Question

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

Answer 1

Answer:

1) $T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }$, 2) $T_{B} \approx 1.137\cdot T_{A}$, where $T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }$.

Explanation:

1) Pendulum A is a simple pendulum, whose period ($T_{A}$) is determined by the following formula:

$T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }$ (1)

Where:

$l$ - Length of the massless bar.

$g$ - Gravitational acceleration.

2) Pendulum B is a physical pendulum, whose period ($T_{B}$) is determined by the following formula:

$T_{B} = 2\pi \cdot \sqrt{\frac{I_{O}}{m\cdot g\cdot l} }$ (2)

Where:

$m$ - Total mass of the pendulum.

$g$ - Gravitational acceleration.

$l$ - Length of the uniform bar.

$I_{O}$ - Moment of inertia of the pendulum with respect to its suspension axis.

The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:

$I_{O} = \frac{1}{2} \cdot m\cdot l^{2}+\frac{1}{24}\cdot m\cdot l^{2} + \frac{3}{4}\cdot m\cdot l^{2}$

$I_{O} = \frac{31}{24}\cdot m\cdot l^{2}$ (3)

By applying (3) in (2) we get the following expression:

$T_{B} = 2\pi \cdot \sqrt{\frac{\frac{31}{24}\cdot m \cdot l^{2} }{m\cdot g \cdot l} }$

$T_{B} = 2\pi \cdot \sqrt{\frac{31\cdot l}{24\cdot g} }$

$T_{B} = \sqrt{\frac{31}{24} } \cdot \left(2\pi \cdot \sqrt{\frac{l}{g} }\right)$

$T_{B} \approx 1.137\cdot T_{A}$