To find the length of the pendulum, would you consider the length from the hinge to the beginning of the sphere or end of the sphere or the centre?

  • 1
    $\begingroup$ From the point of support to the centre of the sphere. $\endgroup$
    – Farcher
    Jun 1, 2022 at 13:15

1 Answer 1


The usual pendulum formulas assume a point mass on the end of a massless rod. If the mass has some non-negligible size or if the rod has non-negligible mass then you have to find the equivalent length.

The equivalent length is not simply the distance from the hinge to the center of mass. Instead, the equivalent length is $$L=\frac{I}{mr}$$ where $I$ is the moment of inertia of the pendulum about the pivot, $m$ is the mass of the pendulum, and $r$ is the distance from the pivot to the center of mass of the pendulum.

  • $\begingroup$ Note that this correction can be pretty small in some circumstances. For the OP's situation (a sphere of radius $a$ whose center is a distance $r$ from the pivot), the fractional difference can be shown to be $\frac25 a^2/r^2$. For a 1-cm ball bearing at the end of a 1-m string, this will be a difference of one part in $10^4$ — which may well be negligible for a home experiment. $\endgroup$ Jun 1, 2022 at 14:14
  • $\begingroup$ But for a sphere of radius $a$ with its centre a distance $r$ from the pivot, the moment of inertia about the pivot is$$I=mr^2+m\tfrac 25 a^2$$ If $r$=1.00 m and $a$=1.00 cm, $I= 1.00004\ mr^2$, so for most purposes we needn't worry about the $m\tfrac 25 a^2$ term. Note to Michael Seifert: How amusing! I wrote this comment before I'd seen yours! $\endgroup$ Jun 1, 2022 at 15:37
  • $\begingroup$ It can also be large in some realistic situations. For example, for a rod hinged on one end the effective length is 2/3 of the actual length $\endgroup$
    – Dale
    Jun 1, 2022 at 16:35

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