Q54E - University Physics with Modern Physics

Question

We want to hang a thin hoop on a horizontal nail and have the hoop make one complete small-angle oscillation each \({\bf{2}}.{\bf{0}}\,{\bf{s}}\). What must the hoop’s radius be?

Step-by-Step Solution

Answer

The hoop’s radius is \(0.496\,{\rm{m}}\)

Step 1: Identification of given data

Time period of oscillation \(T = 2.0\,{\rm{s}}\)  

Step 2: 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.

\(T = 2\pi \sqrt {\frac{I}{{mgR}}} \)                                                                                                         …(i)

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

Step 3: Determining the hoop’s radius

The moment of inertia about the center of mass is expressed as

\({I_{cm}} = m{R^2}\) 

By using parallel axis theorem

\(\begin{aligned}{\underline{\phantom{xx}}}I = {I_{cm}} + m{R^2}\\ = m{R^2} + m{R^2}\\ = 2m{R^2}\end{aligned}\) 

Substitute \(I\) value in equation (i)

 \(\begin{aligned}{\underline{\phantom{xx}}}T = 2\pi \sqrt {\frac{{2m{R^2}}}{{mgR}}} \\ = 2\pi \sqrt {\frac{{2R}}{g}} \\R = \frac{{g{T^2}}}{{8{\pi ^2}}}\end{aligned}\)

Substitute all the values in above equation

\(\begin{aligned}{\underline{\phantom{xx}}}R = \frac{{\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right) \times {{\left( {2.0\,{\rm{s}}} \right)}^2}}}{{8{\pi ^2}}}\\ = 0.496\,{\rm{m}}\end{aligned}\)

Hence, the hoop’s radius is \(0.496\,{\rm{m}}\)