Suppose, a ideal pendulum which has a pendulum lenghth $L$ and a bob of mass $m$, another one whose bob has same mass and same effective length. But the second one's bob is hollow and and the hollow is full of water.
Will the periodic time $T$ of the two pendulums be same?
Answer
If we can neglect friction with air, the formula for the period of a pendulum will be:
$T= 2\pi \sqrt {\frac {l}{g}}$
If $g$ is costant, as in this case, the period of the pendulum will only depend on the lenght of the string.
Because the center of mass of both bobs lays in the middle, the effective lenght (lenght of the string + distance of CM from the top of the bob) will be the same, ergo the two pendula will have the same period.
It is also interesting to see how the period of the pendulum changes when we ad different amounts of water in the bob.
- At first, with no water, the center of mass of the bob lays in the middle (distance of CM from the top of the bob= Radius). The period is T.
2.When we star adding water, the CM of the system will go below the original one (distance of CM from the top of the bob= Radius +y ).The new period of the bob will be $T_f>T$. The period will rise as long as we keep adding water below the original CM of the bob. When the water fills in half the bob, the period will reach a maximum. As we add more water, the period will start to decrease again.
- Eventually,when we fill the bob (completly) with water, the CM of the system will lay in the same positon of the beginning. At this point, the effective lenght will be the same and so will be the period.
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