Question

a simple pendulum consisting of a bob on a 49.4 - cm long string is released from a certain small angle with respect to the vertical. underneath the bob is a vertical disc with a ball glued to the edge rotating at a constant speed.
if the ball glued at appropriate radial distance, what should be the period of the motion of the disc, so that the vertical position of the bob and ball are always in sync?
use g = 10m/s^2
your answer:

Explanation:

Step1: Convert length unit

Convert \( l = 49.4\ cm\) to \( l=0.494\ m\).

Step2: Use pendulum period formula

The period formula for a simple pendulum is \( T = 2\pi\sqrt{\frac{l}{g}}\).
Substitute \( l = 0.494\ m\) and \( g = 10\ m/s^{2}\) into the formula:

$$
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Since the vertical position of the bob and ball are always in sync, the period of the disc's motion is equal to the period of the pendulum.

Answer:

\(1.4\ s\)