Question

Imagine a steel sphere of diameter d is dropped in air with density pf -1.29 kg/m' and
kinematic viscosity coefficient r -1.19x10e-5 m/s. The density of the steel sphere is p - 8000
kg/m and the gravitational acceleration is g -9.81 m/s'.
Employ the fourth order Runge-Kutta method to determine the motion and speed of the steel
sphere. Run the code for 10 seconds. Plot the time history of the displacement and the velocity
for steel spheres of different diameters; d - 0.07, 0.02, 0.01 and 0.001 m. Use time-step sizes
10.1 and 0.05. For each case plot x(t) and v(t) on the same graphics and find Co values vs. Re.​

Answer 1

Answer:

Horizontal component

cos(45^{0})=\frac{V_{2o}}{V_{2ox}} \\ V_{2ox}=V_{2o}cos(45^{0})\\ V_{2ox}=5.94\frac{m}{s}cos(45^{0}) \\V_{2ox}=4.20\frac{m}{s}cos(450)=V2oxV2oV2ox=V2ocos(450)V2ox=5.94smcos(450)V2ox=4.20sm

Vertical component.

sin(45^{0})=\frac{V_{2o}}{V_{2oy}} \\ V_{2oy}=V_{2o}sin(45^{0})\\ V_{2oy}=5.94\frac{m}{s}sin(45^{0}) \\V_{2oy}=4.20\frac{m}{s}sin(450)=V2oyV2oV2oy=V2osin(450)V2oy=5.94smsin(450)V2oy=4.20sm

When there is no gravity.

Marble 1 will follow a horizontal path.

Marble 2 moves from the ground to the height of marble 1 in parallel, therefore it is necessary to calculate the time it takes to travel the height h =0.950m

The time it takes for marble 2 to collide with marble 1 is given by:

v_{oy}=\frac{Y}{t} \\t=\frac{Y}{V_{oy}}voy=tYt=VoyY

Where:

Vertical velocity component V_{oy}=4.20\frac{m}{s}Voy=4.20sm

Distance Y=h--->h=0.950mY=h−−−>h=0.950m

time tt

Numerically evaluating

t=\frac{0.950m}{4.20\frac{m}{s}}\\t=0.226st=4.20sm0.950mt=0.226s

The X coordinate where they collide is given by:

X=X_{o}+V_{1ox}*tX=Xo+V1ox∗t

Where

Marble Horizontal Velocity 1 V_{1ox}=4.20\frac{m}{s}V1ox=4.20sm

Initial position X_{o}=0mXo=0m

time t=0.226st=0.226s

Evaluating numerically. X=0m+4.20\frac{m}{s}*0.226s=0.949mX=0m+4.20sm∗0.226s=0.949m

Finally both marbles collide at a height of h = 0.950m and at a horizontal distance of x =0.949m

The coordinate (x, y) where they collide is equal to \boxed{(0.949,0.950)m}(0.949,0.950)m