Fluid mechanics of a sinking ball

[Modern Algebr]

A cylinder is full of water, a ball is sinking vertically along the central height the cylinder (ie the centre of the ball is along the central axis of the cylinder) with a constant speed V ,if the radius of the ball is r and the radius of the cylinder is R (r<R), find the volume of water displaced by the ball per unit time

the answer is "Volume displaced per unit time = pi*r^2*V''(area of a circle*V)
whereas I think it should be 2*pi*r^2*V ("area of a semi-spherical*V)

How to imagine this correctly?

Thanks very much for any hints!

 

[Shooting Star]

In one second, the ball travels a distance of V. Think of the “envelop” of the path of the ball through water. If we exclude the initial volume of the ball, then it is a cylinder with a concave hemisphere at the starting end and a convex hemisphere at the other end. The area of cross section of the cylinder is πr². If you cut the convex hemisphere and put it in the concave hollow in the starting end, then it becomes a cylinder with length V and area πr². This is the volume displaced per second.

 

[Moetasim]

The volume of water displaced by the sinking ball per unit time can be calculated by considering the shape of the ball and its movement through the water. In this scenario, the ball is sinking vertically with a constant speed, meaning that the water displaced by the ball will form a cylinder with a height equal to the speed multiplied by the time.

To imagine this correctly, think of the ball as creating a column of water as it sinks. The cross-sectional area of this column will be a circle with a radius equal to the radius of the ball. Therefore, the volume of water displaced per unit time is equal to the area of this circle (pi*r^2) multiplied by the speed of the ball (V).

However, this calculation only takes into account the water displaced by the cylindrical portion of the ball. To account for the water displaced by the spherical portion of the ball, we must also consider the volume of a hemisphere with a radius equal to the radius of the ball (2/3*pi*r^3). Therefore, the total volume of water displaced per unit time is equal to the area of the circle (pi*r^2) multiplied by the speed of the ball (V) plus the volume of the hemisphere (2/3*pi*r^3).

In summary, the correct calculation is 2/3*pi*r^3*V + pi*r^2*V, which can also be written as 2*pi*r^2*V, as the 2/3 and pi cancel out. This accounts for both the cylindrical and spherical portions of the ball displacing water as it sinks.