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A hollow sphere is filled with water through a small hole in it. It is hung by a long thread and, as the water flows out of the hole at the bottom, one finds that the period of oscillation first increases and then decreases. Explain.

A hollow sphere is filled with water through a small hole in it. It is hung by a long thread and, as the water flows out of the hole at the bottom, one finds that the period of oscillation first increases and then decreases. Explain.

Grade:upto college level

3 Answers

Jitender Pal
askIITians Faculty 365 Points
6 years ago
Hollow sphere acts as a physical pendulum as the sphere is filled with water. The time period T for a physical pendulum is defined as,
T = 2π√I/mgd
Here, I is the rotational inertia of the sphere, m is the mass of the sphere which is filled with water, g is the acceleration due to gravity and d is the distance from the pivot to the center of mass.
In the above equation T = 2π√I/mgd,
I/mgd = mk2/mgd (Since, I = mk2)
= k2/gd
So, k2/d is the deciding factor here.
As the water flow out from the distance of center of mass from axis, that increases up to a certain limit and radius of gyration increases in square. So there is net increase in period. But after certain limit d decreases, because, center of mass shifted toward heavy portion of sphere, and rate in decrease in d is increases. But the increment in radius of gyration remains constant. Thus after a certain limit k2/d will decrease. Therefore the period of oscillation first increases and then decreases
pa1
357 Points
5 years ago
Hollow sphere acts as a physical pendulum as the sphere is filled with water. The time period T for a physical pendulum is defined as, T = 2π√I/mgd Here, I is the rotational inertia...
ankit singh
askIITians Faculty 614 Points
11 months ago
/d will decrease. Therefore the period of oscillation first increases and then decreases2/d is the deciding factor here.As the water flow out from the distance of center of mass from axis, that increases up to a certain limit and radius of gyration increases in square. So there is net increase in period. But after certain limit d decreases, because, center of mass shifted toward heavy portion of sphere, and rate in decrease in d is increases. But the increment in radius of gyration remains constant. Thus after a certain limit k2/gdSo, k2) = k2/mgd (Since, I = mk2 I is the rotational inertia of the sphere, m is the mass of the sphere which is filled with water, g is the acceleration due to gravity and d is the distance from the pivot to the center of mass.In the above equation T = 2π√I/mgd,I/mgd = mk

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