IIT JEE , AIEEE Forum - Physics , Mechanics

master_purav

A simple pendulum has a hollow bob with a small hole at its bottom. It is filled with water. As the pendulum swings the water drips out of it. Does the Time period of the bob change or does it remain constant??

Doesn't the com change???

vasanth

sorry didnt read the q completely

abhijeet_0201

i think the time period should change.as water drips the com position shifts towards bottom and the effective length increases. so time period increases.after all the water has drained the centre of mass returns to original position time period becomes normal again.

Prakriteesh

Abhijeet is partially correct. When the water is above the centre level, the c.o.m. is going down as the water drips out. So, the time period increases and attains its maximum value when the water level reaches the centre. But as more water drips out and the water level falls below the centre the c.o.m. starts rising up. So, the time period starts decreasing and finally attains its original value when all water is drained out (since the c.o.m. also returns to the initial position)

mpprad

er..
well the time period should not change because when you are draning the water the mass keeps decreasing....but time period of the pendulum doesnt depend on the mass..soo...simple the time period remains constant!!! ;)

umang

Here's the solution -

Initially , when the bob is full of water , the centre of gravity of the bob lies in its geometrical centre . But , when the water falls out , the COG shifts slightly below the centre , due to which the effective length of simple pendulum increases , hence , the time period increases .

I hope I hav clarified everything .I am ready to clear any doubts .If u find the answer useful , then pls rate me !

Thanks !!!!!

umang

In this question , we will talk about centre of gravity , but not centre of mass

edison

Time period of the simple pendulum is denoted by T

The formula for the period, T, is

$T = 2\pi \sqrt\frac{\ell}{g}\,$

where $\ell$ is the length of the pendulum measured from the pivot point to the bob's center of gravity.

Here the hollow bob is filled with water. If at the bottom of the bob there is hole through which water is leaking. This means that center of mass of the bob is shifting and thus effective lenght l, which is the length if the pendulum from point of suspension to center of mass is also changing.

So here the time period is no longer constant but it is changing.

AmanAli

D TIME PERIOD WILL CHANGE

1"S INCRESE THEN DECRESE

BY D EQU T=2pI

l/g

AS WELL AS WATER DROP D CENTER OF SYSTEM ( BOB + WATER) MOVE 2WARD DOUNWARD AS ITS RESULT EFFECTIVE LENTH OF SYSTEM INCRESE THEN (T) OF INCRESE

AFTER SUFFECIENT TIME D CENTER OF SYSTEM WILL MOVE 2WARD UP THEN EFFECTIVE LENGHT OF SYSTEM WILL DECRESE THEN (T) OF SYSTEM DECRESE

IF U SATYSFIDE THEN VOTE 4 ME THK...........

AmanAli

D TIME PERIOD WILL CHANGE

1"S INCRESE THEN DECRESE

BY D EQU T=2pI

l/g

AS WELL AS WATER DROP D CENTER OF SYSTEM ( BOB + WATER) MOVE 2WARD DOUNWARD AS ITS RESULT EFFECTIVE LENTH OF SYSTEM INCRESE THEN (T) OF INCRESE

AFTER SUFFECIENT TIME D CENTER OF SYSTEM WILL MOVE 2WARD UP THEN EFFECTIVE LENGHT OF SYSTEM WILL DECRESE THEN (T) OF SYSTEM DECRESE

IF U SATYSFIDE THEN VOTE 4 ME THK...........

karthik_abiram

to edison sir,

in the formula, $T = 2\pi \sqrt\frac{\ell}{g}\,$ there is no mass term.but sir you have said that mass of the bob changes its effective length .so sir this means that mass of the bob is directly proportional to time period?but in derivation of time period of simple pendulum using dimensional analysis we got the power of mass term as 0.how sir?i cant understand it sir.please help and kindly clear this doubt sir

Prakriteesh

Time period does not change with mass but it changes with the position of the centre of mass. It is becoz the '

' term in the timeperiod formula is defined as d distance between the point of suspension of the pendulum and its centre of mass. So, though it makes no difference whether you take a heavier of lighter sphere if the c.o.m. remains unchanged, it will make difference if the position of the c.o.m. changes. This is what Edison sir meant. Hope u get it this time.

master_purav

Karthik

The time period is independent of mass. Here of course the masss changes but due to that the effective length of the pendulum also changes as the com changes its position. So the Time Period changes too.

master_purav

Thanks a lot guyz...

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