We start with the equation of motion then integrate this equation where the limits of the integration are determined by the initial conditions (t = 0 and v = v0) and final conditions (t and v) Therefore, we can express the velocity v as velocity In every case, the velocity v tends towards the limiting value vT. Plots of the velocity v as functions of time t m = 2.00 kg = 5.00 kg.s-1 g = 9.80 m.s-2 vT = 3.92 m.s-1 Initial values for velocity v0 [m.s-1] blue: 10 red: vT magenta: 2 cyan: 0 The acceleration a as a function of time t is Initial values for velocity v0 [m.s-1] blue: 10 red: vT magenta: 2 cyan: 0
Therefore, the velocity at time tk+2 is to calculate the velocity we need to know the velocity at the two previous time steps and . We know t1 = 0 and v(t1) = v(0) = v0. We estimate the velocity at the second time step t2 where we have assumed a constant acceleration in the first time step. We can improve our estimate of by using an average value of the acceleration in the first time step We can now calculate the velocity v(t) at all from t = t1 to t = tN. The acceleration at each time step is The displacement at each time step is EXAMPLE The mscript mec_fr_mg_bv.m can be used for simulations for the motion of an object upon a resistive force of the form (Model 1). Input parameters m = 2 kg = 10 kg.s-1 v0 = 10 m.s-1 t = 1x10-4 s tmax = 2.00 s Outputs N numerical approach A analytical approach terminal velocity vT = 1.96 m.s-1
We can now calculate the displacement x as a function of velocity v valid only if We can now investigate what happens as time In falling, the object will finally reach a constant velocity vT (a = 0 ) which is known as the terminal velocity. In falling, as time t increases the objects displacement x just gets larger and larger. Example rock dropped from rest: m = 0.010 kg = 1.0010-4 kg.m-1 v0 = 0 m.s-1 vT = 31.3 m.s-1 Example Small rock thrown vertically downward (v < vT) m = 0.010 kg = 1.0010-4 kg.m-1 v0 = +5.00 m.s-1 vT = 31.3 m.s-1 Example Small rock thrown vertically downward (v > vT) m = 0.010 kg = 1.0010-4 kg.m-1 v0 = +40.0 m.s-1 vT = 31.3 m.s-1
For problems in which the object is projected vertically upward, you have to divide the problem into two parts. (1) Calculate the time to reach its maximum height and calculate the maximum height reached for the upward motion. (2) Reset the initial conditions to the position at maximum height where the initial velocity becomes v0 = 0 and do the calculations for the downward movement of the object. Velocity of the object negative and moving up Equation of motion Note: up is the positive direction valid only if The time tup to maximum height occurs when v = 0 The velocity v as a function of time t is The displacement x as a function of velocity v is The maximum height xup reached by the object occurs when v = 0 Example Small rock thrown vertically upward (v0 < 0 v0 = - u u > 0) m = 0.010 kg = 1.0010-4 kg.m-1 v0 = - 12.0 m.s-1 vT = 31.3 m.s-1
The terminal velocity vT is When v = 0 the object reaches its maximum height xup (up is negative) The time tup to the maximum height The calculations agree with the values for tup and xup determined from the graphs. From the graphs: x = 0 time t = 2.275 s velocity v = 11.21 m.s-1 Time to fall from max height to origin tdown = (2.375 – 1.169) s = 1.206 s takes slight longer to fall then rise to and from origin to max height Launch = 12.00 m.s-1 slightly greater than return speed = 11.21 m.s-1 In the absence of any resistive forces At maximum height Numerical Approach We can also find the velocity and displacement of the object by solving Newton’s Second Law of motion using a finite difference method. We start with (4) In the finite difference method we calculate the velocity v and displacement x at N discrete times tk at fixed time intervals t
radius of steel ball R = 20.0x10-3 m density of steel ball = 7800 kg.m-3 mass of table tennis ball m = 12.70x10-3 kg drag coefficient CD = 0.445 density of air F = 1.225 kg.m-3 viscosity of air F = 1.789x10-5 Pa.s max time tMax = 3.00 s No. time steps N = 10000 Output parameters table tennis ball steel ball free fall Fdrag = 0 terminal velocity vT [m.s-1] 8.79 86 t = 3.00 s a [m.s-2] 0.049 8.75 9.80 t = 3.00 s x [m] 20.9 43.3 44.1 t = 3.00 s v [m.s-1] 8.77 28.3 29.4 t = 3.00 s Reynolds number NR 7.8x104 It is “true” that heavier objects do fall faster than lighter ones. table tennis ball steel ball Example A falling skydiver mscript mec_drag.m We can model a falling skydiver falling from rest using the numerical approach where the resistive force or drag force is Input parameters radius (estimate) R = 0.500 m mass m = 70 kg mass entered and not calculated from density drag coefficient CD = 0.5 density of air F = 1.225 kg.m-3 viscosity of air F = 1.789x10-5 Pa.s max time tMax = 20.00 s No. time steps N = 10000 Output parameters cross-sectional area A = 0.7854 m2 terminal velocity |vT| = 54 m.s-1 = 192 km.h-1 time to reach vT t(vT) ~ 18 s