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

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

Step-by-step 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 = (70)/(100) = 0.7 \ m`

Now the period of the pendulum is mathematically represented as

`T = 2 \pi \sqrt{(AP)/(g) }`

`T = 2 \pi \sqrt{(0.7)/(9.8 ) }`

`T = 1.68 \ sec`