Physics 43, Exam 2

7.29 Figure 7.14 shows a simple pendulum (mass m, length l) whose point of support P is attached to the edge of a wheel (center O, radius R) that is forced to rotate at a fixed angular velocity ω. At t=0, the point P is level with O on the right. Write down the Lagrangian and find the equation of motion for the angle φ. [Hint: Be careful writing down the kinetic energy T. A safe way to get the velocity right is to write down the position of the bob at time t, and then differentiate.] Check that your answer makes sense in the special case that ω=0.   θ=ωt L=T-U T=1/2mv^2 r=r1+r2 r1=(Rcosθ+lsinφ)x+(Rsinθ-lcosφ)y r=(Rcosωt+lsinφ)x+(Rsinωt-lcosφ)y v=dR/dt=(-Rωsinωt+lcosφφ')x+(Rωcosωt+lsinφφ')y |v|^2=(-Rωsinωt+lcosφφ')^2+(-Rωcosωt+lsinφφ')^2 =(Rω)^2sin^2ωt+l^2cos^2φφ'^2-2Rωlsinωtcosφφ'+(Rω)^2cos^2ωt+l^2sin^2φφ'^2+2Rωlcosωtsinφφ' =(Rω)^2+l^2φ'^2+2Rlωφ'[cosωtsinφ-sinωtcosφ] =(Rω)^2+l^2φ'^2+2Rlωφ'*sin(φ-ωt) 7-29-2 T = ½m[(lω²) + l²φ̇² + 2Rlωφ̇sin(φ-ωt)] U = mgy = mg[-lcosφ + Rsinωt] L = ½m[(lω²) + l²φ̇² + 2Rlωφ̇sin(φ-ωt)] - mg[-lcosφ + Rsinωt] Lagrange: ∂L/∂φ̇ = ½m[2l²φ̇ + 2lωsin(φ-ωt)] = lm[lφ̇ + Rωsin(φ-ωt)] d/dt(∂L/∂φ̇) = lm[lφ̈ + Rωcos(φ-ωt)(φ̇-ω)] 0 = ∂L/∂φ - d/dt(∂L/∂φ̇) = ml[Rωφ̇cos(φ-ωt) - gsinφ] - ml[lφ̈ + lω(φ̇-ω)cos(φ-ωt)] 0 = -gsinφ - lφ̈ + Rω²cos(φ-ωt) φ̈ = -(g/l)sinφ + (R/l)ω²cos(φ-ωt) for ω = 0, φ̈ = -(g/l)sinφ, which is exactly the equation of motion for a pendulum.