What is its length?

Question

A rod suspended on its end and acting as a physical pendulum swings with a period of 1.03 s. What is its length? (g = 9.80 m/s2)

Answer 1

A physical pendulum is different than a simple pendulum. For a physical pendulum, the rigid body swings unders its own weight about a fixed axis of rotation. Consideration must be given to the geomtery, and thereby the moment of inertia, of the swinging object.

A physical pendulum is governed by the following equation of motion

d2q/dt2 + (m*g*L/I)*sin(q) = 0

For small oscillations, sin(q) may be approximated by q leaving

d2q/dt2 + (m*g*L/I)*q = 0

where m is the mass, g is the gravitational acceleration, L is the distance to the center of mass, and I is the moment of inertia. The solution to this equation yields a period of oscillation of

T = 2*pi*sqrt(I/(m*g*L))

If k^2 = I/m, where k is the radius of gyration, the above equation can be rewritten as

T = 2*pi*sqrt(k^2/(g*L))

For a thin rod of length D, the value of k^2 for rotation about an axis at on end of the rod is

k^2 = D^2/3

Plugging into the equation for the period, and use L = D/2 for the distance to the center of mass yields

T = 2*pi*sqrt(2*D/(3*g))

for the period. Note that this period is the same for a simple pendulum that has a length of 2*D/3.

Solving this equation for D gives

D = 3*g*T^2/(8*pi^2)

For a period of 1.03 sec, the length of the rod is

D = 0.395 m

Answer 2

The period of a pendulum is related to length as:

Period=2*Pi*sqrt(L/g)

For a period of 1.03:

L=(period/(2*Pi))^2*g

In this case the rod is the weight, so this computes L/2
The length of the rod is
.527m

j