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

2024
Views:

264

Pages:

5
Author:

Jabin F.

Vertical circular motion of a bob attached to a string: Radial EOM at lowest point: T - mg = m(v²/r) Radial EOM at highest point: -T - mg = m(-v²/r) Adding eq(1) & eq(2): 2mg + (T₁ - T₂) = m(v₁² - v₂²)/r Now apply conservation of energy at lowest & highest point: (KE + PE)₁ = (KE + PE)₂ ½mv₁² - mgr = ½mv₂² + mgr v₁² - v₂² = 4gr Use this in eq(3): 2mg + (T₁ - T₂) = m(4gr)/r T₁ - T₂ = 6mg From eq(2): -mg - T₂ = m(-v₂²/r) v₂²min = rg From eq(4): v₁² - v₂² = 4gr → v₁² - rg = 4gr v₁²min = 5rg If v < √gl at lowest point, then the bob will stop before reaching the point of suspension level. If √gl < v < √5gl, then object will not stop anywhere but will not complete the circle. If the bob is attached to a rigid wire, then for circular motion to be possible when: v_min (lowest) = √rg v_min (highest) = 0 D - 12 - 20/7/19 → Above conclusion included. Q: A bob of mass m is attached to a string of length l. At the lowest point, bob is given a speed √3gl. At what angle with upward vertical, the string will loose i.e. tension will be zero? Solution: Since v = √3gl > √gl, object above point of suspension. Path will be circular upto the point where tension in the string becomes zero. So radial eom at that point will be: -mg cos θ = m(-v²/l) v² = gl cos θ Now apply conservation of energy: (KE + PE)initial = (KE + PE)final ½m(√3gl)² - mgl = ½mgl cos θ + mgl cos θ 3gl/2 - gl = gl cos θ + gl cos θ ½ = cos θ θ = π/3 Q: A vehicle is moving in a circular path of radius R on the rough horizontal surface (μs) with constant speed. What can be the max velocity? Solution: Since circular path is horizontal: Vertical equilibrium: N = mg Now radial EOM: fs = m(v²/R) μsN = m(v²/R) μsmg = m(v²/R) v² = μs gR v = √(μs gR) vmax = √(μs gR)