University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

394

Pages:

5
Author:

Jabin F.

Terminal Velocity: It is a constant velocity attained by a particle moving under some constant force(s) and a velocity-dependent force. v = Vt (constant) a = dv/dt = 0 Fnet = ma = 0 To calculate terminal velocity, put v = Vt and Fnet = 0. Q: Spherical particles of a given material of density ρ are released from rest inside a liquid medium of lower density. The viscous drag force may be approximated by the Stokes' law, i.e., Fd = 6πηRv, where η is the viscosity of the medium, R the radius of a particle, and v is its instantaneous velocity. If t(m) is the time taken by a particle of mass m to reach half its terminal velocity, then the ratio t(8m)/t(m) is: Solution: Firstly, we have to find out terminal velocity, which is calculated by putting v = Vt and Fnet = 0 in EOM. Fnet = mg - 6πηRv = 0 mg - 6πηRv_t = 0 Vt = mg / 6πηR T(m) = Vt / g = (mg / 6πηR) / g = m / 6πηR (as described in the question) Now EOM for calculating T: Fnet = ma = mg - 6πηRv using a = dv/dt: m * dv/dt = mg - 6πηRv dv/dt = (mg - 6πηRv) / m Integrating both sides: ∫dv = ∫(mg - 6πηRv) / m dt v = (mg/6πηR)t + C At t = 0, v = 0, so C = 0 v = (mg/6πηR)t To find t(m) when v = Vt/2: Vt/2 = (mg/6πηR)t(m) t(m) = (Vt/2) * (6πηR / mg) t(m) = 3πηR / mg Similarly, for t(8m): t(8m) = 3πηR / (8mg) Therefore, t(8m) / t(m) = (3πηR / (8mg)) / (3πηR / mg) = 1/8 For equal density: m / (4/3 πR_m^3) = 8m / (4/3 πR_8m^3) ⇒ R_8m / R_m = 1/2 ∴ t(8m) / t(m) = 8 * R_m / R_8m - 8 * 1 * 4 = 4