A thin, circular hoop with a radius of 0.22 m is hanging from its rim on a nail. When pulled to the side and released, the hoop swings back and forth as a physical pendulum. The moment of inertia of a hoop for a rotational axis passing through its edge is I=2MR2. What is the period of oscillation of the hoop?

Question

Johan

Physics

6 August, 10:00

A thin, circular hoop with a radius of 0.22 m is hanging from its rim on a nail. When pulled to the side and released, the hoop swings back and forth as a physical pendulum. The moment of inertia of a hoop for a rotational axis passing through its edge is I=2MR2. What is the period of oscillation of the hoop?

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Answers (1)

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Tess Bray · Answer 1

Period of oscillation = 1.33 seconds

Explanation:

The period of oscillation is given by:

T = 2π√[I / (MgL) ]

for I = 2MR² and L = R,

Given: L = 0.22m = R

T = 2π√[2R/g]

T = 2 * 3.142 Sqrt[ (2 * 0.22) / 9.8]

T = 6.284 Sqrt (0.44/9.8)

T = 6.284 Sqrt (0.0449)

T = 6.284 * 0.2119

T = 1.33 sec