Pendulums — Question 8

Question 8

A rigid finite bob contributes its own rotational inertia

A solid sphere of mass $M$ and radius $R$ is rigidly attached to a light rod. Its center is a distance $d$ below a horizontal pivot. The sphere rotates with the rod, with $I_{\mathrm{CM}}=2MR^2/5$. Consider small swings.

1. Find its period and the fractional correction to the point-mass-at-center result.

2. For $R/d=.20$, evaluate the period ratio. Would the same inertia correction apply if the attachment allowed the sphere to translate without rotating?

Question 8 - Solution

A rigid finite bob contributes its own rotational inertia

(a) Derivation and calculation

$$I_P=Md^2+I_{\mathrm{CM}}=M\left(d^2+\frac25R^2\right).$$$$I_P\ddot\theta\simeq-Mgd\theta\Rightarrow\boxed{T=2\pi\sqrt{\frac{d^2+(2/5)R^2}{gd}}}.$$$$\frac{T}{T_{\mathrm{point}}}=\sqrt{1+\frac25(R/d)^2}\simeq1+\frac15(R/d)^2.$$

(b) Result and interpretation

$$R/d=.200:\quad\boxed{T/T_{\mathrm{point}}=\sqrt{1.016}=1.007968}.$$$$\frac{\Delta T}{T_{\mathrm{point}}}=.007968=0.7968\%.$$If the attachment permits pure translation without sphere rotation, the $I_{\mathrm{CM}}$ contribution is absent; the rigid-body formula does not apply.

Physical check

The correction tends to zero as $R/d\to0$, recovering the point-bob pendulum.