Q87P
Question
A slender, uniform, metal rod with mass M is pivoted without friction about an axis through its midpoint and perpendicular to the rod. A horizontal spring with force constant k is attached to the lower end of the rod, with the other end of the spring attached to a rigid support. If the rod is displaced by a small angle \(\theta \) from the vertical (Fig. P14.87) and released, show that it moves in angular SHM and calculate the period. (Hint: Assume that the angle \(\theta \) is small enough for the approximations \(\sin \;\theta \approx \theta \;{\rm{and}}\;\cos \;\theta \approx 1\) to be valid. The motion is simple harmonic if \({d^2}\theta /d{t^2} = - {\omega ^2}\theta \) and the period is then \(T = 2\pi /\omega \).)
Step-by-Step Solution
Verified\(T = 2\pi \sqrt {\frac{M}{{3k}}} \)
\(\begin{array}{l}{\rm{force}}\;{\rm{constant}} = k\\{\rm{mass}} = M\\{\rm{Angle}} = \theta \end{array}\)
The time required to complete one cycle is known as the period.
The motion is simple harmonic if the equation of motion for the angular oscillations is of the form \(\frac{{{d^2}\theta }}{{d{t^2}}} = - \frac{K}{I}\theta \) , and in this case the period is,\(T = 2\pi \sqrt {\frac{I}{K}} \) .
For a slender rod pivoted about tis center \(I = \frac{1}{{12}}M{L^2}\)
The torque on the rod about the pivot is \(\tau = \left( {k\frac{L}{2}\theta } \right)\frac{L}{2} = I\alpha = I\frac{{{d^2}\theta }}{{d{t^2}}}\) gives
Is proportional to \(\theta \) and the motion is angular SHM. \(\frac{K}{I} = \frac{{3k}}{M} = {\omega ^2}\)
\(T = 2\pi \sqrt {\frac{M}{{3k}}} \)
The expression we used for the torque \(\tau = - \left( {k\frac{L}{2}\theta } \right)\frac{L}{2}\) is valid only when \(\theta \)is small enough for \(\sin \;\theta \approx \theta \;{\rm{and}}\;\cos \;\theta \approx 1\)
Hence the period is \(T = 2\pi \sqrt {\frac{M}{{3k}}} \)