The resistance of air to the motion of bodies in free fall depends on many facto

Question

The resistance of air to the motion of bodies in free fall depends on many factors, such as the size of the body and its shape, the density and temperature of the air, and the velocity of the body falling through the air. A useful assumption, only approximately true, is that the resisting force F~R = k~v, where k is a constant whose value in any particular case is determined by factors other than velocity. Consider free fall of an object released from rest. (a) Show that Newton’s second law gives ma = mg kv which is equivalent to m d 2y dt2 = mg k dy dt (b) What are the dimensions of k, in terms of mass (M), length (L), and time (T)? (c) Show that the body ceases to accelerate when it reaches a velocity vT = mg/k called the terminal velocity. (c) Prove, by substitution into the equation of part (a), that the velocity varies with time as v = vT 1 e kt/m (e) Make qualitative sketches of v versus t y versus t for this motion. Your sketches should be consistent with an initial acceleration of g and a final acceleration of zero.

Explanation / Answer

a) according to Newton's second law,

Net force = applied force - frictional force

ma= mg - kv----------eq(1)

a= dy^2/ dt^2 , v = dy/dt

m( d^2 y / dt^2) = mg - k ( dy/dt)

b) Dimensions of K,

ma= mg - kv

m(a-g) /v

( M ) ( L/T^2 ) / (L/T) = K

K = M/T---dimensions of k

c)

a= ( mg - kv)/m

plugging a= 0,

mg = kv

VT = mg/k

d) m d/dt ( dy / dt) = mg - k ( dy/dt)

m dv/dt = mg - kv

m ( 1/ mg-kv) dv = dt

(-m/k) ln ( mg - kv)= T

v varies from 0 to V , T varies from 0 to T

( -m/k) [ ln (mg -kv ) - ln ( mg) ] = T

ln ( mg-kv/mg) = -KT/m

( mg -kv/ mg) = e^(-kT/m)

( 1- kv/mg) = e^(-kT/m)

1-  e^(-kT/m) = V/ VT

V =Vt[ 1-  e^(-kT/m)]

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