Question

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?

Answer 1

Answer:

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