Question

A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to oscillate in a vertical plane. Determine the period of oscillation of the pendulum if L = 5.95 m.

Answer 1

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

$T=2\pi \sqrt{\frac{L}{g} }$

Where T is period

           L is length of rod

       g is acceleration due to gravity =     $9.8m/s^{2}$

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this $L_{O}$

$L_{O} = 3/4 * 5.95m$

        = 4.4625m

thus   $T=2\pi \sqrt{\frac{L_{O} }{g} }$

          $T=2\pi \sqrt{\frac{4.4625 }{9.8} }$

          T= 4.24sec