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A thin hoop is supported in a vertical plane by a nail. What should the radius of the hoop be in order for it to have a period of oscillation of 1.00 s

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Answer:

0.124 m

Step-by-step explanation:

the period of a simple pendulum with a small amplitude is given as

T = 2π [√(I/mgd)]

From the above stated formula,

I = moment of inertia

m = mass of the pendulum

g = acceleration due to gravity, 9.8 m/s²

d = distance from rotation axis due to center of gravity

Also, moment of Inertia, I = 2mr², if we substitute this in the above formula, we have

T = 2π [√(2mr²/mgd)]

If we assume that our r = d, then we have

T = 2π [√(2r/g)]

If we make r the subject of the formula in the above equation, we get

r = gT² / 8π²

r = (9.8 * 1²) / 8 * π²

r = 9.8 / 78.98

r = 0.124 m

Thus, the radius of the hoop is 0.124 m

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