# Water escaping from a pendulum bob, how is time period affected? [closed]

well the question is:

A hollow pendulum Bob filled with water has a small hole at bottom, through which water escapes at a constant rate. How is the time period(T) affected as the water flows out?

I have attempted this by assuming ideal simple pendulum... so if water escapes... that must decrease mass of the Bob...and as time period is unaffected by change in mass of the Bob ...I came to the conclusion that T won't change.

the answer provided disagrees with my result...please explain what am I doing wrong and what should be the trend of T.

options provided were : 1) T increases first , then decreases 2) T decreases first then increases 3)remains same 4 ) increases

• The pendulum period doesn't depend on the amount of mass. What does it depend on? Commented Oct 24, 2014 at 7:21
• effective length and acceleration of the Bob...correct me if I am wrong. Commented Oct 24, 2014 at 7:40
• And as the water departs, the distance it is from the axis changes. Commented Oct 24, 2014 at 8:06
• Commented Oct 24, 2014 at 14:06

The formula for the time period of a pendulum (for small angles of displacement from mean position) is

$$T = 2\pi\sqrt{\frac{L}{g}}$$

Now, $L$ here is the length from the point of suspension to the center of mass of the bob. For illustration, assuming the bob is spherical, as the water leaks out, the center of mass will shift downwards, increasing $L$ and hence increasing the time period.

As soon as all the water has leaked out, however, the center of mass will be back at the geometrical center (which was its original position when it was full of water), $L$ will decrease and so will the time period. Therefore, 1 is the answer.

Of course, the effect the leakage of water will have on the change in time period and the point at which the time period again decreases (after the initial increase) depends on the density of the material of the bob and its geometry.

The equation for period of a pendulum is;

$$T=2π\sqrt{\frac{L}{g}}$$

This equation holds for constant lengths, constant gravitation and small angles(such that the string is not horizontal and there is tension in the sting). Assuming these are true, you are not doing anything wrong concerning the experiment. Note that there is no mention of mass in the equation, therefore it won't affect the results. The length of the pendulum is defined as distance between point of suspension to the center of mass of the pendulum. For simplicity the string is assumed to be massless and gravitation invariant with the small elevations.

Since all masses experience the same acceleration, and mass is irrelevant to the equation, the water leakage shouldn't affect the period of oscillation.

However, what is actually observed is that the period initially increases, and decreases later. This is the option 1).

The length of the pendulum is defined as distance between point of suspension to the center of mass of the pendulum. Since the centre of mass is changing(as water leaks out), the original equation does not apply(as length is not constant). Through (more complex) calculations with varying lengths the results observed can be justified. 1) is right due to observational evidence and deeper theory confirming it, you are doing the experiment right.

See http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Small-angle_approximation for an explanation of the equations.

• You also have to take into account the fact that as water leaks out from the sphere it will cause an additional force on the sphere ,just as in the case of rockets and the ejected gases.If you have a hole somewhere in the sphere then water will come out tangential to the trajectory .Therefore it will always increase the restoring torque Commented Oct 24, 2014 at 15:39