An open circular cylinder of 20 cm diameter and 100 cm long contains water upto aheight of 80 cm. It is rotated about its vertical axis. Find the speed of rotation when; (i)no water spills, and (ii) axial depth is zero.
Question
An open circular cylinder of 20 cm diameter and 100 cm long contains water upto aheight of 80 cm. It is rotated about its vertical axis. Find the speed of rotation when; (i)no water spills, and (ii) axial depth is zero.
Solution
This problem involves the concept of centrifugal force and the properties of a cylinder.
(i) When no water spills, the water level is at the same height as the cylinder. The maximum velocity (V) can be calculated using the formula for centrifugal force:
V = sqrt(g*r)
where g is the acceleration due to gravity (9.8 m/s^2) and r is the radius of the cylinder (half of the diameter, so 0.1 m).
V = sqrt(9.8 * 0.1) = 0.99 m/s
(ii) When the axial depth is zero, the water is just about to spill over the top of the cylinder. The maximum velocity in this case can be calculated using the formula for the height of the liquid column in a rotating cylinder:
h = r^2 * (V^2) / (2*g)
Rearranging for V gives:
V = sqrt((2gh) / r^2)
where h is the height of the water column (0.8 m).
V = sqrt((29.80.8) / 0.1^2) = 12.6 m/s
So, the speed of rotation when no water spills is 0.99 m/s, and when the axial depth is zero it is 12.6 m/s.
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