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

2024
Views:

390

Pages:

5
Author:

Jabin F.

Degrees of Freedom (d.o.f.) d.o.f. is the number of independent variables required to describe the dynamics of a system. If a particle moves along a predefined line, then d.o.f. = 1. If a particle moves on a surface, then d.o.f. = 2. If a particle moves in space, then d.o.f. = 3. Example: If 2 particles move on a sphere surface, then d.o.f. of each particle is 2, and total d.o.f. is 4. If there are many independent particles, then total d.o.f. is equal to the sum of d.o.f. of individual particles. If there are geometrical constraints/conditions among the particles, then total d.o.f. is equal to individual d.o.f. minus the number of constraints. Example: If we say that distance between two particles is constant, then there is one geometrical constraint between particles. A spring doesn't put any constraint. Example: Two particles are rigidly connected and allowed to move in space, then d.o.f. (total) = individual d.o.f. - number of constraints = 3 + 3 - 1 = 5. Rigid Bodies If many particles are connected together rigidly, then it forms a rigid body. If a rigid body moves in 3D space, then d.o.f. = 6 (3 for translational motion and 3 for rotational motion). Note: This value of d.o.f. cannot be calculated by considering d.o.f. of individual particles and constraints. If a rigid body rotates about a fixed point, then d.o.f. = 3. If a rigid body rotates about a fixed axis, then d.o.f. = 1. If a rigid body moves freely in a d-dimension space, then d.o.f. = d(d+1)/2. Example: A disk is moving in the x-direction, and its plane lies in the x-y plane. What is its d.o.f.? Solution: Linear and angular velocity, both have one component. So, d.o.f. = 1 + 1 = 2. Problem 1: A tunnel is made along the diameter of Earth (M, R). A particle is thrown into the tunnel with speed √(GM/R). What is the speed of the particle at the center of the tunnel? Solution: x = Rcosθ, y = Rsinθ v_x = -Rωsinθ, v_y = Rωcosθ v^2 = ω^2(R^2 - x^2) At x = 0, v = ωR = √(GM/R) R^2 = A^2 - R^2, A = √2R ωA = √(GM/R) Problem 2: A bob is suspended from a rigid body support symmetrically using two massless inextensible strings, each of length l, making an equilateral triangle. Calculate the time period of oscillation. Solution: T = 2π√(L/g) = 2π√(√3l/g) Problem 3: A pendulum is created by hanging a bob of mass m from a rigid body support symmetrically using two massless inextensible strings, each of length l, making an equilateral triangle. Find the Lagrangian for the angular oscillation of the bob in the plane perpendicular to the paper. Solution: KE = 1/2 m(l^2/4)θ̇^2 = 3/8 ml^2θ̇^2 PE = -mg(l/2)cosθ = -mg(√3l/4)cosθ = -mg(√3l/4)(1 - 2sin^2(θ/2)) L = KE - PE = 3/8 ml^2θ̇^2 + √3mgl sin^2(θ/2) - √3mgl/4