Question

in this lab we will drop a steel ball P steel=8000 kg/m^3, r ball=2.18x10^-3) through the air (nair=1.8x10^-5 Pas, P air= 1.29kg/m^3) and through a viscous soap solution (n soap ~ 9.0 Pa s) you will measure this number in lab, but use this number for prelab.( P soap = 1000kg/m^3) 1Pa= 1N/m

1. (a) assuming that the ball falling through the soap solution experiences visocus drag, find the terminal speed of the ball, the speed at which the weight of the ball and the drag force are of equal strength. ( as part of your answer draw a free body diagram of the forces acting on the ball and recognizing that density p=m/v, where m is the mass of the ball and v is volume of spherical ball)

(b) using the terminal velocity calculated in part (a), calculate the Reynolds number (you may let l=2r in Eq.1 from introduction) to determine if viscous or inertial drag is the dominant contributor to the drag acting on this steel ball falling through a soap solution.

(c) neglecting drag estimate the time it takes for steel ball to reach the terminal speed found in part 1. how far would the ball have travelled in this time interval.

2. calculate terminal speed of ball falling through the air

Answer 1

Msoap @ Terminal velocity of ball through the solution - ve - Bonnie I. Pored Pixab] <g - = 2_x (210x103) [2000-1000]*9.00 = 12 = 2:05X163 m/s 1 6 Reynold number of sphere is given by- Nre = Dx V x Isteel soap = Greball) x V x Isteel 9x9 Noap = 2x(210x103) x (8:05x103)x 8000 inre= 3.12x162

urexing © If we neglect drag force then 2 forces acts on ball- i weight of ball, - W = mg - Art Isteet & 6 of thrust force, уха = 4 To? x Scapig uit ex Isteet & Net downward force- ma = 4mgee free -g - 4 The frapy a = Isted xg - Psoap x g. Ssteel de 8575 (8000 - 1000 x 9.8 Q000 dr = 8.575 dt 29=8-05X103 nt dv = 2.575 I @ 8.05X163 = 8.575 t t = 9.39x1043 v = 2 r all [ Isteet - Pair 3 xg nair = (218X163)(8000 -1:29] *9.8 9x1.8x105 = 4599.1 ms