Question

The figures to the right show a rod with length, \(l\), and mass, \(M\), on a frictionless table rotated an angle \(\theta\) from the horizontal. It is fix to the table by a pin through its center of mass, and can rotate freely about this pin. On the two ends of the rod are connected identical springs with spring constant, \(k\), and the equilibrium position of the springs is when \(\theta=0\). In this problem, \(\theta \ll 1 \mathrm{rad}\), and the spring forces can be approximated as only pointed along the \(y\) -axis. The small angle approximations are \(\sin x \approx x\) and \(\cos x\) \(\approx 1\) for \(x \ll 1\)

a. Initially, a string is attached to between one in of the rod and the lower wall (see Fig. A), and the angle is at \(\theta=\theta_{0}\). What is the tension in this string?

Answer 1

4(a) The system is in equilibrium, hence we can say that the total moment of the system about the rod's center of mass is zero. Let us say that \(T\) is the tension in the string acting away from the rod (Counter clockwise moment about Center of mass). The equilibrium distance for spring is at 0 degrees, hence the spring's are under compression. Therefore both the springs will generate clockwise moment. let the small compression be \(\delta y\) and the equilibrium length is \(y_{0}\). Hence we can write the total moment on the system as, (cos component taken for the perpendicular distance)

\((k \delta y-T)\left(\frac{l}{2} \cos \theta_{0}\right)+k \delta y\left(\frac{l}{2} \cos \theta_{0}\right)=0\)

\(\Rightarrow(k \delta y-T)+k \delta y=0\)

\(\Rightarrow T=2 k \delta y\)

We know that,

\(\sin \theta_{0}=\frac{\delta y}{l / 2}\)

Substituting we get,

\(\Rightarrow T=k l \sin \theta_{0} \simeq k l \theta_{0}\)