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

2024
Views:

173

Pages:

5
Author:

Jabin F.

# Object sliding from the highest point of a fixed smooth sphere/circle Where will it leave the contact? Let, it leaves the contact at angle θ Upto the leaving point, the path is an arc of circle. So let us apply concept of circular motion. Radial EOM: fₙ = maₙ -mg cosθ = m(-v²/r) ⇒ cosθ = v²/gr Now apply conservation of energy: (KE + PE)initial = (KE + PE)leaving point 0 + 0 = ½mv² - mg r(1 - cosθ) ⇒ v² = 2gr(1 - cosθ) But in eqn (i) cosθ = v²/gr ⇒ cosθ = 2gr(1 - cosθ)/gr ⇒ 3cosθ = 2 ⇒ cosθ = ⅔ P-131 Two gravitational bodies A and B with masses mA and mB resp., are moving in circular orbit. Assume that mA << mB and let radius of orbit of A be RA. If the body A is losing mass adiabatically, its orbital radius RA is proportional to: (a) 1/mA (b) 1/mA² (c) mA (d) mA² Solution: Since mA << mB, therefore we can assume that mB is stationary and mA revolves around it in circular path. Now radial EOM: fr = maᵣ -G(mA)(mB)/r² = mA(-v²/r) ⇒ v = √(GmB/r) Note: To find a proportionality relation, look for a quantity which is conserved! Torque: τ = r x F = rFsin180° = 0 Hence angular momentum is conserved. i.e. constant. Now λ = mvr = constant mA * √(GmB/r) * r = l ⇒ r ∝ 1/mA² P-135 Consider circular orbits in a central force potential V(r) = -k/rⁿ, where k > 0 and n ≠ 0. If the time period of the circular orbit of radius R is T, and that of radius 2R is T₂, then T₂/T₁ is: Solution: V(r) = -k/rⁿ ⇒ Fᵣ = -dV/dr = -nk/rⁿ⁺¹ Radial EOM ⇒ Fᵣ = maᵣ ⇒ -nk/rⁿ⁺¹ = m(-v²/r) ⇒ v = √(nk/mrⁿ) v = ωr ⇒ ω = √(nk/mrⁿ⁺¹) ⇒ T = 2πr/ω = 2πr √(mrⁿ/nk) T ∝ r^(n+3/2) ∴ T₂/T₁ = (2R)^(n+3/2) / R^(n+3/2) = (2)^(n+3/2)