Question

Two pendulums have the same dimensions (length {L}) and total mass (m). Pendulum \texttip{A}{A} is a very small ball swinging at the end of a uniform massless bar. In pendulum \texttip{B}{B}, half the mass is in the ball and half is in the uniform bar.

A)Find the period of pendulum A for small oscillations.

B)Find the period of pendulum B for small oscillations.

C)Please be sure to include the specific calculations of the moment of inertia and the distance to the center of mass for pendulum B.

D) By what fraction is the period of one pendulum smaller than the other?

Answer 1

A)

For any gravity pendulum with small amplitude

ω = √[ MgL / I ]

where M is the total mass, g is the gravitational field, L is the distance from the pivot to the center of mass, and I is the rotational inertia about the pivot.

Note that for a simple pendulum, I =ML², so

ω = √[ g / L ]

as expected.

So T_A = 2π √ [ L / g ]

C)

For the second pendulum, the center of mass of the rod is in it's own center, L/2 from the pivot. The center of mass of the pendulum is at

d = [ (m1)L + (m2)(L/2) ] / (m1 + m2) = L [ (m/2) + (m / 4) ] / (m/2 + m/2) = 3L/4

from the pivot.

Since both the ball and rod are pivoting about the same point, the rotational inertia of the combined object is just the sum of their individual moments of inertia.

I = (m1)L² + (1/3)(m2)L² = [ m/2 + m / 6 ] L² = 2mL² /3

B)

ω = √[ MgL / I ]

ω = √[ (m1 + m2) g d / I ]

ω = (mg)(3L/4) / (2mL²/3)

ω = 9g/8L

From here

ω = 2πf = 2π / T

T_B = 2π / ω

T_B = 28L/9g

D)

T_B / T_A = 8/9 = 0.943 .