My Cute Little Experiment
A metal ball is filled with water. It is tied to a string and if oscillated. If a hole is made on the bottom of the ball, how will the time period of the pendulum initially be affected?
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1 solution
The increase stops, because after all the water has emptied out, the distance of centre of mass from the suspension point doesn't change.
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Bingo. That's exactly the answer i was looking for.
I expected it from you though :P
In 2 π × g l , l represents the length from the point of suspension to the point it meets the bob for point mass bob only .But In Rigid bodies (which is the case in question) l represents the effective length which is given by
l = l c o m k 2 + l c o m where k is radius of gyration and l c o m represents distance of C O M from point of suspension .
Further M k 2 = I c o m where is M is mass of rigid body and I c o m is moment of inertia about axis passing through C O M .
@Mehul Arora You missed l c o m k 2 part.
Now answer according to this .
Challenge Student Note : What would happen if I say density of ball >> density of liquid.
The time period of a simple pendulum is calculated by 2 π × g l .
A common misconception is that l represents the length from the point of suspension to the point it meets the bob.
However, l is the distance of the point of suspension, to the Center of mass of the bob
Therefore, When water continuously drips out of the plastic ball (Bob) , It's center of mass shifts downwards. Thus increasing the apparent length of the pendulum. [See diagram]
Because the time period of the pendulum is Directly proportional to the length, thus the time period increases initially.
Follow up: Why does the time period only increase initially?