Question

Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.

Answer 1

Answer:

The period of the pendulum is  $T = 1.68 \ sec$

Explanation:

The diagram illustrating this setup is shown on the first uploaded image

From the question we are told that

     The length of the rod is $L = 80 \ cm$

       The diameter of the ring is $d = 10 \ cm$

       The distance of the hole from the one end  $D = 15cm$

From the diagram we see that point A is the center of the brass ring

 So the length from the axis of  rotation is mathematically evaluated as

          $AP = 80 + 10 -5 -15$  

          $AP = 70 \ cm = \frac{70}{100} = 0.7 \ m$

Now the period of the pendulum is mathematically represented as

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

             $T = 2 \pi \sqrt{\frac{0.7}{9.8 } }$

             $T = 1.68 \ sec$