Question 20.16: A compound pendulum, of mass M and centre of mass G, is free...
A compound pendulum, of mass M and centre of mass G, is free to rotate about a fixed horizontal axis through a point A. AG is perpendicular to the axis of rotation and of length h. The moments of inertia of the pendulum about the fixed axis and a parallel axis through G are I_{A} and I_{G} respectively. Find expressions for
i) the period of small oscillations of the pendulum
ii) the length of a simple pendulum with the same period (called the simple equivalent pendulum)
iii) the period in terms of the radius of gyration, k, of the pendulum about the axis through G.
When such a pendulum has a period of 2 seconds, it is called a seconds pendulum.
iv) Show that, for a seconds pendulum, k cannot be greater than about \frac{1}{2}.
v) Find the value of h for which the period is a minimum.
i) When AG makes an angle θ with the downward vertical as shown, taking moments about the axis gives
\begin{matrix} Mg sin \theta \times h = I_{A} (-\ddot{\theta } ) & \boxed{\ddot{\theta } > 0 \text{ in the direction of increasing θ }} \end{matrix}where I_{A} is the moment of inertia about the fixed axis.
⇒ \ddot{θ} = – \frac{Mgh}{I_{A} } sin θ
Hence, for small θ, \ddot{θ} = – \frac{Mgh}{I_{A} }θ.
This represents simple harmonic motion with period T = 2π \sqrt{\frac{I_{A}}{Mgh}}. ①
By the parallel axes theorem, I_{A} = I_{G} + Mh^{2} . Therefore, the period for small oscillations is T = 2π \sqrt{\frac{I_{G} + Mh^{2}}{Mgh}}.
ii) The period for small oscillations of a simple pendulum of length l is 2π \frac{l}{g} so the length of an equivalent simple pendulum is
l = \frac{I_{G} + Mh^{2}}{Mh} or l = \frac{I_{A}}{Mh} (from ① ).
iii) The radius of gyration about the axis through G is k, so
I_{G} = Mk^{2} ⇒ I_{A} = Mk^{2} + Mh^{2}.
The period is then
T = 2π \sqrt{\frac{M(k^{2} + h^{2})}{Mgh}} (from ① )
⇒ T = 2π \sqrt{\frac{k^{2} + h^{2}}{gh}}
iv) For a given pendulum, the length h, and hence the position of A, can usually be calculated so that it has the correct period. A seconds pendulum has a period of 2 seconds.
In this case, T = 2 and squaring both sides of ② gives
4 = 4π² \frac{(k^{2} + h^{2})}{gh}
⇒ gh = π² (k^{2} + h^{2})
⇒ h^{2} – \left(\frac{g}{π²}\right)h + k^{2} = 0
This has real roots for h provided \left(\frac{g}{π²}\right)^{2} ≥ 4k^{2}
⇒ k ≤ \frac{1}{2}\left(\frac{g}{π²}\right).
But \frac{g}{π²} = 0.994 ≈ 1, so k cannot be greater than about \frac{1}{2}.
v) Consider the function
f(h) = \frac{(k^{2} + h^{2})}{h} = \frac{k^{2}}{h} + h.
The period is least when f(h) is a minimum. Differentiating gives
\acute{f} (h) = -k^{2}h^{2} + 1
so the minimum occurs when k² = h², that is h = k.
The period is least when AG is equal to the radius of gyration about the axis through G. (You can check that f^{"}(k) > 0 to ensure a minimum.)
