MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
IC_W13D3-2 Group Problem Kater’s Pendulum Solutions
A pendulum consists of a rod and two knife-edges separated from the center of mass l1
and l2 respectively (figure below left). The moment of inertia of the rod about the center
of mass is I 0 = mk g2 where k g is a constant called the radius of gyration. When the
pendulum is pivoted about the upper knife-edge (figure below center), the period for
small oscillations is T1 . When the pendulum is turned upside down and pivoted about the
other edge (figure below right), and the period for small oscillations is T2 . The distances
l1 and l2 are adjusted until T1 = T2 ! T . Express your answers to the following questions
in terms of l1 , l2 , and T as needed.
a) What is the radius of gyration k g ?
b) Using your results from part (a), show that the gravitational acceleration g
can be determined by measuring the length between the knife-edges l1 + l2
and the period T .
Solution: We can treat each case as a physical pendulum. The torque equation of motion
when suspended from the upper edge is
!l1mg sin " = I1
d 2"
.
dt 2
The moment on inertia about the upper knife-edge is I1 = I 0 + ml12 = m(k g2 + l12 ) . Using
the small angle approximation sin ! ! ! . The torque equation becomes
d 2"
!l1mg" = m(k + l ) 2 .
dt
2
g
2
1
Then the equation of motion for the pendulum is a simple harmonic oscillator equation of
motion
l1 g
d 2!
+
! = 0.
dt 2 (k g2 + l12 )
The period of oscillation is therefore
T1 = 2! (k g2 + l12 ) / l1 g .
When the pendulum is turned upside down and suspended from the other knife-edge and
similar argument yields for the period
T2 = 2! (k g2 + l22 ) / l2 g .
Because distances l1 and l2 have been adjusted until T1 = T2 ! T , we can solve for k g by
squaring the periods and setting them equal yielding
4! ((k g2 + l12 ) / l1 g) = 4! ((k g2 + l22 ) / l2 g)
A little rearrangement yields
k g2 + l12 = (k g2 + l22 )l1 / l2
k g2 ((l2 ! l1 ) / l2 ) = l1 (l2 ! l1 )
or
k g2 = l1l2 .
Therefore the period is
T = 2! (l1l2 + l12 ) / l1 g = 2! (l2 + l1 ) / g
Hence
g = 4! 2 (l2 + l1 ) / T 2 .
The answer only depends on the distance between the knife-edges l1 + l2 , and the
measured period, T .