Question

A holiday ornament in the shape of a hollow sphere with mass 2.0×10^−2 kg and radius 4.5×10^−2 m is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum.

Calculate its period. (You can ignore friction at the pivot. The moment of inertia of the sphere about the pivot at the tree limb is 5MR^{2}/3.)

Answer 1

0.54938 seconds

Step-by-step explanation:

M = Mass of ornament =
`2* 10^(-2)\ kg`

R = Radius =
`4.5* 10^(-2)\ m`

g = Acceleration due to gravity = 9.81 m/s²

Relation between moment of inertia and time period is given by

`T=2\pi\sqrt{(I)/(MgR)}`

When
`I=(5)/(3)MR^2`

`T=2\pi\sqrt{((5)/(3)MR^2)/(MgR)}\\\Rightarrow T=2\pi\sqrt{((5)/(3)R)/(g)}\\\Rightarrow T=2\pi\sqrt{((5)/(3)4.5* 10^(-2))/(9.81)}\\\Rightarrow T=0.54938\ s`

The period is 0.54938 seconds