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Chapter 1: Q58E (page 463)

A holiday ornament in the shape of a hollow sphere with mass \({\bf{M}} = {\bf{0}}.{\bf{015}}\,{\bf{kg}}\) and radius \({\bf{R}} = {\bf{0}}.{\bf{050}}{\rm{ }}{\bf{m}}\) 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 with negligible friction. Calculate its period. (Hint: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

Short Answer

Expert verified

The time period is \(0.58\,{\rm{s}}\)

Step by step solution

01

Identification of given data

\({\bf{M}} = {\bf{0}}.{\bf{015}}\,{\bf{kg}}\)

\({\bf{R}} = {\bf{0}}.{\bf{050}}\,{\bf{m}}\)

02

Significance of time period of physical pendulum  

The letter "\(T\)" stands for the period of time needed for the pendulum to complete one complete oscillation.

\(T = 2\pi \sqrt {\frac{I}{{MgR}}} \) 鈥�(i)

Where, \(M\) is mass, \(I\) is the moment of inertia, \(R\) is radius of the sphere from and \(g\) is the acceleration due to gravity

03

Determining the time period

The hollow sphere's moment of inertia about the support point is equal to the sum of the sphere's moment of inertia about its axis and its moment of inertia about the support point

\(\begin{aligned}{}I = \frac{2}{3}M{R^2} + M{R^2}\\ = \frac{5}{3}M{R^2}\end{aligned}\)

Substitute all the values in equation (i)

\(\begin{aligned}{}T = 2\pi \sqrt {\frac{{\frac{5}{3}M{R^2}}}{{MgR}}} \\ = 2\pi \sqrt {\frac{{5R}}{{3g}}} \\ = 2\pi \sqrt {\frac{{5 \times 0.05\,{\rm{m}}}}{{3 \times 9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \\ = 0.58\,{\rm{s}}\end{aligned}\)

Hence the time period is \(0.58\,{\rm{s}}\)

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