Discussion
A friction pendulum bearing is to give an isolated period of 3.0 s. The radius of the sliding surface required is nearest to
Answer: A
Correct answer: 2.24 m
Rearranging \( T = 2\pi\sqrt{R/g} \) gives \[ R = \frac{gT^2}{4\pi^2} = \frac{9.81 \times 9}{39.48} = 2.24\ \mathrm{m}. \] The radius grows with the square of the period, so pushing from 2 s to 3 s requires more than double the radius - and the plan size of the bearing grows with it, which is what limits how long a period this device can practically deliver.
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