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  • $\begingroup$ You can edit that into the question itself. $\endgroup$
    – Stuti
    Feb 9 at 15:02
  • $\begingroup$ The tension in the string is the centripetal force, equal to $mg / \cos \alpha$ The real thing you missed in the question, is that the "inertial" equation includes the effects of the rotation of the ball itself, while the "tension" equation doesn't. $\endgroup$
    – fishinear
    Feb 9 at 15:12
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In the "moment of inertia" equation, you assume the string is attached to the top of the ball; it includes the effect of the rotation of the ball of the pendulum itself.

In the "tension" equation, on the other hand, you assume the string is attached to the centre of mass. That is, you assume the ball is itself not rotating.

If you make the same assumption for the "moment of inertia", then the $ {2\over 5} mr^2 $ falls away, and both reduce to the same equation:

$$ 2 \pi \sqrt{d/g} $$ with $d = l+r$, the distance to the centre of mass.

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