The Basset force contribution to total force was determined by the density ratio to a considerable extent, where is the density of the particle. It was found that the Basset force is always important in the calculations for scenarios. For some heavy particles in air such as , the Basset force plays an insignificant role in the total force. The Basset force contribution for depends on the type of particles and on the frequency of the fluid velocity fluctuations [ 24 ].
Particular attention has been devoted to the importance of the Basset force compared to the other hydrodynamic forces [ 25 ]. Lawrence and Weinbaum [ 26 ] investigated the unsteady force on a sphere at low Reynolds number and pointed out that the Basset force for an arbitrary velocity contained a new memory integral whose kernel differed from the classical behaviour derived by Basset [ 15 ]. Sobral et al. However, its computation was far more expensive than other items in the BBO equation because of the dramatic demands of computational time and memory.
In contrast to the numerical solutions, its analytical solution has advantages including simplicity and reliability. However, the analytic solutions of a spherical particle falling in still water were still not found. Attempts to resolve this dilemma were made for the theoretical solution of the Basset force and BBO equation with a simplified method. Guo [ 13 ] assumed to be a constant to separate it from the definite integral and as in the Basset force item.
The terminal fall velocity, , can be deduced with 5 for :. Using 5 with or , and can be expressed as. The fall distance was expressed as. Assuming an initial fall velocity of zero, Guo [ 13 ] derived the following relationship between fall distance and fall velocity: where is the inverse hyperbolic function.
For , the fall distance in 8 approaches an infinite value. In order to simplify the study, the finite values of fall time and distance were expressed with the following assumptions. Applying 6 and to 5 gives the corresponding time as follows:.
Substituting 10 into 8 gives. Substituting 12 into 8 gives. Note that the scenarios of or are not addressed in our study because they are very close to the terminal fall velocity. In order to validate 9 , two data sets of a spherical particle acceleration falling through still fluid from Allen [ 31 ] and Moorman [ 32 ] were used. The water temperature was The fall velocities and distances of the experiments were plotted, and their comparisons with 9 were shown in Figure 1.
As can be seen from Figure 1 , the results from 9 are slightly overestimated according to the experiment with Moorman's scenarios [ 32 ]. In order to further validate 8 , a series of experiments was conducted in the Hydraulic Lab at Ocean University of China. A rectangular glass tank was used with a height of 2. It was filled with tap water to a water depth of 1. The water temperature was measured with a thermometer to be The spherical particles were made of plastic or glass.
The diameters of the spherical particles were measured by Vernier calipers, and the mass of the spherical particles was measured by a balance scale.
As a result, the densities of the spherical particles were calculated. The scale paper was pasted on the outside of the tank to obtain its distance. The spherical particle was released on the surface of the water.
A video camera was used to record the falling process at a speed of 30 frames per second. Video player software was used to read the video frame by frame, and the corresponding time and distance of each frame were obtained. Table 1 presents the detailed scenarios of the fall experiments. Figure 2 illustrates the relationship between fall distance and time for scenarios 1 to 4 with.
Note that the fall in a laminar zone was not measured because of the limitation of camera performance and its very small value. It was observed that results from 8 agree well with experimental data, confirming that the determination of , , and is reliable to compute the falling motion. Previous research shows that the initial velocity plays an important role in the fall time and distance [ 12 , 13 , 31 , 32 ], and it deserves detailed comments. In terms of and , the total fall time and the total acceleration fall distance were discussed in the following nine scenarios.
For scenarios, the fall velocity increases from to. Therefore, For scenarios, the fall velocity decreases from to. The fall velocity decreases from to and , and the corresponding time increases from to and.
Therefore, where can be obtained with , , , , and using The fall velocity decreases from to , , and , and the corresponding time increases from to , , and. The fall velocity increases from to and , and the corresponding time increases from to and. The fall velocity increases from to , , and , and the corresponding time increases from to , , and. Table 2 shows the total acceleration fall distance comparison between 21b and experimental data with.
In Table 2 , is the value of 21b , is the value from the experiments, and the relative error was defined as. It was found that the all of the relative errors between them were smaller than 5. Figure 3 shows the relationship between , , and for and. Note that. Figure 3 a illustrates that decreases with increasing , and its reduction rate is higher for small than for large.
Begin typing your search term above and press enter to search. Press ESC to cancel. Skip to content Home Physics Does the initial velocity affect acceleration? Ben Davis December 22, Does the initial velocity affect acceleration? Does a heavier object hit the ground first? Do heavier objects reach terminal velocity faster? What is the acceleration of a falling object that has reached its terminal velocity? Can a human survive falling at terminal velocity?
How long until an object reaches terminal velocity? How fast is terminal velocity for a human? What is the difference between critical velocity and terminal velocity? What animals can survive terminal velocity? How high can ants fall without dying? How long does it take for a squirrel to starve to death? What animal has the fastest terminal velocity?
What animals can die from falling? What animal can jump 9 feet? Why do insects not die from falling? Do insects fart? Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:. Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.
Step 4. Recombine the two motions to find the total displacement s and velocity v. Figure 2. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. During a fireworks display, a shell is shot into the air with an initial speed of The fuse is timed to ignite the shell just as it reaches its highest point above the ground. Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used.
We can then define x 0 and y 0 to be zero and solve for the desired quantities. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y :. Figure 3. The trajectory of a fireworks shell.
The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of m and m away horizontally.
Because y 0 and v y are both zero, the equation simplifies to. Now we must find v 0 y , the component of the initial velocity in the y -direction. Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding.
In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height. As in many physics problems, there is more than one way to solve for the time to the highest point. Because y 0 is zero, this equation reduces to simply. Note that the final vertical velocity, v y , at the highest point is zero. This time is also reasonable for large fireworks.
When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators.
Once the shell explodes, air resistance has a major effect, and many fragments will land directly below. In solving part a of the preceding example, the expression we found for y is valid for any projectile motion where air resistance is negligible. This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.
Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of The rock strikes the side of the volcano at an altitude Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone.
We will solve for t first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. While the rock is in the air, it rises and then falls to a final position We can find the time for this by using. Substituting known values yields. Its solutions are given by the quadratic formula:.
It is left as an exercise for the reader to verify these solutions. The negative value of time implies an event before the start of motion, and so we discard it. The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of Of course, v x is constant so we can solve for it at any horizontal location.
In this case, we chose the starting point since we know both the initial velocity and initial angle. To find the magnitude of the final velocity v we combine its perpendicular components, using the following equation:. The negative angle means that the velocity is This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is See Figure 4.
Figure 5. Trajectories of projectiles on level ground. How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 , the greater the range, as shown in Figure 5 a.
This is true only for conditions neglecting air resistance. The range also depends on the value of the acceleration of gravity g. The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R of a projectile on level ground for which air resistance is negligible is given by. The proof of this equation is left as an end-of-chapter problem hints are given , but it does fit the major features of projectile range as described.
When we speak of the range of a projectile on level ground, we assume that R is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path.
The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. See Figure 6. If the initial speed is great enough, the projectile goes into orbit. This is called escape velocity. This possibility was recognized centuries before it could be accomplished.
When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls.
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