While calculating the time period of a simple pendulum, i.e $$T=2\pi\sqrt{\frac{L}{g}}$$ Why do we consider the effective length, $L$? It is the distance from the point of suspension to the centre of the bob. Why don't we just use only the length of rod? What importantce does the extra length (radius of the bob) give in its time period?


3 Answers 3


If the bob is small, obviously the two lengths are very similar so it's not too important which length you use. Yes, to first order, the distance to the center of mass is what is relevant, because the dynamics depend on how far away the bob's center of mass is from the center of rotation, not on how long the string is. If the bob is large, we need to be a bit more careful in deriving the oscillation period, and the effective length is a bit larger than $L$: a second order correction due to the moment of inertia of the bob.

I'll assume motion in one direction for simplicity. If the bob is uniform, the weight is equally distributed, and the system behaves as though all the force is applied at the center of mass, which is a distance $L$ away from the point of suspension. If $\theta$ is the angular displacement to the right of vertical, the counterclockwise torque about the point of suspension is $$\tau = -mgL\sin\theta.$$ By the parallel axis theorem, the moment of inertia of the bob about the point of suspension is $$I=mL^2 + I_0$$ where $I_0$ is the moment of inertia of the bob with respect to an axis passing through its center of mass. For a spherical bob of radius $R$, $I_0=\frac{2}{5}mR^2$. The short answer to your question is since $L$ (and not $L-R$) is the quantity that appears in these equations (overlooking $I_0$ for the moment), $L$ is the relevant quantity, not the string length.

The equation of motion is found by equating the torque to the time derivative of angular momentum $\ell=I\dot\theta$. $$\tau=\frac{d\ell}{dt}=\frac{d}{dt}\left(I\dot \theta\right)=I\ddot\theta $$ $$\left(mL^2+I_0\right)\ddot \theta+mgL\sin\theta=0$$ $$\ddot \theta+\frac{mgL}{mL^2+I_0}\sin\theta=0.$$ The oscillation period is $2\pi$ divided by the square root of the coefficient of $\sin \theta$: $$T=2\pi\sqrt{\frac{mL^2+I_0}{mgL}}.$$ Assuming a spherical bob, $$T=2\pi\sqrt{\frac{mL^2+\frac{2}{5}mR^2}{mgL}}=2\pi\sqrt{\frac{L_\text{eff}}{g}}$$ $$L_\text{eff}=L+\frac{2R^2}{5L}.$$ The second term will be quite small in most cases. For example, if $L = 1\text{ m}$ and $R = 0.1\text{ m}$, $L_\text{eff} = 1.004\text{ m}$.


I'll try to get directly to the core of your question.

A simple pendulum is an ideal mechanical system made of a point mass $m$ concentrated at one point, constrained to move at a fixed distance from a point in a uniform gravitational field. In these terms, there is only a relevant length: the distance between the point mass and the suspension point.

The problem of which length should be used starts when we want a real physical system acting as a simple pendulum. In such a case, we face two problems:

  1. if the constraint of a fixed distance is obtained through a rigid rod or a rope (only for angles of oscillations less than $\pi/2$), the mass of the suspension should play a role in the dynamical behavior;
  2. the pointlike approximation for the body's geometry can be poor.

Problem 1 can be minimized by using a suspension device with a mass negligible with respect to the mass of the bob.

In general, problem 2 is unavoidable in practice if the size of the bob is not negligible with respect to the suspension distance. This means that the suspended bob's dynamics must be treated using the full-fledged dynamical theory for a non-pointlike rigid body.

The simplest way of applying the theory is by modeling the physical pendulum (suspension device plus bob) as a unique rigid body rotating around a fixed axis. It is immediate to realize that the relevant information is not the mass but the mass moment of inertia of such a rigid body around the pendulum's axis. However, such moment of inertia can always be reinterpreted as if a pointlike mass were at a fixed distance $\ell$ from the axis. This is the key step where the effective length appears.

In formulas, if $I$ is the mass moment of inertia with respect the suspension point, and $M$ the total mass of the pendulum, the effective distance $\ell$ (radius of gyration) of a simple pendulum corresponding to the real one is given by $$ M \ell^2 = I$$ Notice that even for a spherical and homogeneous bob, the exact effective length is neither the length of the suspension device ($L$) plus the radius of the sphere, nor simply $L$, although, for a fixed radius, it tends to $L+R$ when $L$ increases.


This comes from how we derive the formula for a pendulum.

Imagine a pendulum with lenght $\ell$ hanging from a pivot point. You bring the Pendulum into position $\varphi$, so that it doesn't hang straight to the ground anymore.

Now we can take a look at the forces acting upon our pendulum, the only force we will consider is the gravitational force $F_g$. We can break this force up into two Forces, one that is parallel to our string and one that is perpendicular to it. this perpendicular force is the only relevant force we will consider. We can calculate this force with our given angle $\varphi$, that would look like this $$ F = -F_g\sin{\varphi} $$ Now, we said that this is the total Force in the system, so $F$ is our system force, which we have a formula for: $F=ma$, or $F=m\ddot{x}$. Now we don't actually have an $\ddot{x}$, but only an angle and a length, but we can construct it with the following: $\ddot{x}=\ell\ddot{\varphi}$. We will now set $F = F_g\sin{\varphi}$ and see what happens. $$ m\ell\ddot{\varphi}=-mg\sin{\varphi} $$ we use small angle approximation here, usually, so $\sin{\varphi}=\varphi$ $$ m\ell\ddot{\varphi}=-mg\varphi\\ m\ell\ddot{\varphi}+mg\varphi=0\\ \ell\ddot{\varphi}+g\varphi=0\\ \ddot{\varphi}+\frac{g}{\ell}\varphi=0 $$ this is a harmonic oscillator. We know $\omega_0=\sqrt{\frac{g}{\ell}}$ and with that we also know $T=\frac{2\pi}{\omega_0}=2\pi\sqrt{\frac{\ell}{g}}$

I hope this answers your question. Generally if you're unsure about what forces act on your system in what way you should draw a picture of it and see what variables you're given and what you can derive from those.

  • $\begingroup$ This nicely derives the period of a pendulum in which the bob is a point particle. But I don't see how this gets to the OP's question. $\endgroup$
    – garyp
    Commented May 1, 2023 at 11:52
  • $\begingroup$ I thought that that was what OP meant, as the derivation for a non-point particle pendulum looks the same just involving moment of inertia. $\endgroup$
    – Florian
    Commented May 1, 2023 at 12:36

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