A heavy rod of length I is released from a height h above...

A heavy rod of length $I$ is released from a height $h$ above the water surface as shown in figure. During the fall, the rod is always remaining in vertical position. Assuming the water level remains unchanged due to fall of the rod, find the velocity of the rod when it reaches the bottom ol the tank. Take the specific gravity of the rod $=2$.

Key Concepts

Energy Conservation

This concept involves the transformation of gravitational potential energy into kinetic energy as the rod falls. Conventional energy conservation analysis considers that the potential energy lost by the system is converted into kinetic energy while accounting for any forces, such as buoyant forces, that do work on the rod during its descent.

Specific Gravity

Specific gravity is a dimensionless quantity that represents the ratio of the density of a substance to the density of a reference substance (in this case water). It is crucial in determining the effective weight of the rod when it is immersed in water, as it affects the magnitude of the buoyant force acting on the rod.

Buoyant Force

Buoyant force arises from the displacement of water by the rod and acts upward in opposition to gravity. According to Archimedes’ principle, this force equals the weight of the displaced water. When the rod is partially immersed, the buoyant force must be taken into account to compute the net force and hence the kinetic energy and acceleration of the rod.

Center of Mass

The concept of the center of mass is important in analyzing the motion of extended bodies like rods. Since the rod falls while remaining vertical, its motion can be effectively represented by tracking the movement of its center of mass. This simplifies the energy and force analysis over the length of the rod.

Transcript

00:01 In the given question we have to find out the velocity of this rod which is coming down vertically and will enter into this tank of liquid and its velocity when it finally reaches the bottom most the lower most part of the water being completely emerged.

00:18 So in this case we will see some important values.

00:21 So the specific gravity or we can also call it as relative density has been given to us as to the area is that this is the area of cross -section of the rod we have been given the density of the liquid as row so this is for the liquid or it can be for water as well if this is water and also we've been we can take sigma to differentiate we can take sigma as the density of the rod now firstly here we have to calculate the velocity when this the lower end of this rod hits the water surface.

01:02 This lower end which has been highlighted in blue hits the water surface.

01:06 For that, we'll simply use the formula root 2gh, but in this case, our h will be different.

01:12 So let's take v -dash as that velocity.

01:16 So it will become root 2g and this will become h minus l by 2.

01:21 So h minus l by 2 because the center of mass is located over here.

01:25 So this is at l by 2 distance.

01:28 So we'll subtract that extra distance.

01:31 This is the velocity when the lower most part of the rod touches the water surface.

01:37 Now assuming that this entire rod has reached up till here the surface, when further it goes into the water, let's assume it has immersed to a distance of x and still there is some part yet to be immersed.

01:52 In that case the result in force, there will be some force acting so in that case the resulting force will be equal to l a which is the length of the rod into area into sigma g plus l a here l will be x because x distance is a moist into sigma into row g it's row here because x part is in row liquid and this one is which is the entire rod's density, which is sigma.

02:30 Moving on to the next part, from this we can calculate the acceleration.

02:35 So it will be l -a -sigma g plus xa row g upon.

02:42 We won't write m because we have to write everything in terms of density area and length, so it will become l -a -sigma.

02:49 When using this relation of r -d, which is relative density, we can easily eliminate...

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