Question

A rod of Mass M is performing angular SHM with Time period $T_0$ A flee of mass m sits on the midpt. of rod, find new $T'$

Answer 1

T_0 \sqrt{\frac{4M + 3m}{4(M+m)}}

Answer 2

To find the new time period ($T'$) of angular Simple Harmonic Motion (SHM) for the rod with the flea, we need to compare the initial and final states using the formula for the time period of a physical pendulum:

$T = 2\pi \sqrt{\frac{I}{Mgd}}$

where:

• $I$ is the moment of inertia about the pivot point.
• $M$ is the total mass of the oscillating system.
• $g$ is the acceleration due to gravity.
• $d$ is the distance of the center of mass of the system from the pivot point.

Let the length of the rod be $l$. The problem states the rod is of mass $M$ and the flea of mass $m$ sits at the midpoint of the rod. The diagram implies the rod is pivoted at one end.

1. Initial State (Rod only):

• Mass of the rod: $M_{rod} = M$

• Moment of inertia of the rod about one end: $I_0 = \frac{1}{3}Ml^2$

• Distance of the rod's center of mass from the pivot: $d_0 = \frac{l}{2}$

• Original time period ($T_0$):

  $T_0 = 2\pi \sqrt{\frac{I_0}{M_{rod}gd_0}} = 2\pi \sqrt{\frac{\frac{1}{3}Ml^2}{M g \left(\frac{l}{2}\right)}} = 2\pi \sqrt{\frac{2Ml^2}{3Mgl}} = 2\pi \sqrt{\frac{2l}{3g}}$

2. Final State (Rod + Flea):

• Mass of the rod: $M_{rod} = M$

• Mass of the flea: $m_{flea} = m$

• Position of the flea: At the midpoint of the rod, i.e., at a distance $\frac{l}{2}$ from the pivot.

• New moment of inertia ($I'$):

  The total moment of inertia is the sum of the moment of inertia of the rod and the flea.

  $I' = I_{rod} + I_{flea}$

  $I_{rod} = \frac{1}{3}Ml^2$ (as before)

  $I_{flea} = m \left(\frac{l}{2}\right)^2 = \frac{1}{4}ml^2$ (moment of inertia of a point mass at distance $r$ is $mr^2$)

  So, $I' = \frac{1}{3}Ml^2 + \frac{1}{4}ml^2 = \left(\frac{M}{3} + \frac{m}{4}\right)l^2 = \left(\frac{4M + 3m}{12}\right)l^2$

• New total mass ($M_{total}$):

  $M_{total} = M_{rod} + m_{flea} = M + m$

• New distance of the center of mass from the pivot ($d'$):

  Since both the rod's center of mass and the flea are located at the same distance $\frac{l}{2}$ from the pivot, the combined center of mass will also be at this position.

  $d' = \frac{M_{rod}d_{rod} + m_{flea}d_{flea}}{M_{rod} + m_{flea}} = \frac{M\left(\frac{l}{2}\right) + m\left(\frac{l}{2}\right)}{M + m} = \frac{(M+m)\frac{l}{2}}{M+m} = \frac{l}{2}$

• New time period ($T'$):

  $T' = 2\pi \sqrt{\frac{I'}{M_{total}gd'}} = 2\pi \sqrt{\frac{\left(\frac{4M + 3m}{12}\right)l^2}{(M+m)g\left(\frac{l}{2}\right)}}$

  $T' = 2\pi \sqrt{\frac{(4M + 3m)l^2}{12(M+m)g\frac{l}{2}}} = 2\pi \sqrt{\frac{(4M + 3m)l^2}{6(M+m)gl}} = 2\pi \sqrt{\frac{(4M + 3m)l}{6(M+m)g}}$

3. Relating $T'$ to $T_0$:

We have:

$T_0^2 = (2\pi)^2 \frac{2l}{3g}$

$T'^2 = (2\pi)^2 \frac{(4M + 3m)l}{6(M+m)g}$

Divide $T'^2$ by $T_0^2$:

$\frac{T'^2}{T_0^2} = \frac{(2\pi)^2 \frac{(4M + 3m)l}{6(M+m)g}}{(2\pi)^2 \frac{2l}{3g}}$

$\frac{T'^2}{T_0^2} = \frac{(4M + 3m)l}{6(M+m)g} \times \frac{3g}{2l}$

$\frac{T'^2}{T_0^2} = \frac{(4M + 3m)}{6(M+m)} \times \frac{3}{2}$

$\frac{T'^2}{T_0^2} = \frac{(4M + 3m)}{4(M+m)}$

Taking the square root of both sides:

$T' = T_0 \sqrt{\frac{4M + 3m}{4(M+m)}}$

The final answer is $T_0 \sqrt{\frac{4M + 3m}{4(M+m)}}$.

Explanation of the solution:

The time period of a physical pendulum depends on its moment of inertia, total mass, and the distance of its center of mass from the pivot.

1. Calculate the initial moment of inertia ($I_0$) and center of mass distance ($d_0$) for the rod alone.
2. Calculate the new moment of inertia ($I'$) by adding the moment of inertia of the flea (treated as a point mass) to that of the rod.
3. Calculate the new total mass ($M_{total}$) and the new center of mass distance ($d'$) for the rod-flea system.
4. Use the formula $T = 2\pi \sqrt{\frac{I}{Mgd}}$ to find $T_0$ and $T'$.
5. Express $T'$ in terms of $T_0$ by taking the ratio of their squares.