Question

Define and explain physical pendulum.

Answer 1

Answer on Question #50891, Physics, Mechanics | Kinematics | Dynamics

Define and explain physical pendulum.

Answer:

If a flat, rigid body is pivoted about any point other than its center of mass and displaced by a small angle, the body will execute SHM (Simple Harmonic Motion).

The derivation is similar to that of a simple pendulum since one can consider all the mass $M$ to be located at the body's center of mass. Then a physical pendulum looks like a simple pendulum except that its moment of inertia is found using the parallel axis theorem.

$$I _ {p p} = I _ {c m} + M h ^ {2}$$

$\mathsf{I}_{\mathsf{cm}} =$ the body's moment of inertia about its center of mass.

$\mathbf{h} =$ Distance from pivot point to the center of mass.

$\mathbf{M} =$ Mass of the body.

If the pivot joint is frictionless then the net torque acting on the planar object is given by the force of gravity perpendicular to lever arm, $\mathbf{Mg}\sin (\mathbf{q})$ , times the length of the lever arm, $\mathbf{h}$ :

$$\boldsymbol {\tau} _ {n e t} = - M g \sin (\boldsymbol {0}) h = I _ {p p} \boldsymbol {\alpha}$$

When the angle of oscillation is small then the value of $\sin (q)$ and $q$ or nearly the same provided $q$ is measured in radians. Using this approximation, the above torque equation can be solved for the angular acceleration,

$$\boldsymbol {\alpha} = - \frac {M g h}{I _ {p p}} \boldsymbol {0}$$

Since $a = d^2 q / dt^2$ , this equation is structurally similar to the differential equation for any type of SHM,

$$\frac {d ^ {2} \theta}{d t ^ {2}} = - \frac {M g h}{I _ {p p}} \theta$$

Matching terms with SHM equations,

$$\omega = \sqrt {\frac {M g h}{I _ {p p}}} = \sqrt {\frac {M g h}{I _ {c m} + M h ^ {2}}}$$ $$T = \frac {2 \pi}{\omega} = 2 \pi \sqrt {\frac {I _ {c m} + M h ^ {2}}{M g h}}$$

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