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

2024
Views:

150

Pages:

5
Author:

Jabin F.

Component of Acceleration in Cylindrical Coordinate System T = 1/2 m(ṙ² + r²θ̇² + ż²) From Lagrangian mechanics: (i) Radial component: ar = (1/m)[d/dt(∂T/∂r) - ∂T/∂r] = r̈ - rθ̇² (ii) Transverse component: aθ = (1/mr)[d/dt(∂T/∂θ̇) - ∂T/∂θ] = rθ̈ + 2ṙθ̇ Component of Acceleration in Cylindrical Coordinate System (continued) (iii) Axial component: az = z̈ Example Problem A particle is moving on a plane. Its polar coordinates are given as r = Ae^(wt) and θ = wt, where A and w are constants. Calculate the radial acceleration. Solution: ar = r̈ - rθ̇² = Awe^(wt) - Aw²e^(wt) = 0 Note: If r(t) and θ(t) are not given, then acceleration is calculated using Newton's equations (ar = Fr/m, aθ = Fθ/m, az = Fz/m). Resolving a Vector The component of a vector along a line with which the angle is given is its cosine component. Problem: A particle moves in two dimensions on the ellipse x² + 4y² = 8. At a particular instant, it is at the point (x, y) = (2, 1) and the x-component of its velocity is 6 (in suitable units). Then the y-component of its velocity is: Solution: Given path: x² + 4y² = 8 Differentiating with respect to time: 2xẋ + 8yẏ = 0 Given (x, y) = (2, 1) and ẋ = 6: 2(2)(6) + 8(1)ẏ = 0 Solving for ẏ: ẏ = -3