Time period of a pendulum with $3$ bobs

Question

Here $L$ is the length of th whole massless rod and $3$ balls have been attached to the rod having lengths of $\frac{L}{3}$. We have to find the time period of the pendulum as a whole.

Here is my approach. Since we are used to calculating pendulum related problems for one bob,i tried to reduce the system to an equivalent one ball system.

Let us assume mass of each ball is $m$. Now when the pendulum is displaced with a slight angular displacement,the restoring force acting on each of the ball is $mg\sin \theta=mg\theta$ when $\theta$ is small. Since balls experience $mg\theta$ force in the same direction, in the equivalent one ball system,the net force on that corresponding ball will be $3mg\theta$. Now from hooks law, $F=kx$ where $k$ is the SHM constant and $x$ is displacement. Here the equivalent one ball will travserse the same angle as these $3$ balls, so $\theta_{equivalent ball}$ will be equal to $\theta$. Therefore $3mg\theta=kL\theta$ so $k=\frac{3mg}{L}$. And we know $\omega^2=\frac{k}{m}=\frac{3g}{L}$. And so $T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{L}{3g}}$.

Is this correct?If wrong,please correct my thinking process and kindly post the actual solution.

Answer 1

Rods, unlike ropes, have transverse rigidity. For this reason, the transverse force acting on each ball is not equal to $mg\theta$. Therefore, this problem is difficult to solve based on the idea of an effective ball. To solve this problem, it is better to use the conservation of energy or equation for the angular momentum. Let's consider the later method.

For small angle $\theta$, torque is $$ M = -mg\theta \left(\frac{L}3 + \frac{2L}3 + L \right). $$ Angular momentum is $$ l = I\omega = m\left(\frac{L^2}9 + \frac{4L^2}9 + L^2\right)\dot{\theta} $$ Therefore the equation of motion in this problem has the following form: $$ \dot{l} = M \quad \rightarrow \quad mL^2\frac{14}9 \ddot{\theta} = - mg2L\theta $$ From the last equation, we obtain $$ \omega^2 = \frac{9g}{7L} $$