Question

A cylindrical reservoir with a radius R is hang on a string (so it may freely rotate). The reservoir has two nozzles at its bottom part. The nozzles are oriented tangentially to the reservoir surface. They exert thus a torque giving rise to rotations about the vertical axis. Find the angular velocity of the reservoir rotation. Assume that the reservoir radius is much larger than that of the nozzles. The separation between the nozzles is 2L

R>r Figure 1: water wheel

Answer 1

velocity of the output water = squareroot 2gh where 'h: is the height of water in cylinder F = force = change in momentum volume water flowing out per second is velocity times cross section = squareroot 2gh times pi r^2 mass = volume times density of water = squareroot 2gh times pi r^2 times 1000 F = change in momentum = mass times velocity = 2gh pi r^2 times 1000 Torque = F times L = 2gh pi r^2 L times 1000 so torque due to two nozzle = 2(F times L) z = 4gh pi r^2 L times 1000 I = momement of inertia of cylinder = 1/2 MR^2 = 1/2 (pi R^2 times h times 1000) times R^2 = 1/2 pi R^4 h times 1000 Angular acceleration = alpha = z/I = 4gk pi r62 L times 1000/1/2 times pi R^4 h times 1000 \r\n alpha = 8gLr^2/R^4 alpha = 8gLr^2/R^4 Angular velocity w = w^0