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

2024
Views:

193

Pages:

5
Author:

Jabin F.

+ Elastic force: Y = Stress = (F/A) / (Δl/l) F = Y * (Δl/l) * A Q: Two solid spheres S₁ and S₂ of the same uniform density fell from rest under gravity in a viscous medium and after the same time, each terminal velocity is v₁ and v₂ respectively. If the masses of S₁ and S₂ are m₁ and m₂ respectively, and r₁ and r₂ are their radii, then the ratio of m₁/m₂ is: Solution: For terminal velocity, Fnet = 0: mg - 6πηRv = 0 v = mg / (6πηR) Put in eq (i): v₁ = m₁g / (6πη₁R₁) v₂ = m₂g / (6πη₂R₂) Since densities are the same: m₁ = (4/3)πR₁³ m₂ = (4/3)πR₂³ Therefore: v₁/v₂ = (m₁g / (6πη₁R₁)) / (m₂g / (6πη₂R₂)) v₁/v₂ = (m₁R₂) / (m₂R₁) (m₁/m₂) = (v₁R₁) / (v₂R₂) Q: In Millikan's oil drop experiment, an oil drop of radius r, mass m, and charge q = 6πηr(v₁ + v₀)/E is moving upward with a terminal velocity v₁ due to an applied electric field of magnitude E, where η is the coefficient of viscosity. The acceleration due to gravity is given by: Solution: EOM: Fnet = Fe - mg - F₀ For terminal velocity (v₁), Fnet = 0: Fe - mg - F₀ = 0 qE - mg - 6πηrv₁ = 0 6πηr(v₁ + v₀)/E - mg - 6πηrv₁ = 0 6πηrv₀ = mg q = (6πηrv₀)/E Q: System shown in the figure is in equilibrium and at rest. The spring and string are massless. Now the string is cut. The acceleration of mass m and m just after the string is cut will be: Solution: Equilibrium For mass m: T = mg For mass 3m: kx = T + 3mg Before cut: kx = mg + 3mg kx = 3mg For mass m: Fdown = mg, Fup = 0 a = g downward For mass 3m: Fdown = 3mg, Fup = kx = 3mg EOM: 3ma = 3mg - 3mg a = 0 upward