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Obtain the three differential equations of motion for position (x(t), y(t), z(t)

ID: 2077421 • Letter: O

Question

Obtain the three differential equations of motion for position (x(t), y(t), z(t)) in the inertial frame (use cartesian coordiante system).

Lift Drag mg The different forces acting the spinning on ball are shown in the schematic above Lift force is acting normal to both velocity and angular velocity vector. Drag force is al ways in the direction of velocity. Velocity (V) to be used is the negative of the velocity of the ball in the inertial frame. Magnitude of lift and drag can be calculated using the relations below Lift -,2 Drag where R is the radius of the ball, coo is the angular velocity, p is the density of air (1.225 Kg/m3) and Ca is 0.5 Assumptions Treat the ball as a particle. Angular velocity vector co lies in the blue plane (Y-Z plane as shown in the schematic above at an angle e with the vertical. Angular velocity vector coo remains constant both in magnitude and direction during the entire flight ofthe ball. o Simplified expressions for lift and drag No collision occurs during the trajectory

Explanation / Answer

From the given information
Lift = 4*(4*pi^2*R^3*rho)(w x V)/3 [ w is angular velocity of ball, V is its linear velocity]
Drag = rho*|V^2|(pi*R^2)Cd(V/|V|)/2
Cd = 0.5
rho = 1.225 kg/m^3
Now, netforce = (net acceleration)*mass
mass of ball = m
net force = lift + drag + weight = 4*(4*pi^2*R^3*rho)(w x V)/3 + rho*|V^2|(pi*R^2)Cd(V/|V|)/2 + mg [ V is velocity vector, w is angular velocity vector and g is acceleration due to gravity vector]
m*d(V)/dt = 4*(4*pi^2*R^3*rho)(w x V)/3 + rho*|V^2|(pi*R^2)Cd(V/|V|)/2 + mg

as the angular velocity does not change there is no drag caousing angular acceleration
so the equatiuon is
m*d(V)/dt = 4*(4*pi^2*R^3*rho)(w x V)/3 + rho*|V^2|(pi*R^2)Cd(V/|V|)/2 + mg