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 mas

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.

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Euphemia 4 years 2021-07-13T04:17:59+00:00 1 Answers 248 views 0

Answers ( )

  1. Ladonna
    0
    2021-07-13T04:19:25+00:00
    Reply

    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}

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