Very small objects, such as dust particles, experience a linear drag force, F^ri

Question

Very small objects, such as dust particles, experience a linear drag force, F^rightarrow_drag = (bv, direction opposite the motion), where b is a constant. For a sphere of radius R, the drag constant can be shown to be b = 6 pi eta R, where eta is the viscosity of the gas. (a) Find an expression for the terminal speed upsilon_term of a spherical particle of radius R and mass m falling through a gas of viscosity eta. (b) Suppose a gust of wind has carried a 50-mu m-diameter dust particle to a height of 300m. If the wind suddenly stops, how long will it take the dust particle to settle back to the ground? Dust has a density of 2700kg/m^3, the viscosity of 25 degree C air is 2.0 times 10^?5 Ns/m^2, and you can assume that the falling dust particle reaches terminal speed almost instantly. A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are mu_s and mu_k, respectively. Find an expression for the maximum tension T_max for which the small box rides on top of the large box without slipping. A rubber ball is dropped onto a ramp that is tilted at 20 degree, as shown in the figure. A bouncing ball obeys the law of reflection, which says that the ball leaves the surface at the same angle it approached the surface. The balls next bounce is 3.0 m to the right of its first bounce. What is the balls rebound speed on its first bounce?

Explanation / Answer

Q6.

terminal speed will be achieved when the drag force balances weight of the particle.

part a:

weight of the particle=m*g

drag force=b*v

=6*pi*n*R*v

equating both the forces,

m*g=6*pi*n*R*v

==>v=m*g/(6*pi*n*R)

part b:

diameter=50 um=50*10^(-6) m

then radius R=diameter/2=25*10^(-6) m

height h=300 m

density=2700 kg/m^3

volume of the particle=(4/3)*pi*R^3

=6.545*10^(-14) m^3

then mass=volume*density=1.7671*10^(-10) kg

given viscoity=n=2*10^(-5)

then terminal velocity=m*g/(6*pi*n*R)

=1.7671*10^(-10)*9.8/(6*pi*2*10^(-5)*25*10^(-6))=0.18375 m/s

then time taken to settle back to ground=distance/speed

=300/0.18375=1632.7 seconds

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.