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Chapter 4: Problem 179

\( \mathrm{A}\) disk, of mass \(M,\) is constrained horizontally but is free to move vertically. A jet of water strikes the disk from below. The jet leaves the nozzle at initial speed \(V_{0}\), Obtain a differential equation for the disk height, \(h(t),\) above the jet exit plane if the disk is released from large height, \(H .(\mathrm{You}\) will not be able to solve this ODE, as it is highly nonlinear!) Assume that when the disk reaches equilibrium, its height above the jet exit plane is \(h_{0}\) (a) Sketch \(h(t)\) for the disk released at \(t=0\) from \(H>h_{0}\) (b) Explain why the sketch is as you show it.

Short Answer

Expert verified
The nonlinear ODE that represents the height of the disk \( h(t) \) under the forces present is \( h'' = -g + \frac{\rho A}{M}2gh \). The sketch of \( h(t) \) starts at a height \( h = H \) at time \( t=0 \), decreases until it reaches \( h_0 \), and then stays constant as the disk achieves equilibrium with the jet. This motion occurs because the water jet forces and gravitational force eventually balance out at \( h_0 \).

Step by step solution

01

Understanding the forces involved

Begin by recognizing the forces acting on the disk. These consist of the gravitational force and the force exerted by the water jet from below. Gravity pulls the disk downwards while the jet pushes it upwards. The force exerted by the water jet on the disk is given by the rate of momentum delivered by the water, which is \( \rho A V_{0}^2 \) where \( \rho \) is the water density and \( A \) is the area of the disk. The gravitational force acting on the disk is \( Mg \), where \( M \) is the mass of the disk and \( g \) is gravity's acceleration.
02

Setting up the force balance equation

Next, establish an equation for the forces when in equilibrium. This occurs when the total force acting on the disk is zero. Equilibrium is achieved when the force of the jet equals the gravitational force or when the height is \( h_0 \). Hence, \( \rho A V_{0}^2 = Mg \) which helps to determine \( V_0 \).
03

Differential equation for the disk height

To derive a differential equation for the disk height \( h(t) \), consider the law of conservation of energy. The mechanical energy of the system (the kinetic energy of the disk and the potential energy of the water jet) should be conserved. The kinetic energy of the water is equal to the work done by the water on the disk and can be denoted as \( \frac{1}{2}MV^2 = \frac{1}{2}Mgh \). From this, we can get \( V = \sqrt{2gh} \). The force balance on the disk can now be written as: \( M \frac{dh}{dt^2} = -Mg + \rho AV^2 \). Substitute \( V = \sqrt{2gh} \) in the above equation and rearrange taking into account that \( \frac{dh}{dt} = h' \) and \( \frac{d^2h}{dt^2} = h'' \), to get the nonlinear ODE \( h'' = -g + \frac{\rho A}{M}2gh \). This is the equation of motion for the disk.
04

Sketching the disk height function h(t)

When sketching the function \( h(t) \), be mindful that at the initial time \( t=0 \), \( h = H \) which is greater than \( h_0 \) (the equilibrium height). As time increases, the height decreases till it reaches \( h_0 \) which is where it achieves equilibrium. Here, \( h' = 0 \) and remains constant as there are no more vertical displacements.
05

Explanation of the sketch

The reason h(t) decreases initially is because initially the gravitational force, which pulls the disk downwards, is greater than the force of the jet stream due to the large distance from the jet. As the disk gets closer, the force from the jet increases. Eventually, the disk reaches a point where the forces balance out, and this is represented as \( h_0 \) - the point where the height of the disk stays constant, maintaining dynamic equilibrium with the jet

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Most popular questions from this chapter

Incompressible liquid of negligible viscosity is pumped, at total volume flow rate \(Q,\) through two small holes into the narrow gap between closely spaced parallel disks as shown. The liquid flowing away from the holes has only radial motion. Assume uniform flow across any vertical section and discharge to atmospheric pressure at \(r=R\) Obtain an expression for the pressure variation and plot as a function of radius. Hint: Apply conservation of mass and the momentum equation to a differential control volume of thickness \(d r\) located at radius \(r\)

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