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

2024
Views:

292

Pages:

5
Author:

Jabin F.

Lagrangian of a bead sliding on a wire Case I: If motion of bead is 1D, then use plane polar coordinates. Put ω = angular frequency of wire. Case II: If motion of bead is 3D, then use spherical polar or cylindrical coordinates. Note: The axis of rotation is z-axis, put ω = angular velocity of wire. P-49 Q-3 A very long rod rotates about a pivot with a constant angular velocity 'ω'. A bead is constrained to slide along the rod without friction. At time t=0, the bead is at rest a distance 'd' away from the pivot. Its distance r(t) from the pivot at time t is: At t=0, r=d, r ˙ =0 Motion of bead is in 2D, horizontal plane, let us use polar coordinates. T= 2 1 m( r ˙ 2 +r 2 ω 2 ) V=0 L=T−V= 2 1 m( r ˙ 2 +r 2 ω 2 ) EOM: dt d ( ∂ r ˙ ∂L )− ∂r ∂L =0 ⇒m r ¨ −mrω 2 =0 ⇒ r ¨ −ω 2 r=0 2nd order differential equation ⇒r=C 1 e ωt +C 2 e −ωt (1) at t=0, r=d ⇒d=C 1 +C 2 (2) Now r ˙ =ω(C 1 e ωt −C 2 e −ωt ) at t=0, r ˙ =0 ⇒0=ω(C 1 −C 2 ) ⇒C 1 =C 2 = 2 d From eq. (1) r= 2 d e ωt + 2 d e −ωt =dcosh(ωt) Extra EOM as: r ˙ =ω 2 rdr ⇒ 2 1 r ˙ 2 = 2 ω 2 r 2 +C ⇒ 2 1 r ˙ 2 = 2 ω 2 (r 2 −d 2 ) ⇒ r ˙ =ω (r 2 −d 2 ) r ˙ =ω d 2 cosh 2 (ωt)−d 2 =ωd sinh 2 (ωt) =ωdsinh(ωt) P-71 Q-25 A bead of mass 'm' is sliding on a vertical circular loop of radius 'r'. The loop is rotated with constant angular frequency ω about a tangential axis as shown in figure. Lagrangian of bead is: