Question

If the length of the rod is 2.65 m, and the mass of the bob and the rod are both 1.4 kg, what is the period of this pendulum

Answer 1

T = 5.66 s

Step-by-step explanation:

The system formed by the bar plus ball forms a physical pendulum

w =
`√(mgd/I)`

the moment of inertia of a rod held at one end is

I =
`(1)/(3)` m L²

we substitute

w =
`\sqrt{(d \ d)/( 3 L^2 ) }`

in this case the turning distance and the length of the rod are equal

d = L

w =
`\sqrt{(g)/(3L) }`

angular velocity and period are related

w = 2π / T

2π / T =
`\sqrt{(g)/(3L) }`

T = 2π
`√(3L/g)`

let's calculate

T = 2π
`√(3 \ 2.65 / 9.8)`

T = 5.66 s