Q55E
Question
A \({\bf{1}}.{\bf{80}} - {\bf{kg}}\) connecting rod from a car engine is pivoted about a horizontal knife edge as shown in Fig. E14.55. The center of gravity of the rod was located by balancing and is \({\bf{0}}.{\bf{200}}\,{\bf{m}}\) from the pivot. When the rod is set into small-amplitude oscillation, it makes \(100\) complete swings in \({\bf{120}}\,{\bf{s}}\). Calculate the moment of inertia of the rod about the rotation axis through the pivot.
Step-by-Step Solution
VerifiedThe moment of inertia of the rod about the rotation axis through the pivot is \(0.129\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\)
Mass of the connecting rod \(m = 1.80\,{\rm{kg}}\)
Center of gravity of the rod was located by balancing and is \(d = {\bf{0}}.{\bf{200}}\,{\bf{m}}\) from the pivot
Time taken to complete 100 swings = \(120\,{\rm{s}}\)
The letter "\(T\) " stands for the period of time needed for the pendulum to complete one complete oscillation.
\(T = 2\pi \sqrt {\frac{I}{{mgd}}} \) …(i)
Where, \(m\) is mass, \(I\) is the moment of inertia, \(d\) is the distance of center of gravity from pivot and \(g\) is the acceleration due to gravity
Time period of rod is time taken to complete 1 swing
\(\begin{aligned}{\underline{\phantom{xx}}}T = \frac{{120\,{\rm{s}}}}{{100}}\\ = 1.2\,{\rm{s}}\end{aligned}\)
Rearranging the equation (i)
\(\begin{aligned}{\underline{\phantom{xx}}}\frac{T}{{2\pi }} = \sqrt {\frac{I}{{mgd}}} \\{\left( {\frac{T}{{2\pi }}} \right)^2} = \frac{I}{{mgd}}\\I = mgd{\left( {\frac{T}{{2\pi }}} \right)^2}\end{aligned}\)
Substitute all the values in above equation
\(\begin{aligned}{\underline{\phantom{xx}}}I = \left( {1.80\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {0.2\,{\rm{m}}} \right){\left( {\frac{{1.2{\rm{s}}}}{{2\pi }}} \right)^2}\\ = 0.129\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{aligned}\)
Hence the moment of inertia of the rod about the rotation axis through the pivot is \(0.129\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\)