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.)

Respuesta :

Answer:

0.54938 seconds

Explanation:

M = Mass of ornament = [tex]2\times 10^{-2}\ kg[/tex]

R = Radius = [tex]4.5\times 10^{-2}\ m[/tex]

g = Acceleration due to gravity = 9.81 m/s²

Relation between moment of inertia and time period is given by

[tex]T=2\pi\sqrt{\frac{I}{MgR}}[/tex]

When [tex]I=\frac{5}{3}MR^2[/tex]

[tex]T=2\pi\sqrt{\frac{\frac{5}{3}MR^2}{MgR}}\\\Rightarrow T=2\pi\sqrt{\frac{\frac{5}{3}R}{g}}\\\Rightarrow T=2\pi\sqrt{\frac{\frac{5}{3}4.5\times 10^{-2}}{9.81}}\\\Rightarrow T=0.54938\ s[/tex]

The period is 0.54938 seconds

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