Question

A solid spherical pendulum bob weighing 270 g has a diameter of 13 cm is attached to a light string of length 3.4 cm.

What is the period of the pendulum?

Answer 1

0.68 s

Step-by-step explanation:

Period of a physical pendulum is:

T = 2π √[I / (mgL)]

where I is the moment of inertia about the support,

m is the mass of the pendulum,

g is acceleration due to gravity,

and L is the distance from the support to the center of mass.

Using parallel axis theorem, the moment of inertia is:

I = I₀ + md²

For a solid sphere, I₀ = ⅖ mr².

I = ⅖ mr² + md²

I = m (⅖ r² + d²)

I = (0.270 kg) (⅖ (6.5 cm)² + (6.5 cm + 3.4 cm)²)

I = 31.03 kg cm²

Plugging in:

T = 2π √[I / (mgL)]

T = 2π √[31.03 kg cm² / ((0.270 kg) (981 cm/s²) (6.5 cm + 3.4 cm))]

T = 0.683 s

Rounded to two significant figures, the period is 0.68 s.